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Vectors and Geometry in Two and Three Dimensions

1.2 Vectors

41 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Let a=<2,0>\va=\llt 2,0\rgt and b=<1,1>\vb=\llt 1,1\rgt. Evaluate and sketch a+b, a+2b\va+\vb,\ \va+2\vb and 2a−b2\va-\vb.

Answer

a+b=<3,1>\va+\vb=\llt 3,1\rgt, a+2b=<4,2>\va+2\vb=\llt 4,2\rgt, 2a−b=<3,−1>2\va-\vb=\llt 3,-1\rgt

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Full solution

a+b=<3,1>\va+\vb=\llt 3,1\rgt, a+2b=<4,2>\va+2\vb=\llt 4,2\rgt, 2a−b=<3,−1>2\va-\vb=\llt 3,-1\rgt

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Figure from prob_s1.2, line 19

Q2Stage 1

Determine whether or not the given points are collinear (that is, lie on a common straight line)

  1. (1,2,3), (0,3,7), (3,5,11)(1,2,3),\ (0,3,7),\ (3,5,11)

  2. (0,3,−5), (1,2,−2), (3,0,4)(0,3,-5),\ (1,2,-2),\ (3,0,4)

Hint

If three points are collinear, then the vector from the first point to the second point, and the vector from the first point to the third point must both be parallel to the line, and hence must be parallel to each other (i.e. must be multiples of each other).

Figure from prob_s1.2, line 60

Figure from prob_s1.2, line 60

Answer

(a) not collinear (b) collinear

Full solution

If three points are collinear, then the vector from the first point to the second point, and the vector from the first point to the third point must both be parallel to the line, and hence must be parallel to each other (i.e. must be multiples of each other).

Figure from prob_s1.2, line 60

Figure from prob_s1.2, line 60

(a) The vectors <0,3,7>−<1,2,3>=<−1,1,4>\llt 0,3,7\rgt -\llt 1,2,3\rgt =\llt -1,1,4\rgt and <3,5,11>−<1,2,3>=<2,3,8>\llt 3,5,11\rgt -\llt 1,2,3\rgt =\llt 2,3,8\rgt are not parallel (i.e. are not multiples of each other), so the three points are not on the same line.

(b) The vectors <1,2,−2>−<0,3,−5>=<1,−1,3>\llt 1,2,-2\rgt -\llt 0,3,-5\rgt =\llt 1,-1,3\rgt and <3,0,4>−<0,3,−5>=<3,−3,9>\llt 3,0,4\rgt -\llt 0,3,-5\rgt =\llt 3,-3,9\rgt are parallel (i.e. are multiples of each other), so the three points are on the same line.

Q3Stage 1

Determine whether the given pair of vectors is perpendicular

  1. <1,3,2>, <2,−2,2>\llt 1,3,2\rgt ,\ \llt 2,-2,2\rgt

  2. <−3,1,7>, <2,−1,1>\llt -3,1,7\rgt ,\ \llt 2,-1,1\rgt

  3. <2,1,1>, <−1,4,2>\llt 2,1,1\rgt ,\ \llt -1,4,2\rgt

Hint

Review Theorem 1.2.11 in the CLP-3 text.

Answer

(a) perpendicular (b) perpendicular (c) not perpendicular

Full solution

By property 7 of Theorem 1.2.11 in the CLP-3 text,

<1,3,2>⋅<2,−2,2>=1×2−3×2+2×2=0   ⟹   perpendicular<−3,1,7>⋅<2,−1,1>=−3×2−1×1+7×1=0   ⟹   perpendicular<2,1,1>⋅<−1,4,2>=−2×1+1×4+1×2=4≠0   ⟹   not perpendicular\begin{alignat*}{3} \llt 1,3,2\rgt \cdot\llt 2,-2,2\rgt &=1\times2-3\times 2+2\times2=0\ &&\implies\ \text{perpendicular} \tag{a} \\ \llt -3,1,7\rgt \cdot\llt 2,-1,1\rgt &=-3\times2-1\times 1+7\times1=0\ &&\implies\ \text{perpendicular} \tag{b} \\ \llt 2,1,1\rgt \cdot\llt -1,4,2\rgt &=-2\times1+1\times 4+1\times2=4\ne 0\ &&\implies\ \text{not perpendicular} \tag{c} \end{alignat*}
Q4Stage 1

Consider the vector a=<3,4>\va=\llt 3,4 \rgt.

  1. Find a unit vector in the same direction as a\va.

  2. Find all unit vectors that are parallel to a\va.

  3. Find all vectors that are parallel to a\va and have length 1010.

  4. Find all unit vectors that are perpendicular to a\va.

Hint

Review Definition 1.2.5 and Theorem 1.2.11 in the CLP-3 text.

Answer

(a) 15<3,4>\frac{1}{5}\llt 3,4 \rgt (b) ±15<3,4>\pm\frac{1}{5}\llt 3,4 \rgt (c) ±<6,8>\pm \llt 6,8 \rgt (d) ±15<4,−3>\pm\frac{1}{5}\llt 4,-3 \rgt

Full solution

(a) The vector a\va has length

∣<3,4>∣=32+42=25=5\begin{equation*} |\llt 3,4 \rgt| =\sqrt{3^2+4^2} =\sqrt{25} =5 \end{equation*}

So the vector 15<3,4>\frac{1}{5}\llt 3,4 \rgt has length 11 (i.e. is a unit vector) and is in the same direction as <3,4>\llt 3,4 \rgt.

(b) Recall, from Definition 1.2.5 in the CLP-3 text, that a vector is parallel to a\va if and only if it is of the form sas\va for some nonzero real number ss. Such a vector is a unit vector if and only if

∣sa∣=1  ⟺  ∣s∣ ∣<3,4>∣=1  ⟺  ∣s∣=1∣<3,4>∣=15  ⟺  s=±15\begin{align*} |s\va|=1 &\iff |s|\,|\llt 3,4 \rgt|=1 \iff |s| = \frac{1}{|\llt 3,4 \rgt|}= \frac{1}{5} \\ &\iff s = \pm \frac{1}{5} \end{align*}

So there are two unit vectors that are parallel to a\va, namely ±15<3,4>\pm\frac{1}{5}\llt 3,4 \rgt.

(c) We have already found, in part (b), all vectors that are parallel to a\va and have length 11, namely ±15<3,4>\pm\frac{1}{5}\llt 3,4 \rgt. To increase the lengths of those vectors to 1010, we just need to multiply them by 1010, giving ±105<3,4>=±2<3,4>=±<6,8>\pm\frac{10}{5}\llt 3,4 \rgt=\pm 2\llt 3,4 \rgt=\pm\llt 6,8 \rgt.

(d) A vector <x,y>\llt x,y \rgt is perpendicular to a=<3,4>\va=\llt 3,4\rgt if and only if

0=<x,y>⋅<3,4>=3x+4y  ⟺  y=−34x  ⟺  <x,y>=<x,−34x>=x4<4,−3>\begin{equation*} 0=\llt x,y \rgt\cdot\llt 3,4\rgt = 3x+4y \iff y=-\frac{3}{4}x \iff \llt x,y \rgt = \llt x,-\frac{3}{4}x \rgt = \frac{x}{4} \llt 4,-3\rgt \end{equation*}

Such a vector is a unit vector if and only if

∣x∣4 ∣<4,−3>∣=1  ⟺  ∣x∣4=1∣<4,−3>∣=15  ⟺  x4=±15\begin{align*} \frac{|x|}{4}\,|\llt 4,-3 \rgt|=1 &\iff \frac{|x|}{4} = \frac{1}{|\llt 4,-3 \rgt|}= \frac{1}{5} \\ &\iff \frac{x}{4} = \pm \frac{1}{5} \end{align*}

So there are two unit vectors that are perpendicular to a\va, namely ±15<4,−3>\pm\frac{1}{5}\llt 4,-3 \rgt.

Q5Stage 1

Consider the vector b=<3,4,0>\vb=\llt 3,4,0 \rgt.

  1. Find a unit vector in the same direction as b\vb.

  2. Find all unit vectors that are parallel to b\vb.

  3. Find four different unit vectors that are perpendicular to b\vb.

Hint

Review Definition 1.2.5 and Theorem 1.2.11 in the CLP-3 text.

Answer

(a) 15<3,4,0>\frac{1}{5}\llt 3,4,0 \rgt (b) ±15<3,4,0>\pm\frac{1}{5}\llt 3,4,0 \rgt

(c) A vector is of length one and perpendicular to b\vb if and only if it is of the form <x,−34x,z>\llt x,-\frac{3}{4}x,z \rgt with 2516x2+z2=1\sqrt{\frac{25}{16}x^2+z^2}=1. There are infinitely many such vectors. Four of them are

±<0,0,1>±15<4,−3,0>\begin{equation*} \pm\llt 0,0,1\rgt\qquad \pm\frac{1}{5}\llt 4,-3,0\rgt \end{equation*}
Full solution

(a) The vector b\vb has length

∣<3,4,0>∣=32+42+02=25=5\begin{equation*} |\llt 3,4,0 \rgt| =\sqrt{3^2+4^2+0^2} =\sqrt{25} =5 \end{equation*}

So the vector 15<3,4,0>\frac{1}{5}\llt 3,4,0 \rgt has length 11 (i.e. is a unit vector) and is in the same direction as <3,4,0>\llt 3,4,0 \rgt.

(b) Recall, from Definition 1.2.5 in the CLP-3 text, that a vector is parallel to b\vb if and only if it is of the form sbs\vb for some nonzero real number ss. Such a vector is a unit vector if and only if

∣sb∣=1  ⟺  ∣s∣ ∣<3,4,0>∣=1  ⟺  ∣s∣=1∣<3,4,0>∣=15  ⟺  s=±15\begin{align*} |s\vb|=1 &\iff |s|\,|\llt 3,4,0 \rgt|=1 \iff |s| = \frac{1}{|\llt 3,4,0 \rgt|}= \frac{1}{5} \\ &\iff s = \pm \frac{1}{5} \end{align*}

So there are two unit vectors that are parallel to b\vb, namely ±15<3,4,0>\pm\frac{1}{5}\llt 3,4,0 \rgt.

(c) A vector <x,y,z>\llt x,y,z \rgt is perpendicular to a=<3,4,0>\va=\llt 3,4,0\rgt if and only if

0=<x,y,z>⋅<3,4,0>=3x+4y  ⟺  y=−34x  ⟺  <x,y,z>=<x,−34x,z>\begin{equation*} 0=\llt x,y,z \rgt\cdot\llt 3,4,0\rgt = 3x+4y \iff y=-\frac{3}{4}x \iff \llt x,y,z \rgt = \llt x,-\frac{3}{4}x,z \rgt \end{equation*}

Such a vector is a unit vector if and only if

∣<x,−34x,z>∣=1  ⟺  x2+916x2+z2=1  ⟺  2516x2+z2=1\begin{align*} \left|\llt x,-\frac{3}{4}x,z \rgt\right|=1 &\iff \sqrt{x^2+\frac{9}{16}x^2+z^2}=1 \iff \sqrt{\frac{25}{16}x^2+z^2}=1 \end{align*}

There are infinitely many pairs xx, zz that obey 2516x2+z2=1\sqrt{\frac{25}{16}x^2+z^2}=1. We can easily get two of them by setting x=0x=0 and choosing zz to obey z2=1\sqrt{z^2}=1, i.e. choosing z=±1z=\pm 1. We can easily get two more of them by setting z=0z=0 and choosing xx to obey 2516x2=1\sqrt{\frac{25}{16}x^2}=1, i.e. choosing x=±45x=\pm \frac{4}{5}. This gives us four vectors of length one that are perpendicular to b\vb, namely

±<0,0,1>±<45 , −34 45 , 0>=±15<4,−3,0>\begin{equation*} \pm\llt 0,0,1\rgt\qquad \pm\llt \frac{4}{5}\,,\,-\frac{3}{4}\,\frac{4}{5}\,,\,0\rgt =\pm\frac{1}{5}\llt 4,-3,0\rgt \end{equation*}
Q6Stage 1

Let a=<a1,a2>\va=\llt a_1,a_2\rgt. Compute the projection of a\va on ı^\hi and ȷ^\hj.

