Three Dimensional Geometry — Practice

Question 1

A line makes equal angles with all three axes. Find that angle.

Accepted Answer

$l=m=n$ and $3l^{2}=1$, so $l=\pm\dfrac{1}{\sqrt3}$.

The angle is $\cos^{-1}\dfrac{1}{\sqrt3}\approx54.7^{\circ}$, or its supplement for the opposite sense.

Question 2

Show that the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-2}{3}=\dfrac{y-3}{4}=\dfrac{z-4}{5}$ intersect, and find the point.

Accepted Answer

$\mathbf a_2-\mathbf a_1=(1,1,1)$ and $\mathbf b_1	imes\mathbf b_2=(2,3,4)	imes(3,4,5)=(-1,2,-1)$.

The triple product is $-1+2-1=0$, so they are coplanar and not parallel: they intersect.

Setting $1+2s=2+3t$ and $2+3s=3+4t$ gives $s=t=-1$, and both give the point $(-1,-1,-1)$.

Question 3

Find the shortest distance between $\dfrac{x}{1}=\dfrac{y}{-1}=\dfrac{z}{1}$ and $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z}{1}$.

Accepted Answer

$\mathbf a_2-\mathbf a_1=(1,-1,0)$, $\mathbf b_1=(1,-1,1)$, $\mathbf b_2=(2,1,1)$.

$\mathbf b_1	imes\mathbf b_2=(-1-1,\ 2-1,\ 1+2)=(-2,1,3)$.

Numerator: $\left|(1,-1,0)\cdot(-2,1,3)ight|=\left|-2-1ight|=3$.

$d=\dfrac{3}{\sqrt{14}}$.

Question 4

Find the plane through $(2,2,-1)$, $(3,4,2)$ and $(7,0,6)$.

Accepted Answer

Edges: $(1,2,3)$ and $(5,-2,7)$.

Cross product: $(14+6,\ 15-7,\ -2-10)=(20,8,-12)$, or $(5,2,-3)$ after dividing by $4$.

$5(x-2)+2(y-2)-3(z+1)=0$, that is $5x+2y-3z=17$.

All three points give $17$.

Question 5

Find the angle between the line $\dfrac{x+1}{2}=\dfrac{y}{3}=\dfrac{z-3}{6}$ and the plane $10x+2y-11z=3$.

Accepted Answer

$\mathbf b=(2,3,6)$ with $\left|\mathbf bight|=7$; $\mathbf n=(10,2,-11)$ with $\left|\mathbf night|=15$.

$\mathbf b\cdot\mathbf n=20+6-66=-40$.

$\sin	heta=\dfrac{40}{105}=\dfrac{8}{21}$, so $	heta=\sin^{-1}\dfrac{8}{21}$.

Question 6

Show that $\dfrac{x-1}{3}=\dfrac{y+2}{-1}=\dfrac{z}{1}$ is parallel to the plane $x+2y-z=5$ but does not lie in it.

Accepted Answer

$(3,-1,1)\cdot(1,2,-1)=3-2-1=0$, so the direction is perpendicular to the normal and the line is parallel to the plane.

The point $(1,-2,0)$ gives $1-4-0=-3
eq5$.

So the line is parallel to the plane and misses it entirely.

Question 7

Find the distance between the parallel planes $2x-2y+z=4$ and $4x-4y+2z=1$.

Accepted Answer

Halve the second: $2x-2y+z=	frac12$.

$\sqrt{4+4+1}=3$.

Distance $=\dfrac{\left|4-	frac12ight|}{3}=\dfrac{7}{6}$.

Question 8

Find the foot of the perpendicular from $(1,2,3)$ to the plane $x-2y+2z=9$.

Accepted Answer

$ax_1+by_1+cz_1-d=1-4+6-9=-6$ and $a^{2}+b^{2}+c^{2}=9$.

Common ratio $=\dfrac{6}{9}=\dfrac23$.

Foot $=\left(1+	frac23,\ 2-	frac43,\ 3+	frac43ight)=\left(	frac53,	frac23,	frac{13}{3}ight)$.

Check: $	frac53-	frac43+	frac{26}{3}=9$.

Question 9

Find the distance of $(1,-2,3)$ from the line $\dfrac{x}{2}=\dfrac{y-1}{-1}=\dfrac{z+1}{2}$.

Accepted Answer

$\mathbf a=(0,1,-1)$, $\mathbf b=(2,-1,2)$, $\mathbf p-\mathbf a=(1,-3,4)$.

$(1,-3,4)	imes(2,-1,2)=(-6+4,\ 8-2,\ -1+6)=(-2,6,5)$.

Magnitude $\sqrt{4+36+25}=\sqrt{65}$; $\left|\mathbf bight|=3$.

$d=\dfrac{\sqrt{65}}{3}$.

Question 10

Find the plane through the line of intersection of $x+2y+3z=4$ and $2x+y-z=-5$ that passes through the origin.

Accepted Answer

$\left(x+2y+3z-4ight)+\lambda\left(2x+y-z+5ight)=0$.

At the origin: $-4+5\lambda=0$, so $\lambda=	frac45$.

Multiplying by $5$: $5x+10y+15z-20+8x+4y-4z+20=0$, that is $13x+14y+11z=0$.

Question 11

Are the points $(1,2,3)$, $(2,3,4)$, $(3,4,5)$ and $(4,5,7)$ coplanar?

Accepted Answer

From the first point, the edges are $(1,1,1)$, $(2,2,2)$ and $(3,3,4)$.

The first two are proportional, so the determinant is zero.

The four points are coplanar — indeed the first three are collinear, which forces it.

Question 12

Find the line of intersection of the planes $x+y-z=1$ and $x-2y+3z=2$.

Accepted Answer

Direction: $(1,1,-1)	imes(1,-2,3)=(3-2,\ -1-3,\ -2-1)=(1,-4,-3)$.

Set $z=0$: $x+y=1$ and $x-2y=2$ give $y=-	frac13$, $x=	frac43$.

$\dfrac{x-\frac43}{1}=\dfrac{y+\frac13}{-4}=\dfrac{z}{-3}$.

Practice — Three Dimensional Geometry

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