Chapter 3: Analytical Solid Geometry

Exercise Solutions (3.1 to 3.5)

Exercise 3.1

Problem 1

Find the equation of the plane containing the point $\displaystyle (1,-2,3)$ and parallel to the

(a) $\displaystyle xy$-plane
(b) $\displaystyle yz$-plane
(c) $\displaystyle zx$-plane.
Solution

(a) The equation of a plane parallel to the $\displaystyle xy$-plane is of the form $\displaystyle z = c$.
Since it passes through $\displaystyle (1, -2, 3)$, the equation is:

$\displaystyle \begin{aligned} z &= 3 \end{aligned}$

(b) The equation of a plane parallel to the $\displaystyle yz$-plane is of the form $\displaystyle x = a$.
Since it passes through $\displaystyle (1, -2, 3)$, the equation is:

$\displaystyle \begin{aligned} x &= 1 \end{aligned}$

(c) The equation of a plane parallel to the $\displaystyle zx$-plane is of the form $\displaystyle y = b$.
Since it passes through $\displaystyle (1, -2, 3)$, the equation is:

$\displaystyle \begin{aligned} y &= -2 \end{aligned}$

Problem 2

Find the equation of the line through the point $\displaystyle (2,3,-4)$ and perpendicular to

(a) $\displaystyle xy$-plane
(b) $\displaystyle yz$-plane
(c) $\displaystyle zx$-plane.

Find the point of intersection of the line and plane.

Solution

(a) The line perpendicular to the $\displaystyle xy$-plane has the same direction as the $\displaystyle z$-axis. Its directed values are $\displaystyle \langle 0, 0, 1 \rangle$.
The equation of the line through $\displaystyle (2,3,-4)$ is:

$\displaystyle \begin{aligned} x &= 2, \quad y = 3, \quad z = t \end{aligned}$

The line intersects the $\displaystyle xy$-plane where $\displaystyle z = 0$. The point of intersection is $\displaystyle (2, 3, 0)$.

(b) The line perpendicular to the $\displaystyle yz$-plane has the same direction as the $\displaystyle x$-axis. Its directed values are $\displaystyle \langle 1, 0, 0 \rangle$.
The equation of the line through $\displaystyle (2,3,-4)$ is:

$\displaystyle \begin{aligned} x &= t, \quad y = 3, \quad z = -4 \end{aligned}$

The line intersects the $\displaystyle yz$-plane where $\displaystyle x = 0$. The point of intersection is $\displaystyle (0, 3, -4)$.

(c) The line perpendicular to the $\displaystyle zx$-plane has the same direction as the $\displaystyle y$-axis. Its directed values are $\displaystyle \langle 0, 1, 0 \rangle$.
The equation of the line through $\displaystyle (2,3,-4)$ is:

$\displaystyle \begin{aligned} x &= 2, \quad y = t, \quad z = -4 \end{aligned}$

The line intersects the $\displaystyle zx$-plane where $\displaystyle y = 0$. The point of intersection is $\displaystyle (2, 0, -4)$.

Problem 3

Find the distance between the points $\displaystyle (2,-3, 5)$ and $\displaystyle (7, 5,-2)$.

Solution

Let $\displaystyle P(2, -3, 5)$ and $\displaystyle Q(7, 5, -2)$. Using the distance formula:

$\displaystyle \begin{aligned} PQ &= \sqrt{(7-2)^2 + (5 - (-3))^2 + (-2-5)^2} \\ &= \sqrt{5^2 + 8^2 + (-7)^2} \\ &= \sqrt{25 + 64 + 49} \\ &= \sqrt{138} \text{ units.} \end{aligned}$

Problem 4

Show that the points $\displaystyle (-1, 2, 5)$, $\displaystyle (1, 1, 6)$ and $\displaystyle (0, 5, 6)$ form a right triangle.

Solution

Let $\displaystyle A(-1, 2, 5)$, $\displaystyle B(1, 1, 6)$, and $\displaystyle C(0, 5, 6)$. We calculate the square of the distance between each pair of points.