Answer

projı^a=a1ı^\text{proj}_{\hi}\va=a_1\hi projȷ^a=a2ȷ^\text{proj}_{\hj}\va=a_2\hj.

Full solution

projı^a=(a⋅ı^)ı^=a1ı^\text{proj}_{\hi}\va=(\va\cdot\hi)\hi =a_1\hi and projȷ^a=(a⋅ȷ^)ȷ^=a2ȷ^\text{proj}_{\hj}\va=(\va\cdot\hj)\hj =a_2\hj.

Q7Stage 1

Does the triangle with vertices (1,2,3), (4,0,5)(1,2,3),\ (4,0,5) and (3,6,4)(3,6,4) have a right angle?

Answer

Yes.

Full solution

The vector from (1,2,3)(1,2,3) to (4,0,5)(4,0,5) is <3,−2,2>\llt 3,-2,2\rgt. The vector from (1,2,3)(1,2,3) to (3,6,4)(3,6,4) is <2,4,1>\llt 2,4,1\rgt. The dot product between these two vectors is <3,−2,2>⋅<2,4,1>=0\llt 3,-2,2\rgt\cdot\llt 2,4,1\rgt=0, so the vectors are perpendicular and the triangle does contain a right angle.

Q8Stage 1

Show that the area of the parallelogram determined by the vectors a\va and b\vb is ∣a×b∣|\va\times \vb|.

Figure from prob_s1.2, line 346

Figure from prob_s1.2, line 346

Answer

See the solution.

Full solution

The area of a parallelogram is the length of its base time its height.

Figure from prob_s1.2, line 363

Figure from prob_s1.2, line 363

We can choose the base to be a\va. Then, if θ\theta is the angle between its sides a\va and b\vb, its height is ∣b∣sin⁡θ|\vb|\sin\theta.
So

area=∣a∣∣b∣sin⁡θ=∣a×b∣\begin{equation*} \text{area} = |\va||\vb|\sin\theta=|\va\times\vb| \end{equation*}
Q9Stage 1

Show that the volume of the parallelepiped determined by the vectors a, b\va,\ \vb and c\vc is

∣a⋅(b×c)∣\begin{equation*} |\va\cdot(\vb\times\vc)| \end{equation*}

Figure from prob_s1.2, line 381

Figure from prob_s1.2, line 381

Answer

See the solution

Full solution

The volume of a parallelepiped is the area of its base time its height. We can choose the base to be the parallelogram determined by the vectors b\vb and c\vc. It has area ∣b×c∣|\vb\times\vc|. The vector b×c\vb\times\vc is perpendicular to the base.

Figure from prob_s1.2, line 401

Figure from prob_s1.2, line 401

Denote by θ\theta the angle between a\va and the perpendicular b×c\vb\times\vc. The height of the parallelepiped is ∣a∣∣cos⁡θ∣|\va| |\cos\theta|. So

volume=∣a∣ ∣cos⁡θ∣ ∣b×c∣=∣a⋅(b×c)∣\begin{equation*} \text{volume} = |\va|\, |\cos\theta|\, |\vb\times\vc| =|\va\cdot(\vb\times\vc)| \end{equation*}
Q10Stage 1

Verify by direct computation that

  1. ı^×ȷ^=k^\hi\times\hj=\hk, ȷ^×k^=ı^\hj\times\hk=\hi, k^×ı^=ȷ^\hk\times\hi=\hj

  2. a⋅(a×b)=b⋅(a×b)=0\va\cdot(\va\times\vb)=\vb\cdot(\va\times\vb)=\vZero

Answer

See the solution.

Full solution

(a)

ı^×ȷ^=det⁡[ı^ȷ^k^100010]=ı^(0×0−0×1)−ȷ^(1×0−0×0)+k^(1×1−0×0)=k^ȷ^×k^=det⁡[ı^ȷ^k^010001]=ı^(1×1−0×0)−ȷ^(0×1−0×0)+k^(0×0−1×0)=ı^k^×ı^=det⁡[ı^ȷ^k^001100]=ı^(0×0−1×0)−ȷ^(0×0−1×1)+k^(0×0−0×1)=ȷ^\begin{alignat*}{5} \hi\times\hj&=\det\left[\begin{matrix}\hi&\hj &\hk\\ 1&0&0\\ 0&1&0\end{matrix}\right] &&=\hi(0\times 0-0\times 1) -\hj(1\times 0-0\times 0) +\hk(1\times 1-0\times 0) \\ &=\hk\\[0.1in] \hj\times\hk&=\det\left[\begin{matrix}\hi&\hj &\hk\\ 0&1&0\\ 0&0&1\end{matrix}\right] &&=\hi(1\times 1-0\times 0) -\hj(0\times 1-0\times 0) +\hk(0\times 0-1\times 0)\\ &=\hi\\[0.1in] \hk\times\hi&=\det\left[\begin{matrix}\hi&\hj &\hk\\ 0&0&1\\ 1&0&0\end{matrix}\right] &&=\hi(0\times 0-1\times 0) -\hj(0\times 0-1\times 1) +\hk(0\times 0-0\times 1)\\ &=\hj \end{alignat*}

(b)

a⋅(a×b)=a1(a2b3−a3b2)−a2(a1b3−a3b1)+a3(a1b2−a2b1)=0b⋅(a×b)=b1(a2b3−a3b2)−b2(a1b3−a3b1)+b3(a1b2−a2b1)=0\begin{alignat*}{5} \va\cdot(\va\times\vb) &=a_1\big(a_2b_3-a_3b_2\big) -a_2\big(a_1b_3-a_3b_1\big) +a_3\big(a_1b_2-a_2b_1\big) &&=0\\ \vb\cdot(\va\times\vb) &=b_1\big(a_2b_3-a_3b_2\big) -b_2\big(a_1b_3-a_3b_1\big) +b_3\big(a_1b_2-a_2b_1\big) &&=0 \end{alignat*}
Q11Stage 1

Consider the following statement: “If a≠0\va\ne\vZero and if a⋅b=a⋅c\va\cdot\vb=\va\cdot\vc then b=c\vb=\vc.” If the statment is true, prove it. If the statement is false, give a counterexample.

Answer

This statement is false. One counterexample is a=<1,0,0>\va=\llt 1,0,0\rgt, b=<0,1,0>, c=<0,0,1>\vb=\llt 0,1,0\rgt ,\ \vc=\llt 0,0,1\rgt. Then a⋅b=a⋅c=0\va\cdot\vb=\va\cdot\vc=0, but b≠c\vb\ne\vc. There are many other counterexamples.

Full solution

This statement is false. The two numbers a⋅b\va\cdot\vb, a⋅c\va\cdot\vc are equal if and only if a⋅(b−c)=0\va\cdot(\vb-\vc)= 0. This in turn is the case if and only if a\va is perpendicular to b−c\vb-\vc (under the convention that 0\vZero is perpendicular to all vectors). For example, if a=<1,0,0>\va=\llt 1,0,0\rgt, b=<0,1,0>, c=<0,0,1>\vb=\llt 0,1,0\rgt ,\ \vc=\llt 0,0,1\rgt, then b−c=<0,1,−1>\vb-\vc=\llt 0,1,-1\rgt is perpendicular to a\va so that a⋅b=a⋅c\va\cdot\vb=\va\cdot\vc.

Q12Stage 1

Consider the following statement: “The vector a×(b×c)\va\times(\vb\times\vc) is of the form αb+βc\al\vb+\be\vc for some real numbers α\al and β\be.” If the statement is true, prove it. If the statement is false, give a counterexample.

Answer

True.

Full solution

This statement is true. In the event that b\vb and c\vc are parallel, b×c=0\vb\times\vc=\vZero so that a×(b×c)=0=0b+0c\va\times(\vb\times\vc)=\vZero=0\vb+0\vc, so we may assume that b\vb and c\vc are not parallel. Then as α\al and β\be run over R\bbbr, the vector αb+βc\al\vb+\be \vc runs over the plane that contains the origin and the vectors b\vb and c\vc. Call this plane PP. Because d=b×c\vd=\vb\times\vc is nonzero and perpendicular to both b\vb and c\vc, PP is the plane that contains the origin and is perpendicular to d\vd. As a×(b×c)=a×d\va\times(\vb\times\vc)=\va\times\vd is always perpendicular to d\vd, it lies in PP.

Q13Stage 1

What geometric conclusions can you draw from a⋅(b×c)=<1,2,3>\va\cdot(\vb\times\vc)=\llt 1,2,3\rgt?

Answer

None. The given equation is nonsense.

Full solution

None. The given equation is nonsense. The left hand side is a number while the right hand side is a vector.

Q14Stage 1

What geometric conclusions can you draw from a⋅(b×c)=0\va\cdot(\vb\times\vc)=0?

Answer

If b\vb and c\vc are parallel, then a⋅(b×c)=0\va\cdot(\vb\times\vc)=0 for all a\va. If b\vb and c\vc are not parallel, then a\va must be of the form
αb+βc\al\vb+\be\vc with α\al and β\be real numbers.

Full solution

If b\vb and c\vc are parallel, then b×c=0\vb\times\vc=\vZero and a⋅(b×c)=0\va\cdot(\vb\times\vc)=0 for all a\va. If b\vb and c\vc are not parallel, a⋅(b×c)=0\va\cdot(\vb\times\vc)=0 if and only if a\va is perpendicular to d=b×c\vd=\vb\times\vc. As we saw in question 12, the set of all vectors perpendicular to d\vd is the plane consisting of all vectors of the form
αb+βc\al\vb+\be\vc with α\al and β\be real numbers. So a\va must be of this form.

Q15Stage 1

Consider the three points O=(0,0)O=(0,0), A=(a,0)A=(a,0) and B=(b,c)B=(b,c).

  1. Sketch, in a single figure,

    • the triangle with vertices OO, AA and BB, and

    • the circumscribing circle for the triangle (i.e. the circle that goes through all three vertices), and

    • the vectors

      • OA→\overrightarrow{OA}, from OO to AA,

      • OB→\overrightarrow{OB}, from OO to BB,

      • OC→\overrightarrow{OC}, from OO to CC, where CC is the centre of the circumscribing circle.

    Then add to the sketch and evaluate, from the sketch,

    • the projection of the vector OC→\overrightarrow{OC} on the vector OA→\overrightarrow{OA}, and

    • the projection of the vector OC→\overrightarrow{OC} on the vector OB→\overrightarrow{OB}.

  2. Determine CC.

  3. Evaluate, using the formula (1.2.14) in the CLP-3 text,

    • the projection of the vector OC→\overrightarrow{OC} on the vector OA→\overrightarrow{OA}, and

    • the projection of the vector OC→\overrightarrow{OC} on the vector OB→\overrightarrow{OB}.

Hint

(a) The three line segments from CC to OO, from CC to AA and from CC to BB all have exactly the same length, namely the radius of the circumscribing circle.

(b) Let (xˉ,yˉ)(\bar x,\bar y) be the coordinates of CC. Write down the equations that say that (xˉ,yˉ)(\bar x,\bar y) is equidistant from the three vertices OO, AA and BB.

Answer

(a), (c)

Figure from prob_s1.2, line 645

Figure from prob_s1.2, line 645

projOA→ OC→=OPA→=<a/2,0>\text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC} =\overrightarrow{OP_A}=\llt a/2,0\rgt projOB→ OC→=OPA→=<b/2,c/2>\text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC} =\overrightarrow{OP_A}=\llt b/2,c/2\rgt

(b) The centre of the circumscribing circle is (xˉ,yˉ)(\bar x,\bar y) with xˉ=a2\bar x=\frac{a}{2} and yˉ=b2+c2−ab2c\bar y =\frac{b^2+c^2-ab}{2c}.

