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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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

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
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
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
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
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
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
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
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
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
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
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

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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.