"Three Dimensional Geometry" — Practice

Question 1

Find the direction cosines of the line joining $(1,2,3)$ and $(4,5,7)$.

Accepted Answer

The direction ratios are the coordinate differences, $(3,3,4)$. Their magnitude is the square root of $9+9+16$, which is root 34. Dividing each ratio by that gives direction cosines $3/\sqrt{34}$, $3/\sqrt{34}$ and $4/\sqrt{34}$. Check: the squares are $9/34$, $9/34$ and $16/34$, summing to 1 as required. The opposite triple, with all three signs reversed, is equally valid and corresponds to traversing the line the other way.

Question 2

A line makes $60^\circ$ with the $x$-axis and $45^\circ$ with the $z$-axis. Find the angle it makes with the $y$-axis.

Accepted Answer

Use the identity that the squares of the three direction cosines sum to 1. Here $\cos^2 60 = 1/4$ and $\cos^2 45 = 1/2$, so $\cos^2\beta = 1 - 1/4 - 1/2 = 1/4$ and $\cos\beta = \pm 1/2$. The angle is therefore $60^\circ$ or $120^\circ$, and both are genuine answers corresponding to the two directions along the line. Quoting only $60^\circ$ is the standard omission.

Question 3

Show that the lines with direction ratios $(1,-1,0)$ and $(1,1,0)$ are perpendicular, and find the direction ratios of a line perpendicular to both.

Accepted Answer

The dot product is $1(1) + (-1)(1) + 0(0) = 0$, so they are perpendicular and no denominators are needed. A line perpendicular to both has direction ratios given by the cross product: $(1,-1,0) 	imes (1,1,0)$ has $i$ component $(-1)(0) - (0)(1) = 0$, $j$ component $-[(1)(0)-(0)(1)] = 0$, and $k$ component $(1)(1) - (-1)(1) = 2$. So $(0,0,2)$, or equivalently $(0,0,1)$: the $z$-axis, which is correct since both given lines lie in the $xy$-plane.

Question 4

Find the shortest distance between $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$.

Accepted Answer

The joining vector is $(1,2,2)$ and the directions are $(2,3,4)$ and $(3,4,5)$. Their cross product is $(15-16,\ -(10-12),\ 8-9)$, that is $(-1,2,-1)$, with magnitude root 6. The triple product is $1(-1) + 2(2) + 2(-1) = 1$, which is non-zero, so the lines are skew. The shortest distance is the modulus of that, divided by root 6, giving $1/\sqrt{6}$, about 0.408.

Question 5

Show that $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}$ intersect, and find the point.

Accepted Answer

The joining vector is $(3,7,11)$ and the cross product of the directions $(3,5,7)$ and $(1,3,5)$ is $(25-21,\ -(15-7),\ 9-5)$, that is $(4,-8,4)$. The triple product is $12 - 56 + 44 = 0$, so the lines are coplanar and, not being parallel, they meet. Setting general points equal gives $3t-1 = s+2$ and $5t-3 = 3s+4$; the first gives $s = 3t-3$, and substituting gives $5t-3 = 9t-5$, so $t = 1/2$ and $s = -3/2$. The third equation checks: $7(1/2)-5 = -3/2$ and $5(-3/2)+6 = -3/2$. The point is $(1/2,\ -1/2,\ -3/2)$.

Question 6

Write the equation of the line through $(1,2,3)$ and $(4,5,6)$ in symmetric form.

Accepted Answer

The direction ratios are the coordinate differences, $(3,3,3)$, which reduce to $(1,1,1)$ since only the proportion matters. Taking the first point as the anchor gives $(x-1)/1 = (y-2)/1 = (z-3)/1$. Using the second point instead gives $(x-4)/1 = (y-5)/1 = (z-6)/1$, which is the same line written differently, and either is a complete answer.

Question 7

Find the equation of the plane through $(1,1,1)$ whose normal has direction ratios $(2,-3,4)$.

Accepted Answer

The point-normal form writes the normal dotted with the displacement from the known point as zero: $2(x-1) - 3(y-1) + 4(z-1) = 0$. Expanding gives $2x - 3y + 4z - 2 + 3 - 4 = 0$, that is $2x - 3y + 4z = 3$. Check the point: $2 - 3 + 4 = 3$. Notice that the coefficients of the final equation are exactly the normal, which is the reading that makes most plane questions short.

Question 8

Find the plane through $(1,0,-1)$ and $(3,2,2)$ that is parallel to the line $\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}$.

Accepted Answer

The normal must be perpendicular to the vector joining the two points, $(2,2,3)$, and also to the line's direction $(1,-2,3)$, since a plane parallel to a line is perpendicular to nothing about it except through its normal. Their cross product is $(6+6,\ -(6-3),\ -4-2)$, that is $(12,-3,-6)$, which reduces to $(4,-1,-2)$. The plane is $4(x-1) - (y-0) - 2(z+1) = 0$, that is $4x - y - 2z = 6$. Check the second point: $12 - 2 - 4 = 6$.

Question 9

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

Write the family as $(x+2y+3z-4) + \lambda(2x+y-z+5) = 0$, which avoids finding the line of intersection. Substituting the origin gives $-4 + 5\lambda = 0$, so $\lambda = 4/5$. Multiplying through by 5 gives $5x+10y+15z-20 + 8x+4y-4z+20 = 0$, that is $13x + 14y + 11z = 0$. The constant vanishing is the visible confirmation that the plane passes through the origin.

Question 10

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

The direction is $(2,3,6)$ with magnitude 7, and the normal is read off the coefficients as $(10,2,-11)$ with magnitude 15. Their dot product is $20 + 6 - 66 = -40$. Because the normal is perpendicular to the plane, the formula uses sine: $\sin	heta = 40/105 = 8/21$, so $	heta$ is about $22.4^\circ$. Using cosine instead would give about $67.6^\circ$, which is the angle to the normal rather than to the plane.

Question 11

Find the image of $(1,3,4)$ in the plane $2x-y+z+3=0$.

Accepted Answer

The common value is $k = -(2 - 3 + 4 + 3)/(4+1+1) = -6/6 = -1$. Stepping once along the normal gives the foot $(1+2(-1),\ 3-(-1),\ 4+(-1))$, that is $(-1,4,3)$, which checks on the plane since $-2 - 4 + 3 + 3 = 0$. Stepping twice gives the image $(1+4(-1),\ 3-2(-1),\ 4+2(-1))$, that is $(-3,5,2)$. As a check, the distance from the point to the plane is $6/\sqrt{6} = \sqrt6$, and the distance from the point to its image is the magnitude of $(4,-2,2)$, which is $2\sqrt6$, exactly twice.

Question 12

Find the equation of the line through $(1,2,3)$ perpendicular to both $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$ and $\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{1}$.

Accepted Answer

A direction perpendicular to two given directions is their cross product. For $(1,2,3)$ and $(3,2,1)$ this is $(2-6,\ -(1-9),\ 2-6)$, that is $(-4,8,-4)$, which reduces to $(1,-2,1)$. The line is therefore $(x-1)/1 = (y-2)/(-2) = (z-3)/1$. Check both perpendicularity conditions: $(1,-2,1)$ dotted with $(1,2,3)$ is $1-4+3 = 0$, and with $(3,2,1)$ is $3-4+1 = 0$.

Practice — "Three Dimensional Geometry"

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