Full solution

(a) The sketch for part (a) is on the left below. To sketch the projections, we dropped perpendiculars

  • from CC to the line from OO to AA, and

  • from CC to the line from OO to BB.

By definition,

  • projOA→ OC→\text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC} is the vector OPA→\overrightarrow{OP_A} from OO to the point PAP_A, where the perpendicular from CC to the line from OO to AA hits the line, and

  • projOB→ OC→\text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC} is the vector OPB→\overrightarrow{OP_B} from OO to the point PBP_B, where the perpendicular from CC to the line from OO to BB hits the line.

Figure from prob_s1.2, line 645

Figure from prob_s1.2, line 645

Figure from prob_s1.2, line 660

Figure from prob_s1.2, line 660

To evaluate the projections we observe that the three lines from CC to OO, from CC to AA and from CC to BB all have exactly the same length (namely the radius of the circumscribing circle). Consequently (see the figure on the right above),

  • the triangle OCAOCA is an isoceles triangle, so that PAP_A is exactly the midpoint of the line segement from OO to AA. That is, PAP_A is (a/2,0)(a/2,0) and

    projOA→ OC→=OPA→=<a/2,0>\begin{equation*} \text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC} =\overrightarrow{OP_A}=\llt a/2,0\rgt \end{equation*}

  • Similarly, the triangle OCBOCB is an isoceles triangle, so that PBP_B is exactly the midpoint of the line segement from OO to BB. That is PAP_A is (b/2,c/2)(b/2,c/2) and

    projOB→ OC→=OPB→=<b/2,c/2>\begin{equation*} \text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC} =\overrightarrow{OP_B}=\llt b/2,c/2\rgt \end{equation*}

(b) Call the centre of the circumscribing circle (xˉ,yˉ)(\bar x,\bar y). This centre must be equidistant from the three vertices. So

xˉ2+yˉ2=(xˉ−a)2+yˉ2=(xˉ−b)2+(yˉ−c)2\begin{align*} \bar x^2+\bar y^2=(\bar x-a)^2+\bar y^2=(\bar x-b)^2+(\bar y-c)^2 \end{align*}

or, subtracting xˉ2+yˉ2\bar x^2+\bar y^2 from all three expression,

0=a2−2axˉ=b2−2bxˉ+c2−2cyˉ\begin{align*} 0=a^2-2a\bar x=b^2-2b\bar x+c^2-2c\bar y \end{align*}

which implies

xˉ=a2yˉ=b2+c2−2bxˉ2c=b2+c2−ab2c\begin{equation*} \bar x=\frac{a}{2}\qquad\qquad \bar y =\frac{b^2+c^2-2b\bar x}{2c}=\frac{b^2+c^2-ab}{2c} \end{equation*}

(c) From part (b), we have

OA→⋅OC→=<a,0>⋅<a2 , b2+c2−ab2c>=a22=12∣OA→∣2OB→⋅OC→=<b,c>⋅<a2 , b2+c2−ab2c>=ab2+b2+c2−ab2=b2+c22=12∣OB→∣2\begin{align*} \overrightarrow{OA}\cdot\overrightarrow{OC} &=\llt a,0\rgt\cdot\llt\frac{a}{2}\,,\,\frac{b^2+c^2-ab}{2c}\rgt =\frac{a^2}{2}=\frac{1}{2}|\overrightarrow{OA}|^2\\ \overrightarrow{OB}\cdot\overrightarrow{OC} &=\llt b,c\rgt\cdot\llt\frac{a}{2}\,,\,\frac{b^2+c^2-ab}{2c}\rgt =\frac{ab}{2}+\frac{b^2+c^2-ab}{2}=\frac{b^2+c^2}{2} =\frac{1}{2}|\overrightarrow{OB}|^2 \end{align*}

So, by Equation (1.2.14) in the CLP-3 text,

projOA→ OC→=OA→⋅OC→∣OA→∣2OA→=12OA→=<a/2,0>projOB→ OC→=OB→⋅OC→∣OB→∣2OB→=12OB→=<b/2,c/2>\begin{align*} \text{proj}_{\overrightarrow{\scriptstyle OA}}\,\overrightarrow{OC} &=\frac{\overrightarrow{OA}\cdot\overrightarrow{OC}}{|\overrightarrow{OA}|^2} \overrightarrow{OA} =\frac{1}{2}\overrightarrow{OA} =\llt a/2,0\rgt \\ \text{proj}_{\overrightarrow{\scriptstyle OB}}\,\overrightarrow{OC} &=\frac{\overrightarrow{OB}\cdot\overrightarrow{OC}}{|\overrightarrow{OB}|^2} \overrightarrow{OB} =\frac{1}{2}\overrightarrow{OB} =\llt b/2,c/2\rgt \end{align*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q16Stage 2

Find the equation of a sphere if one of its diameters has end points (2,1,4)(2,1,4) and (4,3,10)(4,3,10).

Hint

The centre of the sphere is the midpoint of the diameter.

Answer

(x−3)2+(y−2)2+(z−7)2=11(x-3)^2+(y-2)^2+(z-7)^2=11

Full solution

The center of the sphere is 12{(2,1,4)+(4,3,10)}=(3,2,7)\half\big\{(2,1,4)+(4,3,10)\big\}=(3,2,7). The diameter (i.e. twice the radius) is ∣(2,1,4)−(4,3,10)∣=∣(−2,−2,−6)∣=2∣(1,1,3)∣=211|(2,1,4)-(4,3,10)|=|(-2,-2,-6)|=2|(1,1,3)|=2\sqrt{11}. So the radius of the sphere is 11\sqrt{11} and the equation of the sphere is

(x−3)2+(y−2)2+(z−7)2=11\begin{equation*} (x-3)^2+(y-2)^2+(z-7)^2=11 \end{equation*}
Q17Stage 2

Use vectors to prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Hint

Draw a sketch. Call the vertices of the triangle AA, BB and CC with CC being the vertex that joins the two sides. Let a\va be the vector from CC to AA and b\vb be the vector from CC to BB. Determine, in terms of a\va and b\vb,

  • the vector from AA to BB,

  • the two vectors from CC to the two midpoints and finally

  • the vector joining the two midpoints.

Answer

See the solution.

Full solution

Call the vertices of the triangle AA, BB and CC with CC being the vertex that joins the two sides. We can always choose our coordinate system so that CC is at the origin. Let a\va be the vector from CC to AA and b\vb be the vector from CC to BB.

Figure from prob_s1.2, line 815

Figure from prob_s1.2, line 815

  • Then the vector from CC to the midpoint of the side from CC to AA is 12a\half\va and

  • the vector from CC to the midpoint of the side from CC to BB is 12b\half\vb so that

  • the vector joining the two midpoints is 12b−12a\half\vb-\half\va.

As the vector from AA to BB is b−a=2[12b−12a]\vb-\va=2\big[\half\vb-\half\va\big], the line joining the midpoints is indeed parallel to the third side and half its length.

Q18Stage 2

Compute the areas of the parallelograms determined by the following vectors.

  1. <−3,1>, <4,3>\llt -3,1\rgt,\ \llt 4,3\rgt

  2. <4,2>, <6,8>\llt 4,2\rgt,\ \llt 6,8\rgt

Hint

Review §1.2.4 in the CLP-3 text.

Answer

(a) 1313 (b) 2020

Full solution

(a) By (1.2.17) in the CLP-3 text, the area is

∣det⁡[−3143]∣=∣−3×3−1×4∣=∣−13∣=13\begin{align*} \left| \det\left[\begin{matrix}-3&1 \\ 4&3 \end{matrix}\right] \right| &=\big|-3\times 3-1\times 4\big| = |-13| = 13 \end{align*}

(b) By (1.2.17) in the CLP-3 text, the area is

∣det⁡[4268]∣=∣4×8−2×6∣=20\begin{align*} \left|\det\left[\begin{matrix} 4&2 \\ 6&8 \end{matrix}\right]\right| &=\big|4\times 8-2\times 6\big| = 20 \end{align*}
Q19Stage 2Past exam · M200 2014A

Consider the plane WW, defined by:

W : −x+3y+3z=6,\begin{equation*} W\ :\ -x + 3y + 3z = 6,\qquad \end{equation*}

Find the area of the parallelogram on WW defined by 0≤x≤30 \le x \le 3, 0≤y≤20 \le y \le 2.

Hint

Determine the four corners of the parallelogram.

Answer

2192\sqrt{19}

Full solution

Note that

  • the point on WW with x=0x=0, y=0y=0 obeys −0+3(0)+3z=6-0+3(0)+3z=6 and so has z=2z=2

  • the point on WW with x=0x=0, y=2y=2 obeys −0+3(2)+3z=6-0+3(2)+3z=6 and so has z=0z=0

  • the point on WW with x=3x=3, y=0y=0 obeys −3+3(0)+3z=6-3+3(0)+3z=6 and so has z=3z=3

  • the point on WW with x=3x=3, y=2y=2 obeys −3+3(2)+3z=6-3+3(2)+3z=6 and so has z=1z=1

So the four corners of the parallelogram are (0,0,2)(0,0,2), (0,2,0)(0,2,0), (3,0,3)(3,0,3) and (3,2,1)(3,2,1). The vectors

d1=<0−0 , 2−0 , 0−2>=<0 , 2 , −2>d2=<3−0 , 0−0 , 3−2>=<3 , 0 , 1>\begin{align*} \vd_1&=\llt 0-0 \,,\, 2-0 \,,\, 0-2 \rgt = \llt 0 \,,\, 2 \,,\, -2\rgt \\ \vd_2&=\llt 3-0 \,,\, 0-0 \,,\, 3-2 \rgt = \llt 3 \,,\, 0 \,,\, 1\rgt \end{align*}

form two sides of the paralleogram. So the area of the parallelogram is

∣d1×d2∣=∣det⁡[ı^ȷ^k^02−2301]∣=∣2 ı^−6 ȷ^−6k^∣=76=219\begin{align*} \big|\vd_1\times\vd_2\big| =\left|\det\left[\begin{matrix} \hi & \hj & \hk \\ 0 & 2 & -2 \\ 3 & 0 & 1 \end{matrix}\right]\right| =\left| 2\,\hi - 6\,\hj -6\hk \right| =\sqrt{76} =2\sqrt{19} \end{align*}
Q20Stage 2

Compute the volumes of the parallelepipeds determined by the following vectors.

  1. <4,1,−1>, <−1,5,2>, <1,1,6>\llt 4,1,-1\rgt,\ \llt -1,5,2\rgt,\ \llt 1,1,6\rgt

  2. <−2,1,2>, <3,1,2>, <0,2,5>\llt -2,1,2\rgt,\ \llt 3,1,2\rgt,\ \llt 0,2,5\rgt

Hint

Review §1.2.4 in the CLP-3 text.

Answer

(a) 126126 (b) 55

Full solution

(a) By (1.2.18) in the CLP-3 text, the volume is

∣det⁡[41−1−152116]∣=∣4det⁡[5216]−1det⁡[−1216]+(−1)det⁡[−1511]∣=∣4(30−2)−1(−6−2)−1(−1−5)∣=4×28+8+6=126\begin{align*} \left| \det\left[\begin{matrix} 4&1&-1 \\ -1&5&2 \\ 1&1&6\end{matrix}\right] \right| &=\left| 4\det\left[\begin{matrix} 5&2 \\ 1&6 \end{matrix}\right] -1\det\left[\begin{matrix} -1&2 \\ 1&6 \end{matrix}\right] +(-1)\det\left[\begin{matrix} -1&5\\1&1\end{matrix}\right]\right| \\ &= \big|4(30-2)-1(-6-2)-1(-1-5)\big| = 4\times28+8+6\\ &=126 \end{align*}

(b) By (1.2.18) in the CLP-3 text, the volume is

∣det⁡[−212312025]∣=∣−2det⁡[1225]−1det⁡[3205]+2det⁡[3102]∣=∣−2(5−4)−1(15−0)+2(6−0)∣=∣−2−15+12∣=∣−5∣=5\begin{align*} \left|\det\left[\begin{matrix} -2&1&2\\ 3&1&2\\ 0&2&5\end{matrix}\right] \right| &=\left|-2\det\left[\begin{matrix} 1&2\\ 2&5\end{matrix}\right] -1\det\left[\begin{matrix} 3&2\\ 0&5\end{matrix}\right] +2\det\left[\begin{matrix} 3&1\\ 0&2\end{matrix}\right] \right|\\ &=\big|-2(5-4)-1(15-0)+2(6-0)\big| =\big|-2-15+12\big|=\big|-5\big|\\ &=5 \end{align*}
Q21Stage 2

Compute the dot product of the vectors a\va and b\vb. Find the angle between them.