$\displaystyle \begin{aligned} AB^2 &= (1 - (-1))^2 + (1 - 2)^2 + (6 - 5)^2 \\ &= 2^2 + (-1)^2 + 1^2 \\ &= 4 + 1 + 1 = 6 \end{aligned}$
$\displaystyle \begin{aligned} BC^2 &= (0 - 1)^2 + (5 - 1)^2 + (6 - 6)^2 \\ &= (-1)^2 + 4^2 + 0^2 \\ &= 1 + 16 + 0 = 17 \end{aligned}$
$\displaystyle \begin{aligned} AC^2 &= (0 - (-1))^2 + (5 - 2)^2 + (6 - 5)^2 \\ &= 1^2 + 3^2 + 1^2 \\ &= 1 + 9 + 1 = 11 \end{aligned}$

Since $\displaystyle AB^2 + AC^2 = 6 + 11 = 17 = BC^2$, the points satisfy the Pythagorean theorem. Therefore, they form a right triangle with the right angle at $\displaystyle A$.

Problem 5

Show that the points $\displaystyle (1,-1,2)$, $\displaystyle (3,-2,3)$ and $\displaystyle (5,-3,4)$ are collinear.

Solution

Let $\displaystyle A(1, -1, 2)$, $\displaystyle B(3, -2, 3)$, and $\displaystyle C(5, -3, 4)$. We calculate the distances $\displaystyle AB$, $\displaystyle BC$, and $\displaystyle AC$.

$\displaystyle \begin{aligned} AB &= \sqrt{(3-1)^2 + (-2 - (-1))^2 + (3-2)^2} \\ &= \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \end{aligned}$
$\displaystyle \begin{aligned} BC &= \sqrt{(5-3)^2 + (-3 - (-2))^2 + (4-3)^2} \\ &= \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \end{aligned}$
$\displaystyle \begin{aligned} AC &= \sqrt{(5-1)^2 + (-3 - (-1))^2 + (4-2)^2} \\ &= \sqrt{4^2 + (-2)^2 + 2^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6} \end{aligned}$

Since $\displaystyle AB + BC = \sqrt{6} + \sqrt{6} = 2\sqrt{6} = AC$, the points $\displaystyle A, B,$ and $\displaystyle C$ lie on the same straight line. Hence, they are collinear.

Exercise 3.2

Problem 1

Given $\displaystyle P(3,1,5)$ and $\displaystyle Q(-3,7,-2)$, find the coordinates of the point $\displaystyle R(x,y,z)$ on the line $\displaystyle PQ$ with respect to the point $\displaystyle P$ and the following parameters.

(a) $\displaystyle k=\frac{1}{2}$
(b) $\displaystyle k=3$
(c) $\displaystyle k=-2$
Solution

The directed values of $\displaystyle PQ$ are:

$\displaystyle \begin{aligned} \text{directed values of } PQ &= \langle l,m,n \rangle \\ &= \langle -3-3, 7-1, -2-5 \rangle \\ &= \langle -6, 6, -7 \rangle \end{aligned}$

The coordinates of $\displaystyle R$ are given by $\displaystyle (x_1+kl, y_1+km, z_1+kn)$, so $\displaystyle (x,y,z) = (3 - 6k, 1 + 6k, 5 - 7k)$.

(a) For $\displaystyle k=\frac{1}{2}$:

$\displaystyle \begin{aligned} (x,y,z) &= \left(3 - 6\left(\frac{1}{2}\right), 1 + 6\left(\frac{1}{2}\right), 5 - 7\left(\frac{1}{2}\right)\right) \\ &= \left(3 - 3, 1 + 3, 5 - \frac{7}{2}\right) \\ &= \left(0, 4, \frac{3}{2}\right) \end{aligned}$

(b) For $\displaystyle k=3$:

$\displaystyle \begin{aligned} (x,y,z) &= (3 - 6(3), 1 + 6(3), 5 - 7(3)) \\ &= (3 - 18, 1 + 18, 5 - 21) \\ &= (-15, 19, -16) \end{aligned}$

(c) For $\displaystyle k=-2$:

$\displaystyle \begin{aligned} (x,y,z) &= (3 - 6(-2), 1 + 6(-2), 5 - 7(-2)) \\ &= (3 + 12, 1 - 12, 5 + 14) \\ &= (15, -11, 19) \end{aligned}$

Problem 2

Given $\displaystyle P(-2,1,3)$ and $\displaystyle Q(4,4,-3)$ determine whether or not the following points are on the line $\displaystyle PQ$. If the point is on the line $\displaystyle PQ$, find the corresponding parameter with respect to the point $\displaystyle P$.

