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Vectors and Geometry in Two and Three Dimensions

1.4 Equations of Planes in 3d

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

1Stage 1Conceptual

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

Q1Stage 1

The vector k^\hk is a normal vector (i.e. is perpendicular) to the plane z=0z=0. Find another nonzero vector that is normal to z=0z=0.

Hint

You are looking for a vector that is perpendicular to z=0z=0 and hence is parallel to k^\hk.

Answer

Any vector of the form c k^c\,\hk with c≠0c\ne 0 and c≠1c\ne 1 works. Three possible choices are −k^-\hk, 2 k^2\,\hk, 7.12345 k^7.12345\,\hk.

Full solution
Q2Stage 1

Consider the plane PP with equation 3x+12y+z=43x+\frac{1}{2}y+z=4.

  1. Find the intersection of PP with the yy-axis.

  2. Find the intersection of PP with the zz-axis.

  3. Sketch the part of the intersection of PP with the yzyz-plane that is in the first octant. (That is, with x,y,z≥0x,y,z\ge 0.)

Hint

A point (x,y,z)(x,y,z) is on the yy-axis if and only if x=z=0x=z=0. Similarly, a point (x,y,z)(x,y,z) is on the zz-axis if and only if x=y=0x=y=0.

Answer

(a) (0,8,0)(0,8,0) (b) (0,0,4)(0,0,4)

(c)

Figure from prob_s1.4, line 52

Figure from prob_s1.4, line 52

Full solution
Q3Stage 1

  1. Find the equation of the plane that passes through the origin and has normal vector <1,2,3>\llt 1,2,3\rgt.

  2. Find the equation of the plane that passes through the point (0,0,1)(0,0,1) and has normal vector <1,1,3>\llt 1,1,3\rgt.

  3. Find, if possible, the equation of a plane that passes through both (1,2,3)(1,2,3) and (1,0,0)(1,0,0) and has normal vector <4,5,6>\llt 4,5,6\rgt.

  4. Find, if possible, the equation of a plane that passes through both (1,2,3)(1,2,3) and (0,3,4)(0,3,4) and has normal vector <2,1,1>\llt 2,1,1\rgt.

Answer

(a) x+2y+3z=0x+2y+3z=0 (b) x+y+3z=3x+y+3z=3

(c) There is no plane that passes through both (1,2,3)(1,2,3) and (1,0,0)(1,0,0) and has normal vector <4,5,6>\llt 4,5,6\rgt.

(d) 2x+y+z=72x+y+z=7

Full solution
Q4Stage 1Past exam · M200 2008A

Find the equation of the plane that contains (1,0,0)(1,0,0), (0,1,0)(0,1,0) and (0,0,1)(0,0,1).

Hint

Guess.

Answer

x+y+z=1x+y+z=1

Full solution
Q5Stage 1

  1. Find the equation of the plane containing the points (1,0,1)(1,0,1), (1,1,0)(1,1,0) and (0,1,1)(0,1,1).

  2. Is the point (1,1,1)(1,1,1) on the plane?

  3. Is the origin on the plane?

  4. Is the point (4,−1,−1)(4,-1,-1) on the plane?

Hint

(a) See Question 4 — or just have a guess!

Answer

(a) x+y+z=2x+y+z=2 (b) No. (c) No. (d) Yes.

Full solution
Q6Stage 1

What's wrong with the following exercise? “Find the equation of the plane containing (1,2,3)(1,2,3), (2,3,4)(2,3,4) and (3,4,5)(3,4,5).”

Hint

Three points don't always determine a plane — why?

Answer

All three points (1,2,3), (2,3,4)(1,2,3),\ (2,3,4) and (3,4,5)(3,4,5) are on the line x(t)=(1,2,3)+t(1,1,1)\vx(t)=(1,2,3)+t(1,1,1). There are many planes through that line.

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q7Stage 2

Find the plane containing the given three points.

  1. (1,0,1), (2,4,6), (1,2,−1)(1,0,1),\ (2,4,6),\ (1,2,-1)

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

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

Answer

(a) 9x−y−z=89x-y-z=8 (b) 14x−7y−8z=5214x-7y-8z=52

(c) For any real numbers aa and bb, the plane
ax+by−(a+b)z=4a+bax+by-(a+b)z=4a+b contains the three given points.

Full solution
Q8Stage 2

Find the distance from the given point to the given plane.

  1. point (−1,2,3)(-1,2,3), plane x+y+z=7x+y+z=7

  2. point (1,−4,3)(1,-4,3), plane x−2y+z=5x-2y+z=5

Answer

(a) 3\sqrt{3} (b) 7/67/\sqrt{6}

Full solution
Q9Stage 2Past exam · M200 2007A

A plane Π\Pi passes through the points A=(1,1,3)A = (1, 1, 3), B=(2,0,2)B = (2, 0, 2) and C=(2,1,0)C = (2, 1, 0) in R3\bbbr^3.

  1. Find an equation for the plane Π\Pi.

  2. Find the point EE in the plane Π\Pi such that the line LL through D=(6,1,2)D = (6, 1, 2) and EE is perpendicular to Π\Pi.