  1. a=<1,2>, b=<−2,3>\va=\llt 1,2\rgt ,\ \vb=\llt -2,3\rgt

  2. a=<−1,1>, b=<1,1>\va=\llt -1,1\rgt ,\ \vb=\llt 1,1\rgt

  3. a=<1,1>, b=<2,2>\va=\llt 1,1\rgt ,\ \vb=\llt 2,2\rgt

  4. a=<1,2,1>, b=<−1,1,1>\va=\llt 1,2,1\rgt ,\ \vb=\llt -1,1,1\rgt

  5. a=<−1,2,3>, b=<3,0,1>\va=\llt -1,2,3\rgt ,\ \vb=\llt 3,0,1\rgt

Answer
a⋅b=4θ=60.25∘a⋅b=0θ=90∘a⋅b=4θ=0∘a⋅b=2θ=61.87∘a⋅b=0θ=90∘\begin{alignat*}{3} \va\cdot\vb&=4\qquad & \theta &= 60.25^\circ\hskip4in \tag{a}\\ \va\cdot\vb&=0 & \theta &= 90^\circ \tag{b}\\ \va\cdot\vb&=4 & \theta &= 0^\circ \tag{c}\\ \va\cdot\vb&=2 & \theta &= 61.87^\circ \tag{d}\\ \va\cdot\vb&=0 & \theta &= 90^\circ \tag{e} \end{alignat*}
Full solution
a⋅b=<1,2>⋅<−2,3>=4cos⁡θ=4513=.4961θ=60.25∘a⋅b=<−1,1>⋅<1,1>=0cos⁡θ=022=0θ=90∘a⋅b=<1,1>⋅<2,2>=4cos⁡θ=428=1θ=0∘a⋅b=<1,2,1>⋅<−1,1,1>=2cos⁡θ=263=.4714θ=61.87∘a⋅b=<−1,2,3>⋅<3,0,1>=0cos⁡θ=01410=0θ=90∘\begin{align*} \va\cdot\vb&=\llt 1,2\rgt\cdot\llt -2,3\rgt=4 & \cos\theta&=\frac{4}{\sqrt{5}\sqrt{13}}=.4961 & \theta &= 60.25^\circ \tag{a}\\ \va\cdot\vb&=\llt -1,1\rgt\cdot\llt 1,1\rgt=0 & \cos\theta&=\frac{0}{\sqrt{2}\sqrt{2}}=0 & \theta &= 90^\circ \tag{b}\\ \va\cdot\vb&=\llt 1,1\rgt\cdot\llt 2,2\rgt=4 & \cos\theta&=\frac{4}{\sqrt{2}\sqrt{8}}=1 & \theta &= 0^\circ \tag{c}\\ \va\cdot\vb&=\llt 1,2,1\rgt\cdot\llt -1,1,1\rgt=2 & \cos\theta&=\frac{2}{\sqrt{6}\sqrt{3}}=.4714 & \theta &= 61.87^\circ \tag{d}\\ \va\cdot\vb&=\llt -1,2,3\rgt\cdot\llt 3,0,1\rgt=0 & \cos\theta&=\frac{0}{\sqrt{14}\sqrt{10}}=0 & \theta &= 90^\circ \tag{e} \end{align*}
Q22Stage 2

Determine the angle between the vectors a\va and b\vb if

  1. a=<1,2>, b=<3,4>\va=\llt 1,2\rgt,\ \vb=\llt 3,4\rgt

  2. a=<2,1,4>, b=<4,−2,1>\va=\llt 2,1,4\rgt,\ \vb=\llt 4,-2,1\rgt

  3. a=<1,−2,1>, b=<3,1,0>\va=\llt 1,-2,1\rgt,\ \vb=\llt 3,1,0\rgt

Answer

(a) 10.3∘10.3^\circ (b) 61.6∘61.6^\circ (c) 82.6∘82.6^\circ

Full solution

By property 6 of Theorem 1.2.11 in the CLP-3 text,

cos⁡θ=a⋅b∣a∣ ∣b∣=1×3+2×41+49+16=1155=.9839  ⟹  θ=10.3∘cos⁡θ=a⋅b∣a∣ ∣b∣=2×4−1×2+4×14+1+1616+4+1=1021=.4762  ⟹  θ=61.6∘cos⁡θ=a⋅b∣a∣ ∣b∣=1×3−2×1+1×01+4+19+1=160=.1291  ⟹  θ=82.6∘\begin{alignat*}{3} &\cos\theta=\frac{\va\cdot\vb}{|\va|\,|\vb|} =\frac{1\times 3+2\times 4}{\sqrt{1+4}\sqrt{9+16}} =\frac{11}{5\sqrt{5}}= .9839 \qquad &&\implies\quad \theta=10.3^\circ \tag{a} \\ &\cos\theta=\frac{\va\cdot\vb}{|\va|\,|\vb|} =\frac{2\times 4-1\times 2+4\times 1}{\sqrt{4+1+16}\sqrt{16+4+1}} =\frac{10}{21}= .4762 \qquad &&\implies\quad \theta=61.6^\circ \tag{b} \\ &\cos\theta=\frac{\va\cdot\vb}{|\va|\,|\vb|} =\frac{1\times 3-2\times 1+1\times 0}{\sqrt{1+4+1}\sqrt{9+1}} =\frac{1}{\sqrt{60}}= .1291 \qquad &&\implies\quad \theta=82.6^\circ \tag{c} \end{alignat*}
Q23Stage 2

Determine all values of yy for which the given vectors are perpendicular.

  1. <2,4>, <2,y>\llt 2,4\rgt ,\ \llt 2,y\rgt

  2. <4,−1>, <y,y2>\llt 4,-1\rgt ,\ \llt y,y^2\rgt

  3. <3,1,1>, <2,5y,y2>\llt 3,1,1\rgt ,\ \llt 2,5y,y^2\rgt

Answer

(a) −1-1 (b) 00, 44 (c) −2-2, −3-3

Full solution
<2,4>⋅<2,y>=2×2+4×y=4+4y=0   ⟺   y=−1<4,−1>⋅<y,y2>=4×y−1×y2=4y−y2=0   ⟺   y=0,4<3,1,1>⋅<2,5y,y2>=6+5y+y2=0   ⟺   y=−2,−3\begin{alignat*}{3} &\llt 2,4\rgt \cdot\llt 2,y\rgt =2\times2+4\times y=4+4y=0 &&\ \iff\ y=-1 \tag{a} \\ &\llt 4,-1\rgt \cdot\llt y,y^2\rgt =4\times y-1\times y^2=4y-y^2=0 &&\ \iff\ y=0,4 \tag{b} \\ &\llt 3,1,1\rgt \cdot\llt 2,5y,y^2\rgt =6+5y+y^2=0 &&\ \iff\ y=-2,-3 \tag{c} \end{alignat*}
Q24Stage 2

Let u=−2ı^+5ȷ^\vu=-2\hi+5\hj and v=αı^−2ȷ^\vv=\al\hi-2\hj. Find α\al so that

  1. u⊥v\vu\perp\vv

  2. u∥v\vu \| \vv

  3. The angle between u\vu and v\vv is 60∘60^\circ.

Answer

(a) −5-5 (b) 0.80.8 (c) none

Full solution

(a) We want 0=u⋅v=−2α−100=\vu\cdot\vv=-2\al-10 or α=−5\al=-5.

(b) We want −2/α=5/(−2)-2/\al=5/(-2) or α=0.8\al=0.8.

(c) We want u⋅v=−2α−10=∣u∣ ∣v∣ cos⁡60∘=29 α2+4 12\vu\cdot\vv=-2\al-10 =|\vu|\,|\vv|\,\cos 60^\circ =\sqrt{29}\,\sqrt{\al^2+4}\,\half. Squaring both sides gives

4α2+40α+100=294(α2+4)  ⟹  13α2−160α−284=0  ⟹  α=160±1602+4×13×28426≈13.88 or −1.574\begin{alignat*}{3} & & 4\al^2+40\al+100&=\frac{29}{4}(\al^2+4) \\ &\implies\quad & 13\al^2-160\al-284&=0 \\ &\implies\quad & \al &=\frac{160\pm\sqrt{160^2+4\times13\times284}}{26} \approx 13.88\text{ or }-1.574 \end{alignat*}

Both of these α\al's give u⋅v<0\vu\cdot\vv<0 so no α\al works.

Q25Stage 2

Define a=<1,2,3>\va=\llt 1,2,3\rgt and b=<4,10,6>\vb=\llt 4,10,6\rgt.

  1. Find the component of b\vb in the direction a\va.

  2. Find the projection of b\vb on a\va.

  3. Find the projection of b\vb perpendicular to a\va.

Answer

(a) 4214\frac{42}{\sqrt{14}} (b) <3,6,9>\llt 3,6,9\rgt (c) <1,4,−3>\llt 1,4,-3\rgt

Full solution

(a) The component of b\vb in the direction a\va is

b⋅a∣a∣=1×4+2×10+3×61+4+9=4214\vb\cdot\frac{\va}{|\va|} =\frac{1\times 4+2\times 10+3\times 6}{\sqrt{1+4+9}} =\frac{42}{\sqrt{14}}

(b) The projection of b\vb on a\va is a vector of length 42/1442/\sqrt{14} in direction a/∣a∣\va/|\va|, namely 4214<1,2,3>=<3,6,9>\frac{42}{14}\llt 1,2,3\rgt=\llt 3,6,9\rgt.

(c) The projection of b\vb perpendicular to a\va is b\vb minus its projection on a\va, namely <4,10,6>−<3,6,9>=<1,4,−3>\llt 4,10,6\rgt-\llt 3,6,9\rgt=\llt 1,4,-3\rgt.

Q26Stage 2

Compute <1,2,3>×<4,5,6>\llt 1,2,3\rgt\times\llt 4,5,6\rgt.

Answer

−3ı^+6ȷ^−3k^-3\hi+6\hj-3\hk

Full solution
<1,2,3>×<4,5,6>=det⁡[ı^ȷ^k^123456]=ı^ (2×6−3×5)−ȷ^ (1×6−3×4)+k^ (1×5−2×4)=−3 ı^+6 ȷ^−3 k^\begin{align*} \llt 1,2,3\rgt\times\llt 4,5,6\rgt &=\det\left[\begin{matrix}\hi&\hj &\hk \\ 1&2&3 \\ 4&5&6\end{matrix}\right] =\hi\,(2\times 6-3\times 5) -\hj\,(1\times 6-3\times 4) +\hk\,(1\times 5-2\times 4) \\ &=-3\,\hi+6\,\hj-3\,\hk \end{align*}
Q27Stage 2

Calculate the following cross products.

  1. <1,−5,2>×<−2,1,5>\llt 1,-5,2\rgt \times\llt -2,1,5\rgt

  2. <2,−3,−5>×<4,−2,7>\llt 2,-3,-5\rgt \times\llt 4,-2,7\rgt

  3. <−1,0,1>×<0,4,5>\llt -1,0,1\rgt \times\llt 0,4,5\rgt

Answer

(a) <−27,−9,−9>\llt -27,-9,-9\rgt (b) <−31,−34,8>\llt -31,-34,8\rgt (c) <−4,5,−4>\llt -4,5,-4\rgt

Full solution
det⁡[ı^ȷ^k^1−52−215]=ı^det⁡[−5215]−ȷ^det⁡[12−25]+k^det⁡[1−5−21]=ı^(−25−2)−ȷ^(5+4)+k^(1−10)=<−27,−9,−9>det⁡[ı^ȷ^k^2−3−54−27]=ı^det⁡[−3−5−27]−ȷ^det⁡[2−547]+k^det⁡[2−34−2]=ı^(−21−10)−ȷ^(14+20)+k^(−4+12)=<−31,−34,8>det⁡[ı^ȷ^k^−101045]=ı^det⁡[0145]−ȷ^det⁡[−1105]+k^det⁡[−1004]=ı^(0−4)−ȷ^(−5−0)+k^(−4−0)=<−4,5,−4>\begin{align} \det\left[\begin{matrix}\hi&\hj&\hk\cr1&-5&2\cr-2&1&5\end{matrix}\right] &=\hi\det\left[\begin{matrix}-5&2\cr1&5\end{matrix}\right] -\hj\det\left[\begin{matrix}1&2\cr-2&5\end{matrix}\right] +\hk\det\left[\begin{matrix}1&-5\cr-2&1\end{matrix}\right]\tag{a}\\ &=\hi(-25-2)-\hj(5+4)+\hk(1-10) = \llt -27,-9,-9\rgt \notag\\ \det\left[\begin{matrix}\hi&\hj&\hk\cr2&-3&-5\cr4&-2&7\end{matrix}\right] &=\hi\det\left[\begin{matrix}-3&-5\cr-2&7\end{matrix}\right] -\hj\det\left[\begin{matrix}2&-5\cr4&7\end{matrix}\right] +\hk\det\left[\begin{matrix}2&-3\cr4&-2\end{matrix}\right]\tag{b} \\ &=\hi(-21-10)-\hj(14+20)+\hk(-4+12) = \llt -31,-34,8\rgt \notag\\ \det\left[\begin{matrix}\hi&\hj&\hk\cr-1&0&1\cr0&4&5\end{matrix}\right] &=\hi\det\left[\begin{matrix}0&1\cr4&5\end{matrix}\right] -\hj\det\left[\begin{matrix}-1&1\cr0&5\end{matrix}\right] +\hk\det\left[\begin{matrix}-1&0\cr0&4\end{matrix}\right] \tag{c} \\ &=\hi(0-4)-\hj(-5-0)+\hk(-4-0) = \llt -4,5,-4\rgt \notag \end{align}
Q28Stage 2

Let p=<−1,4,2>, q=<3,1,−1>, r=<2,−3,−1>\vp=\llt -1,4,2\rgt ,\ \vq=\llt 3,1,-1\rgt ,\ \vr=\llt 2,-3,-1\rgt. Check, by direct computation, that

  1. p×p=0\vp\times\vp=\vZero

  2. p×q=−q×p\vp\times\vq=-\vq\times\vp

  3. p×(3r)=3(p×r)\vp\times(3\vr)=3(\vp\times\vr)

  4. p×(q+r)=p×q+p×r\vp\times(\vq+\vr) = \vp\times\vq+\vp\times\vr

  5. p×(q×r)≠(p×q)×r\vp\times(\vq\times\vr) \ne (\vp\times\vq)\times\vr

Answer

(a) See the solution.

(b) p×q=−q×p=<−6,5,−13>\vp\times\vq = -\vq\times\vp = \llt -6,5,-13\rgt

(c) p×(3r)=3(p×r)=<6,9,−15>\vp \times (3\vr) = 3(\vp\times\vr) = \llt 6,9,-15\rgt

(d) p×(q+r)=p×q+p×r=<−4,8,−18>\vp\times(\vq+\vr) = \vp\times\vq+\vp\times\vr = \llt -4,8,-18\rgt

(e) p×(q×r)=<−46,−19,15>\vp\times(\vq\times\vr) = \llt -46,-19,15\rgt, (p×q)×r=<−44,−32,8>(\vp\times\vq)\times\vr = \llt -44,-32,8\rgt

Full solution
p×p=det⁡[ı^ȷ^k^−142−142]=ı^(4×2−2×4)−ȷ^(2−(−2))+k^(−4−(−4))=<0,0,0>p×q=det⁡[ı^ȷ^k^−14231−1]=ı^(−4−2)−ȷ^(1−6)+k^(−1−12)=<−6,5,−13>q×p=det⁡[ı^ȷ^k^31−1−142]=ı^(2+4)−ȷ^(6−1)+k^(12+1)=<6,−5,13>p ⁣× ⁣(3r)=det⁡[ı^ȷ^k^−1426−9−3]=ı^(−12+18)−ȷ^(3−12)+k^(9−24)=<6,9,−15>3(p×r)=3det⁡[ı^ȷ^k^−1422−3−1]=3(ı^(−4+6)−ȷ^(1−4)+k^(3−8))=<6,9,−15>\begin{align} \vp\times\vp &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr-1&4&2\end{matrix}\right] =\hi(4\times2-2\times4)-\hj(2-(-2)) +\hk(-4-(-4)) \tag{a} \\ &= \llt 0,0,0\rgt \notag \\ \vp\times\vq &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr3&1&-1\end{matrix}\right] =\hi(-4-2)-\hj(1-6) +\hk(-1-12) = \llt -6,5,-13\rgt \tag{b} \\ \vq\times\vp &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr3&1&-1\cr-1&4&2\end{matrix}\right] =\hi(2+4)-\hj(6-1) +\hk(12+1) = \llt 6,-5,13\rgt \notag \\ \vp\!\times\!(3\vr) &= \det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr6&-9&-3\end{matrix}\right] =\hi(-12+18)-\hj(3-12) +\hk(9-24) = \llt 6,9,-15\rgt \tag{c} \\ 3(\vp\times\vr) &= 3\det\left[ \begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr2&-3&-1\end{matrix}\right] =3\Big(\hi(-4+6)-\hj(1-4) +\hk(3-8) \Big) = \llt 6,9,-15\rgt \notag \end{align}

(d) As q+r=<5,−2,−2>\vq+\vr=\llt 5,-2,-2\rgt

p×(q+r)=det⁡[ı^ȷ^k^−1425−2−2]=ı^(−8+4)−ȷ^(2−10)+k^(2−20)=<−4,8,−18>\begin{align*} \vp\times(\vq+\vr) = \det\left[\begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr5&-2&-2\end{matrix}\right] =\hi(-8+4)-\hj(2-10) +\hk(2-20) = \llt -4,8,-18\rgt \end{align*}

Using the values of p×q\vp\times\vq and 3(p×r)3(\vp\times\vr) computed in parts (b) and (c)

p×q+p×r=<−6,5,−13>+13<6,9,−15>=<−4,8,−18>\begin{equation*} \vp\times\vq+\vp\times\vr=\llt -6,5,-13\rgt +\frac{1}{3}\llt 6,9,-15\rgt = \llt -4,8,-18\rgt \end{equation*}
q×r=det⁡[ı^ȷ^k^31−12−3−1]=ı^(−1−3)−ȷ^(−3+2)+k^(−9−2)=<−4,1,−11>p×(q×r)=det⁡[ı^ȷ^k^−142−41−11]=ı^(−44−2)−ȷ^(11+8)+k^(−1+16)=<−46,−19,15>(p×q)×r=det⁡[ı^ȷ^k^−65−132−3−1]=ı^(−5−39)−ȷ^(6+26)+k^(18−10)=<−44,−32,8>\begin{align} \vq\times\vr &= \det\left[\begin{matrix}\hi&\hj&\hk\cr3&1&-1\cr2&-3&-1\end{matrix}\right] =\hi(-1-3)-\hj(-3+2) +\hk(-9-2) = \llt -4,1,-11\rgt \tag{e} \\ \vp\times(\vq\times\vr) &= \det\left[\begin{matrix}\hi&\hj&\hk\cr-1&4&2\cr-4&1&-11\end{matrix}\right] =\hi(-44-2)-\hj(11+8) +\hk(-1+16) = \llt -46,-19,15\rgt \notag \\ (\vp\times\vq)\times\vr &= \det\left[\begin{matrix}\hi&\hj&\hk\cr-6&5&-13\cr2&-3&-1\end{matrix}\right] =\hi(-5-39)-\hj(6+26) +\hk(18-10) = \llt -44,-32,8\rgt \notag \end{align}
Q29Stage 2

Calculate the area of the triangle with vertices (0,0,0)(0,0,0), (1,2,3)(1,2,3) and (3,2,1)(3,2,1).

Answer

262\sqrt{6}

Full solution

Denote by θ\theta the angle between the two vectors a=<1,2,3>\va=\llt 1,2,3\rgt and b=<3,2,1>\vb=\llt 3,2,1\rgt. The area of the triangle is one half times the length, ∣a∣|\va|, of its base times its height h=∣b∣sin⁡θh=|\vb|\sin\theta.

Figure from prob_s1.2, line 1409

Figure from prob_s1.2, line 1409

Thus the area of the triangle is 12∣a∣ ∣b∣ sin⁡θ\half|\va|\,|\vb|\,\sin\theta. By property 2 of the cross product in Theorem 1.2.23 of the CLP-3 text, ∣a×b∣=∣a∣ ∣b∣ sin⁡θ|\va\times\vb|=|\va|\,|\vb|\,\sin\theta. So

area=12∣a×b∣=12∣<1,2,3>×<3,2,1>∣=12∣ı^ (2−6)−ȷ^ (1−9)+k^ (2−6)∣=1216+64+16=26\begin{align*} \text{area} &= \half|\va\times\vb| =\half|\llt 1,2,3\rgt\times\llt 3,2,1\rgt| \\ &=\half | \hi\,(2-6)-\hj\,(1-9) +\hk\,(2-6)| \\ &=\half\sqrt{16+64+16} \\ &=2\sqrt{6} \end{align*}
Q30Stage 2Past exam · M200 2003D

A particle PP of unit mass whose position in space at time tt is r(t)\vr(t) has angular momentum L(t)=r(t)×r′(t)L(t)=\vr(t)\times\vr'(t). If r′′(t)=ρ(t)r(t)\vr''(t)=\rho(t)\vr(t) for a scalar function ρ\rho, show that LL is constant, i.e. does not change with time. Here ′' denotes ddt\diff{}{t}.

Hint

Evaluate dLdt\diff{L}{t} by differentiating r(t)×r′(t)\vr(t)\times\vr'(t).

Answer

See the solution.

Full solution

The derivative of LL is

dLdt=ddt(r(t)×r′(t))=r′(t)×r′(t)+r(t)×r′′(t)=r′(t)×r′(t)+r(t)×(ρ(t)r(t))\begin{align*} \diff{L}{t}=\diff{}{t}\big(\vr(t)\times\vr'(t)\big) =\vr'(t)\times\vr'(t)+\vr(t)\times\vr''(t) =\vr'(t)\times\vr'(t)+\vr(t)\times\big(\rho(t)\vr(t)\big) \end{align*}

Both terms vanish because the cross product of any two parallel vectors is zero. So dLdt=0\diff{L}{t}=0 and L(t)L(t) is independent of tt.

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q31Stage 3

Show that the diagonals of a parallelogram bisect each other.

Answer

See the solution.