Answer

(a) 3x+2y+z=83x+2y+z = 8 (b) (3 , −1 , 1)\big(3\,,\,-1\,,\,1\big)

Full solution
Q10Stage 2Past exam · M200 2011A

Let A=(2,3,4)A = (2, 3, 4) and let LL be the line given by the equations x+y=1x + y = 1 and x+2y+z=3x + 2y + z = 3.

  1. Write an equation for the plane containing AA and perpendicular to LL.

  2. Write an equation for the plane containing AA and LL.

Answer

(a) x−y+z=3x-y+z=3 (b) 5x+y−4z=−35x+y-4z=-3

Full solution
Q11Stage 2Past exam · M200 2015D

Consider the plane 4x+2y−4z=34x + 2y - 4z = 3. Find all parallel planes that are distance 22 from the above plane. Your answers should be in the following form: 4x+2y−4z=C4x + 2y - 4z = C.

Answer

4x+2y−4z=154x + 2y - 4z=15 and 4x+2y−4z=−94x + 2y - 4z=-9

Full solution
Q12Stage 2Past exam · M200 2004A

Find the distance from the point (1,2,3)(1,2,3) to the plane that passes through the points (0,1,1)(0,1,1), (1,−1,3)(1,-1,3) and (2,0,−1)(2,0,-1).

Answer

22

Full solution

3Stage 3Application

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

Q13Stage 3Past exam · M200 2014A

Consider two planes W1W_1, W2W_2, and a line MM defined by:

W1 : −2x+y+z=7,W2 : −x+3y+3z=6,M : x2=2y−44=z+5.\begin{equation*} W_1\ :\ -2x + y + z = 7,\qquad W_2\ :\ -x + 3y + 3z = 6,\qquad M\ :\ \frac{x}{2} = \frac{2y-4}{4} = z + 5. \end{equation*}

  1. Find a parametric equation of the line of intersection LL of W1W_1 and W2W_2.

  2. Find the distance from L to M .

Answer

(a) (x,y,z)=(−3,1,0)+t<0,−1,1>(x,y,z) = (-3,1,0) +t \llt 0,-1,1\rgt (b) 17\sqrt{17}

Full solution
Q14Stage 3

Find the equation of the sphere which has the two planes x+y+z=3, x+y+z=9x+y+z=3,\ x+y+z=9 as tangent planes if the center of the sphere is on the planes 2x−y=0, 3x−z=02x-y=0,\ 3x-z=0.

Answer

(x−1)2+(y−2)2+(z−3)2=3(x-1)^2+(y-2)^2+(z-3)^2=3

Full solution
Q15Stage 3

Find the equation of the plane that passes through the point (−2,0,1)(-2,0,1) and through the line of intersection of 2x+3y−z=0, x−4y+2z=−52x+3y-z=0,\ x-4y+2z=-5.

Answer

3x−y+z=−53x-y+z=-5

Full solution
Q16Stage 3

Find the distance from the point p\vp to the plane n⋅x=c\vn\cdot \vx= c.

Answer

∣c−n⋅p∣/∣n∣|c-\vn\cdot\vp|/|\vn|

Full solution
Q17Stage 3

Describe the set of points equidistant from (1,2,3)(1,2,3) and (5,2,7)(5,2,7).

Answer

It is the plane x+z=8x+z=8, which is the plane through (3,2,5)=12(1,2,3)+12(5,2,7)(3,2,5)=\half(1,2,3)+\half(5,2,7) with normal <1,0,1>=14(<5,2,7>−<1,2,3>)\llt 1,0,1\rgt=\frac{1}{4}\big(\llt 5,2,7\rgt-\llt 1,2,3\rgt \big).

Full solution
Q18Stage 3

Describe the set of points equidistant from a\va and b\vb.

Answer

It is the plane 2(b−a)⋅x=∣b∣2−∣a∣22(\vb-\va)\cdot\vx=|\vb|^2-|\va|^2, which is the plane through 12a+12b\half\va+\half\vb with normal vector b−a\vb-\va.

Full solution
Q19Stage 3Past exam · M200 2003D

Consider a point P(5,−10,2)P(5,-10,2) and the triangle with vertices A(0,1,1)A(0,1,1), B(1,0,1)B(1,0,1) and C(1,3,0)C(1,3,0).

  1. Compute the area of the triangle ABCABC.

  2. Find the distance from the point PP to the plane containing the triangle.

Answer

(a) 1211≈1.658\half\sqrt{11}\approx 1.658 (b) 311≈0.9045\frac{3}{\sqrt{11}}\approx 0.9045

Full solution
Q20Stage 3Past exam · M200 2003A

Consider the sphere given by

(x−1)2+(y−2)2+(z+1)2=2(x-1)^2+(y-2)^2+(z+1)^2=2

Suppose that you are at the point (2,2,0)(2,2,0) on SS, and you plan to follow the shortest path on SS to (2,1,−1)(2,1,-1). Express your initial direction as a cross product.

Answer

Any positive constant times <1,1,−1>×<1,0,1>=<1,−2,−1>\llt 1,1,-1\rgt\times\llt 1,0,1\rgt =\llt 1, -2, -1\rgt

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.