Full solution

The parallelogram determined by the vectors a\va and b\vb has vertices 0, a, b\vZero,\ \va,\ \vb and a+b\va+\vb. As tt varies from 00 to 11, t(a+b)t(\va+\vb) traverses the diagonal from 0\vZero to a+b\va+\vb. As ss varies from 00 to 11, a+s(b−a)\va+s(\vb-\va) traverses the diagonal from a\va to b\vb. These two straight lines meet when ss and tt are such that

t(a+b)=a+s(b−a)\begin{equation*} t(\va+\vb)=\va+s(\vb-\va) \end{equation*}

or

(t+s−1)a=(s−t)b\begin{equation*} (t+s-1)\va=(s-t)\vb \end{equation*}

Assuming that a\va and b\vb are not parallel (i.e. the parallelogram has not degenerated to a line segment), this is the case only when t+s−1=0t+s-1=0 and s−t=0s-t=0. That is, s=t=12s=t=\half. So the two lines meet at their midpoints.

Q32Stage 3

Consider a cube such that each side has length ss. Name, in order, the four vertices on the bottom of the cube A,B,C,DA, B, C, D and the corresponding four vertices on the top of the cube A′,B′,C′,D′A', B', C', D'.

  1. Show that all edges of the tetrahedron A′C′BDA'C'BD have the same length.

  2. Let EE be the center of the cube. Find the angle between EAEA and ECEC.

Answer

(a) All 66 edges have length 2s\sqrt{2}s. (b) 109.5∘109.5^\circ

Full solution

We may choose our coordinate axes so that A=(0,0,0)A=(0,0,0), B=(s,0,0)B=(s,0,0), C=(s,s,0)C=(s,s,0), D=(0,s,0)D=(0,s,0) and A′=(0,0,s)A'=(0,0,s), B′=(s,0,s)B'=(s,0,s), C′=(s,s,s)C'=(s,s,s), D′=(0,s,s)D'=(0,s,s).

(a) Then

∣A′C′∣=∣<s,s,s>−<0,0,s>∣=∣<s,s,0>∣=2 s∣A′B∣=∣<s,0,0>−<0,0,s>∣=∣<s,0,−s>∣=2 s∣A′D∣=∣<0,s,0>−<0,0,s>∣=∣<0,s,−s>∣=2 s∣C′B∣=∣<s,0,0>−<s,s,s>∣=∣<0,−s,−s>∣=2 s∣C′D∣=∣<0,s,0>−<s,s,s>∣=∣<−s,0,−s>∣=2 s∣BD∣=∣<0,s,0>−<s,0,0>∣=∣<−s,s,0>∣=2 s\begin{alignat*}{5} |A'C'|&=\big|\llt s,s,s\rgt -\llt 0,0,s\rgt \big| &&=\big|\llt s,s,0\rgt \big|&=\sqrt{2}\,s\\ |A'B|&=\big|\llt s,0,0\rgt -\llt 0,0,s\rgt \big| &&=\big|\llt s,0,-s\rgt \big|&=\sqrt{2}\,s\\ |A'D|&=\big|\llt 0,s,0\rgt -\llt 0,0,s\rgt \big| &&=\big|\llt 0,s,-s\rgt \big|&=\sqrt{2}\,s\\ |C'B|&=\big|\llt s,0,0\rgt -\llt s,s,s\rgt \big| &&=\big|\llt 0,-s,-s\rgt \big|&=\sqrt{2}\,s\\ |C'D|&=\big|\llt 0,s,0\rgt -\llt s,s,s\rgt \big| &&=\big|\llt -s,0,-s\rgt \big|&=\sqrt{2}\,s\\ |BD|&=\big|\llt 0,s,0\rgt -\llt s,0,0\rgt \big| &&=\big|\llt -s,s,0\rgt \big|&=\sqrt{2}\,s \end{alignat*}

(b) E=12(s,s,s)E=\half(s,s,s) so that EA=<0,0,0>−12<s,s,s>=−12<s,s,s>EA=\llt 0,0,0\rgt -\half\llt s,s,s\rgt =-\half\llt s,s,s\rgt and EC=<s,s,0>−12<s,s,s>=12<s,s,−s>EC=\llt s,s,0\rgt -\half\llt s,s,s\rgt =\half\llt s,s,-s\rgt.

cos⁡θ=−<s,s,s>⋅<s,s,−s>∣<s,s,s>∣ ∣<s,s,−s>∣=−s23s2=−13  ⟹  θ=109.5∘\cos\theta =\frac{-\llt s,s,s\rgt \cdot\llt s,s,-s\rgt} {|\llt s,s,s\rgt |\,|\llt s,s,-s\rgt |} =\frac{-s^2}{3s^2} =-\frac{1}{3} \qquad \implies\quad \theta=109.5^\circ
Q33Stage 3

Find the angle between the diagonal of a cube and the diagonal of one of its faces.

Answer

35.26∘35.26^\circ or 90∘90^\circ or 144.74∘144.74^\circ

Full solution

Suppose that the cube has height, length and width ss. We may choose our coordinate axes so that the vertices of the cube are at (0,0,0)(0,0,0), (s,0,0)(s,0,0), (0,s,0)(0,s,0), (0,0,s)(0,0,s), (s,s,0)(s,s,0), (0,s,s)(0,s,s), (s,0,s)(s,0,s) and (s,s,s)(s,s,s).

We'll start with a couple of examples. The diagonal from (0,0,0)(0,0,0) to (s,s,s)(s,s,s) is <s,s,s>\llt s,s,s\rgt. One face of the cube has vertices (0,0,0)(0,0,0), (s,0,0)(s,0,0), (0,s,0)(0,s,0) and (s,s,0)(s,s,0). One diagonal of this face runs from (0,0,0)(0,0,0) to (s,s,0)(s,s,0) and hence is <s,s,0>\llt s,s,0\rgt. The angle between <s,s,s>\llt s,s,s\rgt and <s,s,0>\llt s,s,0\rgt is

cos⁡−1(<s,s,s>⋅<s,s,0>∣<s,s,s>∣ ∣<s,s,0>∣)=cos⁡−1(2s23s 2s)=cos⁡−1(26)≈35.26∘\begin{align*} \cos^{-1}\left(\frac{\llt s,s,s\rgt\cdot\llt s,s,0\rgt} {|\llt s,s,s\rgt|\,|\llt s,s,0\rgt|}\right) =\cos^{-1}\left(\frac{2s^2}{\sqrt{3}s\,\sqrt{2}s}\right) =\cos^{-1}\left(\frac{2}{\sqrt{6}}\right)\approx 35.26^\circ \end{align*}

A second diagonal for the face with vertices (0,0,0)(0,0,0), (s,0,0)(s,0,0), (0,s,0)(0,s,0) and (s,s,0)(s,s,0) is that running from (s,0,0)(s,0,0) to (0,s,0)(0,s,0). This diagonal is <−s,s,0>\llt -s,s,0\rgt. The angle between <s,s,s>\llt s,s,s\rgt and <−s,s,0>\llt -s,s,0\rgt is

cos⁡−1(<s,s,s>⋅<−s,s,0>∣<s,s,s>∣ ∣<−s,s,0>∣)=cos⁡−1(03s 2s)=cos⁡−1(0)=90∘\begin{align*} \cos^{-1}\left(\frac{\llt s,s,s\rgt\cdot\llt -s,s,0\rgt} {|\llt s,s,s\rgt|\,|\llt -s,s,0\rgt|}\right) =\cos^{-1}\left(\frac{0}{\sqrt{3}s\,\sqrt{2}s}\right) =\cos^{-1}(0)=90^\circ \end{align*}

Now we'll consider the general case. Note that every component of every vertex of the cube is either 00 or ss. In general, two vertices of the cube are at opposite ends of a diagonal of the cube if all three components of the two vertices are different. For example, if one end of the diagonal is (s,0,s)(s,0,s), the other end is (0,s,0)(0,s,0). The diagonals of the cube are all of the form <±s,±s,±s>\llt \pm s,\pm s,\pm s\rgt. All of these diagonals are of length 3s\sqrt{3}s. Two vertices are on the same face of the cube if one of their components agree. They are on opposite ends of a diagonal for the face if their other two components differ. For example (0,s,s)(0,s,s) and (s,0,s)(s,0,s) are both on the face with z=sz=s. Because the xx components 0, s0,\ s are different and the yy components s, 0s,\ 0 are different, (0,s,s)(0,s,s) and (s,0,s)(s,0,s) are the ends of a diagonal of the face with z=sz=s. The diagonals of the faces with z=0z=0 or z=sz=s are <±s,±s,0>\llt \pm s,\pm s,0\rgt. The diagonals of the faces with y=0y=0 or y=sy=s are <±s,0,±s>\llt \pm s,0, \pm s\rgt. The diagonals of the faces with x=0x=0 or x=sx=s are <0,±s,±s>\llt 0,\pm s,\pm s\rgt. All of these diagonals have length 2s\sqrt{2}s. The dot product of one the cube diagonals <±s,±s,±s>\llt \pm s,\pm s,\pm s\rgt with one of the face diagonals <±s,±s,0>\llt \pm s,\pm s,0\rgt, <±s,0,±s>\llt \pm s,0, \pm s\rgt, <0,±s,±s>\llt 0,\pm s,\pm s\rgt is of the form ±s2±s2+0\pm s^2\pm s^2+0 and hence must be either 2s22s^2 or 00 or −2s2-2s^2. In general, the angle between a cube diagonal and a face diagonal is

cos⁡−1(2s2 or 0 or −2s23s 2s)=cos⁡−1(2 or 0 or −26)≈35.26∘ or 90∘ or 144.74∘.\begin{align*} \cos^{-1}\left(\frac{2s^2\text{ or 0 or }-2s^2}{\sqrt{3}s\,\sqrt{2}s}\right) =\cos^{-1}\left(\frac{2\text{ or }0\text{ or }-2}{\sqrt{6}}\right)\approx 35.26^\circ\text{ or }90^\circ\text{ or }144.74^\circ. \end{align*}
Q34Stage 3

Consider a skier who is sliding without friction on the hill y=h(x)y=h(x) in a two dimensional world. The skier is subject to two forces. One is gravity. The other acts perpendicularly to the hill. The second force automatically adjusts its magnitude so as to prevent the skier from burrowing into the hill. Suppose that the skier became airborne at some (x0,y0)(x_0,y_0) with y0=h(x0)y_0=h(x_0). How fast was the skier going?

Answer

1+h′(x0)2−g/h′′(x0)\sqrt{1+h'\big(x_0\big)^2}\sqrt{-g/h''\big(x_0\big)}

Full solution

Denote by (x(t),y(t))\big(x(t),y(t)\big) the position of the skier at time tt. As long as the skier remains on the surface of the hill

y(t)=h(x(t))  ⟹  y′(t)=h′(x(t)) x′(t)  ⟹  y′′(t)=h′′(x(t)) x′(t)2+h′(x(t)) x′′(t)\begin{align*} y(t)&=h\big(x(t)\big) \\ \implies y'(t)&=h'\big(x(t)\big)\,x'(t) \\ \implies y''(t)&=h''\big(x(t)\big)\, x'(t)^2+h'\big(x(t)\big)\,x''(t) \end{align*}

So the velocity and acceleration vectors of the skier are

v(t)=<1,h′(x(t))>x′(t)a(t)=<1,h′(x(t))>x′′(t)+<0,h′′(x(t))>x′(t)2\begin{align*} \vv(t)&=\llt 1,h'\big(x(t)\big)\rgt x'(t) \\ \va(t)&=\llt 1,h'\big(x(t)\big)\rgt x''(t) +\llt 0,h''\big(x(t)\big)\rgt x'(t)^2 \end{align*}

The skier is subject to two forces. One is gravity. The other acts perpendicularly to the hill and has a magnitude such that the skier remains on the surface of the hill. From the velocity vector of the skier (which remain tangential to the hill as long as the skier remains of the surface of the hill),we see that one vector normal to the hill at (x(t),y(t))\big(x(t),y(t)\big) is

n(t)=<−h′(x(t)),1>\vn(t)=\llt-h'\big(x(t)\big),1\rgt

This vector is not a unit vector, but that's ok. By Newton's law of motion

ma=−mg ȷ^+p(t) n(t)\begin{equation*} m\va=-mg\,\hj+p(t)\,\vn(t) \end{equation*}

for some function p(t)p(t). Dot both sides of this equation with n(t)\vn(t).

ma(t)⋅n(t)=−mgȷ^⋅n(t)+p(t)∣n(t)∣2\begin{equation*} m\va(t)\cdot\vn(t)=-mg\hj\cdot\vn(t)+p(t)|\vn(t)|^2 \end{equation*}

Substituting in

mh′′(x(t)) x′(t)2=−mg+p(t)[1+h′(x(t))2]  ⟹  p(t)[1+h′(x(t))2]=m(g+h′′(x(t)) x′(t)2)\begin{align*} mh''\big(x(t)\big)\,x'(t)^2&=-mg+p(t)\left[1+h'\big(x(t)\big)^2\right] \\ \implies p(t)\left[1+h'\big(x(t)\big)^2\right] &=m\Big(g+h''\big(x(t)\big)\,x'(t)^2\Big) \end{align*}

As long as p(t)≥0p(t)\ge 0, the hill is pushing up in order to keep the skier on the surface. When p(t)p(t) becomes negative, the hill has to pull on the skier in order to keep her on the surface. But the hill can't pull, so the skier becomes airborne instead. This happens when

g+h′′(x(t))x′(t)2=0\begin{equation*} g+h''\big(x(t)\big)x'(t)^2=0 \end{equation*}

That is when x′(t)=−g/h′′(x(t))x'(t)=\sqrt{-g/h''\big(x(t)\big)}. At this time x(t)=x0x(t)=x_0, y(t)=y0y(t)=y_0 and the speed of the skier is

x′(t)2+y′(t)2=1+h′(x0)2−g/h′′(x0)\begin{equation*} \sqrt{x'(t)^2+y'(t)^2} =\sqrt{1+h'\big(x_0\big)^2}\sqrt{-g/h''\big(x_0\big)} \end{equation*}
Q35Stage 3

A marble is placed on the plane ax+by+cz=dax+by+cz=d. The coordinate system has been chosen so that the positive zz–axis points straight up. The coefficient cc is nonzero and the coefficients aa and bb are not both zero. In which direction does the marble roll? Why were the conditions “c≠0c\ne 0” and “a,ba,b not both zero” imposed?

Answer

The marble rolls in the directionn<ac,bc,−a2−b2>\llt ac,bc,-a^2-b^2\rgt. If c=0c=0, the plane is vertical. In this case, the marble doesn't roll – it falls straight down. If a=b=0a=b=0, the plane is horizontal. In this case, the marble doesn't roll — it remains stationary.

Full solution

The marble is subject to two forces. The first, gravity, is −mg k^-mg\,\hk with mm being the mass of the marble. The second is the normal force imposed by the plane. This forces acts in a direction perpendicular to the plane. One vector normal to the plane is a ı^+b ȷ^+c k^a\,\hi+b\,\hj+c\,\hk. So the force due to the plane is T<a,b,c>T\llt a,b,c\rgt with TT determined by the property that the net force perpendicular to the plane must be exactly zero, so that the marble remains on the plane, neither digging into nor flying off of it. The projection of the gravitational force onto the normal vector <a,b,c>\llt a,b,c\rgt is

−mg<0,0,1>⋅<a,b,c>∣<a,b,c>∣2<a,b,c>=−mgca2+b2+c2<a,b,c>\begin{equation*} \frac{-mg\llt 0,0,1\rgt\cdot\llt a,b,c\rgt}{|\llt a,b,c\rgt|^2}\llt a,b,c\rgt =\frac{-mgc}{a^2+b^2+c^2}\llt a,b,c\rgt \end{equation*}

The condition that determines TT is thus

T<a,b,c>+−mgca2+b2+c2<a,b,c>=0  ⟹  T=mgca2+b2+c2\begin{equation*} T\llt a,b,c\rgt+\frac{-mgc}{a^2+b^2+c^2}\llt a,b,c\rgt =0 \implies T=\frac{mgc}{a^2+b^2+c^2} \end{equation*}

The total force on the marble is then (ignoring friction – which will have no effect on the direction of motion)

T<a,b,c>−mg<0,0,1>=mgca2+b2+c2<a,b,c>−mg<0,0,1>=mgc<a,b,c>−<0,0,a2+b2+c2>a2+b2+c2=mg<ac,bc,−a2−b2>a2+b2+c2\begin{align*} T\llt a,b,c\rgt-mg\llt 0,0,1\rgt&=\frac{mgc}{a^2+b^2+c^2}\llt a,b,c\rgt-mg\llt 0,0,1\rgt\cr &=mg\frac{c\llt a,b,c\rgt-\llt 0,0,a^2+b^2+c^2\rgt}{a^2+b^2+c^2}\cr &=mg\frac{\llt ac,bc,-a^2-b^2\rgt}{a^2+b^2+c^2}\cr \end{align*}

The direction of motion <ac,bc,−a2−b2>\llt ac,bc,-a^2-b^2\rgt. If you want to turn this into a unit vector, just divide by (a2+b2)(a2+b2+c2)\sqrt{(a^2+b^2)(a^2+b^2+c^2)}. Note that the direction vector in perpendicular <a,b,c>\llt a,b,c\rgt and hence is parallel to the plane. If c=0c=0, the plane is vertical. In this case, the marble doesn't roll – it falls straight down. If a=b=0a=b=0, the plane is horizontal. In this case, the marble doesn't roll — it remains stationary.

Q36Stage 3

Show that a⋅(b×c)=(a×b)⋅c\va\cdot(\vb\times\vc) =(\va\times\vb)\cdot\vc.

Answer

See the solution.

Full solution

By definition, the left and right hand sides are

a⋅(b×c)=<a1,a2,a3>⋅<b2c3−b3c2,b3c1−b1c3,b1c2−b2c1>=a1b2c3−a1b3c2+a2b3c1−a2b1c3+a3b1c2−a3b2c1(a×b)⋅c=<a2b3−a3b2,a3b1−a1b3,a1b2−a2b1>⋅<c1,c2,c3>=a2b3c1−a3b2c1+a3b1c2−a1b3c2+a1b2c3−a2b1c3\begin{align} \va\cdot(\vb\times\vc) &=\llt a_1,a_2,a_3\rgt \cdot\llt b_2c_3-b_3c_2, b_3c_1-b_1c_3, b_1c_2-b_2c_1\rgt \notag\\ &=a_1b_2c_3 - a_1b_3c_2 + a_2b_3c_1 - a_2b_1c_3 + a_3b_1c_2 - a_3b_2c_1 \tag{lhs}\\ \notag\\ (\va\times\vb)\cdot\vc &=\llt a_2b_3-a_3b_2, a_3b_1-a_1b_3, a_1b_2-a_2b_1\rgt \cdot\llt c_1,c_2,c_3\rgt \notag\\ &=a_2b_3c_1 - a_3b_2c_1 + a_3b_1c_2 - a_1b_3c_2 + a_1b_2c_3 - a_2b_1c_3 \tag{rhs} \end{align}

(lhs) and (rhs) are the same.

Q37Stage 3

Show that a×(b×c)=(a⋅c)b−(a⋅b)c\va\times(\vb\times\vc) =(\va\cdot\vc)\vb-(\va\cdot\vb)\vc.

Answer

See the solution.

Full solution

By definition,

b×c = (b2c3−b3c2)ı^−(b1c3−b3c1)ȷ^+(b1c2−b2c1)k^\begin{align*}\vb\times\vc \ =\ &(b_2c_3-b_3c_2)\hi-(b_1c_3-b_3c_1)\hj +(b_1c_2-b_2c_1)\hk\end{align*}

so that the left and right hand sides are

a×(b×c) = det⁡[ı^ȷ^k^a1a2a3b2c3−b3c2−b1c3+b3c1b1c2−b2c1] = ı^ [a2(b1c2−b2c1)−a3(−b1c3+b3c1)]−ȷ^ [a1(b1c2−b2c1)−a3(b2c3−b3c2)]+k^ [a1(−b1c3+b3c1)−a2(b2c3−b3c2)](lhs)(a⋅c)b−(a⋅b)c = (a1c1+a2c2+a3c3)(b1ı^+b2ȷ^+b3k^)−(a1b1+a2b2+a3b3)(c1ı^+c2ȷ^+c3k^)= ı^ [a1b1c1+a2b1c2+a3b1c3−a1b1c1−a2b2c1−a3b3c1]+ȷ^ [a1b2c1+a2b2c2+a3b2c3−a1b1c2−a2b2c2−a3b3c2]+k^ [a1b3c1+a2b3c2+a3b3c3−a1b1c3−a2b2c3−a3b3c3]= ı^ [a2b1c2+a3b1c3−a2b2c1−a3b3c1]+ȷ^ [a1b2c1+a3b2c3−a1b1c2−a3b3c2]+k^ [a1b3c1+a2b3c2−a1b1c3−a2b2c3](rhs)\begin{align*}\va\times(\vb\times\vc) \ =\ &\det\left[\begin{matrix}\hi&\hj &\hk\\ a_1&a_2&a_3\\ b_2c_3-b_3c_2&-b_1c_3+b_3c_1&b_1c_2-b_2c_1\end{matrix}\right]\\ \ =\ &\hi\,[a_2(b_1c_2-b_2c_1)-a_3(-b_1c_3+b_3c_1)]\\ {-}&\hj\,[a_1(b_1c_2-b_2c_1)-a_3(b_2c_3-b_3c_2)]\\ {+}&\hk\,[a_1(-b_1c_3+b_3c_1)-a_2(b_2c_3-b_3c_2)]&{\rm (lhs)}\\ {} (\va\cdot\vc)\vb-(\va\cdot\vb)\vc \ =\ &(a_1c_1+a_2c_2+a_3c_3)(b_1\hi+b_2\hj+b_3\hk) -(a_1b_1+a_2b_2+a_3b_3)(c_1\hi+c_2\hj+c_3\hk)\\ {=}\ & \hi\,[a_1b_1c_1+a_2b_1c_2+a_3b_1c_3-a_1b_1c_1-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hj\,[a_1b_2c_1+a_2b_2c_2+a_3b_2c_3-a_1b_1c_2-a_2b_2c_2-a_3b_3c_2] \\ {+}&\hk\,[a_1b_3c_1+a_2b_3c_2+a_3b_3c_3-a_1b_1c_3-a_2b_2c_3-a_3b_3c_3]\cr {=}\ & \hi\,[a_2b_1c_2+a_3b_1c_3-a_2b_2c_1-a_3b_3c_1] \\ {+}&\hj\,[a_1b_2c_1+a_3b_2c_3-a_1b_1c_2-a_3b_3c_2] \\ {+}&\hk\,[a_1b_3c_1+a_2b_3c_2-a_1b_1c_3-a_2b_2c_3] &{\rm (rhs)}\end{align*}

(lhs) and (rhs) are the same.

Q38Stage 3

Derive a formula for (a×b)⋅(c×d)(\va\times\vb)\cdot(\vc\times\vd) that involves dot but not cross products.

Answer

(a×b)⋅(c×d)=(a⋅c)(b⋅d)−(a⋅d)(b⋅c)(\va\times\vb)\cdot(\vc\times\vd)= (\va\cdot\vc)(\vb\cdot\vd) -(\va\cdot\vd)(\vb\cdot\vc)

Full solution

By properties 9 and 10 of Theorem 1.2.23 in the CLP-3 text,

(a×b)⋅(c×d)=a⋅[b×(c×d)](by property 9 with c→(c×d))=a⋅[(b⋅d)c−(b⋅c)d](by property 10)=(a⋅c)(b⋅d)−(a⋅d)(b⋅c)\begin{align*} (\va\times\vb)\cdot(\vc\times\vd) &=\va\cdot[\vb\times(\vc\times\vd)] \hskip.5in&\text{(by property 9 with }\vc\rightarrow (\vc\times\vd)\text{)} \\ &=\va\cdot[(\vb\cdot\vd)\vc-(\vb\cdot\vc)\vd] \hskip.25in&\text{(by property 10)}\cr &=(\va\cdot\vc)(\vb\cdot\vd) -(\va\cdot\vd)(\vb\cdot\vc) \end{align*}

So

(a×b)⋅(c×d)=(a⋅c)(b⋅d)−(a⋅d)(b⋅c)\begin{equation*} (\va\times\vb)\cdot(\vc\times\vd)= (\va\cdot\vc)(\vb\cdot\vd) -(\va\cdot\vd)(\vb\cdot\vc) \end{equation*}
Q39Stage 3

A prism has the six vertices

A=(1,0,0)A′=(5,0,1)B=(0,3,0)B′=(4,3,1)C=(0,0,4)C′=(4,0,5)\begin{alignat*}{3} A&=(1,0,0)\qquad & A'&=(5,0,1) \\ B&=(0,3,0) & B'&=(4,3,1) \\ C&=(0,0,4) & C'&=(4,0,5) \end{alignat*}

  1. Verify that three of the faces are parallelograms. Are they rectangular?

  2. Find the length of AA′AA'.

  3. Find the area of the triangle ABCABC.

  4. Find the volume of the prism.

Answer

(a) AA′B′BAA'B'B is a parallelogram, but not a rectangle.
AA′C′CAA'C'C is a rectangle.
BB′C′CBB'C'C is a parallelogram, but not a rectangle.

(b) 17\sqrt{17} (c) 132\frac{13}{2} (d) 512\frac{51}{2}

Full solution

(a) AA′=<4,0,1>AA'=\llt 4,0,1\rgt and BB′=<4,0,1>BB'=\llt 4,0,1\rgt are opposite sides of the quadrilateral AA′B′BAA'B'B. They have the same length and direction. The same is true for AB=<−1,3,0>AB=\llt -1,3,0\rgt and A′B′=<−1,3,0>A'B'=\llt -1,3,0\rgt. So AA′B′BAA'B'B is a parallelogram. Because, AA′⋅AB=<4,0,1>⋅<−1,3,0>=−4≠0AA'\cdot AB=\llt 4,0,1\rgt \cdot\llt -1,3,0\rgt =-4\ne 0, the neighbouring edges of AA′B′BAA'B'B are not perpendicular and so AA′B′BAA'B'B is not a rectangle.

Similarly, the quadilateral ACC′A′ACC'A' has opposing sides AA′=<4,0,1>=CC′=<4,0,1>AA'=\llt 4,0,1\rgt =CC'=\llt 4,0,1\rgt and AC=<−1,0,4>=A′C′=<−1,0,4>AC=\llt -1,0,4\rgt =A'C'=\llt -1,0,4\rgt and so is a parallelogram. Because AA′⋅AC=<4,0,1>⋅<−1,0,4>=0AA'\cdot AC=\llt 4,0,1\rgt \cdot\llt -1,0,4\rgt = 0, the neighbouring edges of ACC′A′ACC'A' are perpendicular, so ACC′A′ACC'A' is a rectangle.

Finally, the quadilateral BCC′B′BCC'B' has opposing sides BB′=<4,0,1>=CC′=<4,0,1>BB'=\llt 4,0,1\rgt =CC'=\llt 4,0,1\rgt and BC=<0,−3,4>=B′C′=<0,−3,4>BC=\llt 0,-3,4\rgt =B'C'=\llt 0,-3,4\rgt and so is a parallelogram. Because BB′⋅BC=<4,0,1>⋅<0,−3,4>=4≠0BB'\cdot BC=\llt 4,0,1\rgt \cdot\llt 0,-3,4\rgt = 4\ne 0, the neighbouring edges of BCC′B′BCC'B' are not perpendicular, so BCC′B′BCC'B' is not a rectangle.

(b) The length of AA′AA' is ∣<4,0,1>∣=16+1=17|\llt 4,0,1\rgt |=\sqrt{16+1}=\sqrt{17}.

(c) The area of a triangle is one half its base times its height. That is, one half times ∣AB∣|AB| times ∣AC∣sin⁡θ|AC|\sin\theta, where θ\theta is the angle between ABAB and ACAC. This is precisely 12∣AB×AC∣=12∣<−1,3,0>×<−1,0,4>∣=12∣<12,4,3>∣=132\half |AB\times AC|=\half|\llt -1,3,0\rgt \times\llt -1,0,4\rgt | =\half |\llt 12,4,3\rgt|=\frac{13}{2}.

(d) The volume of the prism is the area of its base ABCABC, times its height, which is the length of AA′AA' times the cosine of the angle between AA′AA' and the normal to ABCABC. This coincides with 12<12,4,3>⋅<4,0,1>=12(48+3)=512\half \llt 12,4,3\rgt \cdot\llt 4,0,1\rgt =\half(48+3)=\frac{51}{2}, which is one half times the length of <12,4,3>\llt 12,4,3\rgt (the area of ABCABC) times the length of <4,0,1>\llt 4,0,1\rgt (the length of AA′AA') times the cosine of the angle between <12,4,3>\llt 12,4,3\rgt and <4,0,1>\llt 4,0,1\rgt (the angle between the normal to ABCABC and AA′AA').

Q40Stage 3

(Three dimensional Pythagorean Theorem) A solid body in space with exactly four vertices is called a tetrahedron. Let AA, BB, CC and DD be the areas of the four faces of a tetrahedron. Suppose that the three edges meeting at the vertex opposite the face of area DD are perpendicular to each other. Show that D2=A2+B2+C2D^2=A^2+B^2+C^2.

Figure from prob_s1.2, line 1940

Figure from prob_s1.2, line 1940

Hint

Choose coordinate axes so that the vertex opposite the face of area DD is at the origin. Denote by a\va, b\vb and c\vc the vertices opposite the sides of area AA, BB and CC respectively. Express AA, BB, CC and DD, which are areas of triangles, as one half times cross products of vectors built from a\va, b\vb and c\vc.

Answer

See the solution.

Full solution

Choose our coordinate axes so that the vertex opposite the face of area DD is at the origin. Denote by a\va, b\vb and c\vc the vertices opposite the sides of area AA, BB and CC respectively. Then the face of area AA has edges b\vb and c\vc so that A=12∣b×c∣A=\half |\vb\times\vc|. Similarly B=12∣c×a∣B=\half|\vc\times\va| and C=12∣a×b∣C=\half|\va\times \vb|. The face of area DD is the triangle spanned by b−a\vb-\va and c−a\vc-\va so that

D=12∣(b−a)×(c−a)∣=12∣b×c−a×c−b×a∣=12∣b×c+c×a+a×b∣\begin{align*} D&=\half|(\vb-\va)\times(\vc-\va)|\cr &=\half|\vb\times \vc-\va\times\vc-\vb\times\va|\\ &=\half|\vb\times \vc+\vc\times\va+\va\times\vb| \end{align*}

By hypothesis, the vectors a\va, b\vb and c\vc are all perpendicular to each other. Consequently the vectors b×c\vb\times \vc (which is a scalar times a\va), c×a\vc\times\va (which is a scalar times b\vb) and a×b\va\times\vb (which is a scalar times c\vc) are also mutually perpendicular. So, when we multiply out

D2=14[b×c+c×a+a×b]⋅[b×c+c×a+a×b]\begin{equation*} D^2=\frac{1}{4}\big[\vb\times \vc+\vc\times\va+\va\times\vb\big]\cdot\big[\vb\times \vc+\vc\times\va+\va\times\vb\big] \end{equation*}

all the cross terms vanish, leaving

D2=14[(b×c)⋅(b×c)+(c×a)⋅(c×a)+(a×b)⋅(a×b)]=A2+B2+C2\begin{equation*} D^2=\frac{1}{4}\big[(\vb\times \vc)\cdot(\vb\times \vc) +(\vc\times\va)\cdot(\vc\times\va) +(\va\times\vb)\cdot(\va\times\vb)\big]=A^2+B^2+C^2 \end{equation*}
Q41Stage 3

(Three dimensional law of cosines) Let AA, BB, CC and DD be the areas of the four faces of a tetrahedron. Let α\al be the angle between the faces with areas BB and CC, β\be be the angle between the faces with areas AA and CC and γ\ga be the angle between the faces with areas AA and BB. (By definition, the angle between two faces is the angle between the normal vectors to the faces.) Show that

D2=A2+B2+C2−2BCcos⁡α−2ACcos⁡β−2ABcos⁡γ\begin{equation*} D^2=A^2+B^2+C^2-2BC\cos\al-2AC\cos\be-2AB\cos\ga \end{equation*}
Hint

Do problem 40 first.

Answer

See the solution.

Full solution

As in problem 40,

D2=14[b×c+c×a+a×b]⋅[b×c+c×a+a×b]\begin{equation*} D^2=\frac{1}{4} \big[\vb\times \vc+\vc\times\va+\va\times\vb\big]\cdot \big[\vb\times \vc+\vc\times\va+\va\times\vb\big] \end{equation*}

But now (b×c)⋅(a×c)(\vb\times \vc)\cdot(\va\times\vc), instead of vanishing, is ∣b×c∣=2A|\vb\times \vc|=2A times ∣a×c∣=2B|\va\times\vc|=2B times the cosine of the angle between b×c\vb\times \vc (which is perpendicular to the face of area AA) and a×c\va\times\vc (which is perpendicular to the face of area BB). That is

(b×c)⋅(a×c)=4ABcos⁡γ(a×b)⋅(c×b)=4ACcos⁡β(b×a)⋅(c×a)=4BCcos⁡α\begin{align*} (\vb\times \vc)\cdot(\va\times\vc)&=4 AB\cos \ga\\ (\va\times \vb)\cdot(\vc\times\vb)&=4 AC\cos \be\\ (\vb\times \va)\cdot(\vc\times\va)&=4 BC\cos \al \end{align*}

(If you're worried about the signs, that is, if you are worried about why (b×c)⋅(a×c)=4ABcos⁡γ(\vb\times \vc)\cdot(\va\times\vc)=4 AB\cos \ga rather than (b×c)⋅(c×a)=4ABcos⁡γ(\vb\times \vc)\cdot(\vc\times\va)=4 AB\cos \ga, note that when a≈b\va\approx\vb, (b×c)⋅(a×c)≈∣b×c∣2(\vb\times \vc)\cdot(\va\times\vc)\approx|\vb\times\vc|^2 is positive and (b×c)⋅(c×a)≈−∣b×c∣2(\vb\times \vc)\cdot(\vc\times\va) \approx -|\vb\times\vc|^2 is negative.) Now, expanding out

D2 = 14[b×c+c×a+a×b]⋅[b×c+c×a+a×b]= 14[(b×c)⋅(b×c)+(c×a)⋅(c×a)+(a×b)⋅(a×b)+2(b×c)⋅(c×a)+2(b×c)⋅(a×b)+2(c×a)⋅(a×b)]= A2+B2+C2−2ABcos⁡γ−2ACcos⁡β−2BCcos⁡α\begin{align*} D^2\ =\ &\frac{1}{4} \big[\vb\times \vc+\vc\times\va+\va\times\vb\big]\cdot \big[\vb\times \vc+\vc\times\va+\va\times\vb\big] \\ =\ &\frac{1}{4}\big[(\vb\times \vc)\cdot(\vb\times \vc) +(\vc\times\va)\cdot(\vc\times\va) +(\va\times\vb)\cdot(\va\times\vb) \\ &+2(\vb\times \vc)\cdot(\vc\times \va) +2(\vb\times \vc)\cdot(\va\times \vb) +2(\vc\times \va)\cdot(\va\times \vb)\big] \\ =\ &A^2+B^2+C^2-2 AB\cos \ga-2 AC\cos \be-2 BC\cos \al \end{align*}

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From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.