(a) $\displaystyle (6,3,-6)$
(b) $\displaystyle (6,5,-5)$
(c) $\displaystyle (-4,0,5)$
(d) $\displaystyle (7,8,-2)$
Solution

By given $\displaystyle P(-2,1,3)$ and $\displaystyle Q(4,4,-3)$,

$\displaystyle \begin{aligned} \text{directed values of } PQ &= \langle l,m,n \rangle \\ &= \langle 4 - (-2), 4 - 1, -3 - 3 \rangle \\ &= \langle 6, 3, -6 \rangle \end{aligned}$

Thus, the equation of the line $\displaystyle PQ$ is $\displaystyle \frac{x+2}{6} = \frac{y-1}{3} = \frac{z-3}{-6} = k$.

(a) If $\displaystyle (x,y,z) = (6,3,-6)$, then

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{6+2}{6} = \frac{8}{6} = \frac{4}{3} \\ \frac{y-1}{3} &= \frac{3-1}{3} = \frac{2}{3} \end{aligned}$

Since $\displaystyle \frac{4}{3} \ne \frac{2}{3}$, the point $\displaystyle (6,3,-6)$ is not on the line $\displaystyle PQ$.

(b) If $\displaystyle (x,y,z) = (6,5,-5)$, then

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{6+2}{6} = \frac{8}{6} = \frac{4}{3} \\ \frac{y-1}{3} &= \frac{5-1}{3} = \frac{4}{3} \\ \frac{z-3}{-6} &= \frac{-5-3}{-6} = \frac{-8}{-6} = \frac{4}{3} \end{aligned}$

So, the point $\displaystyle (6,5,-5)$ is on the line $\displaystyle PQ$ with corresponding parameter $\displaystyle k = \frac{4}{3}$.

(c) If $\displaystyle (x,y,z) = (-4,0,5)$, then

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{-4+2}{6} = -\frac{1}{3} \\ \frac{y-1}{3} &= \frac{0-1}{3} = -\frac{1}{3} \\ \frac{z-3}{-6} &= \frac{5-3}{-6} = -\frac{1}{3} \end{aligned}$

So, the point $\displaystyle (-4,0,5)$ is on the line $\displaystyle PQ$ with corresponding parameter $\displaystyle k = -\frac{1}{3}$.

(d) If $\displaystyle (x,y,z) = (7,8,-2)$, then

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{7+2}{6} = \frac{9}{6} = \frac{3}{2} \\ \frac{y-1}{3} &= \frac{8-1}{3} = \frac{7}{3} \end{aligned}$

Since $\displaystyle \frac{3}{2} \ne \frac{7}{3}$, the point $\displaystyle (7,8,-2)$ is not on the line $\displaystyle PQ$.

Problem 3

Given $\displaystyle P(-2,1,3)$ and $\displaystyle Q(4,4,-3)$, determine whether or not the following points are on the line $\displaystyle PQ$. If the point is on the line $\displaystyle PQ$, find the corresponding real number with respect to point $\displaystyle P$.

(a) $\displaystyle (5, 2, 11)$
(b) $\displaystyle (2,2,-7)$
(c) $\displaystyle \left(\frac{7}{2},2,2\right)$
(d) $\displaystyle (6, 2, 10)$
Solution

As found above, the directed values are $\displaystyle \langle 6, 3, -6 \rangle$ and the equation is $\displaystyle \frac{x+2}{6} = \frac{y-1}{3} = \frac{z-3}{-6} = k$.

(a) If $\displaystyle (x,y,z) = (5, 2, 11)$:

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{5+2}{6} = \frac{7}{6} \\ \frac{y-1}{3} &= \frac{2-1}{3} = \frac{1}{3} \end{aligned}$

Since $\displaystyle \frac{7}{6} \ne \frac{1}{3}$, the point $\displaystyle (5, 2, 11)$ is not on the line $\displaystyle PQ$.

(b) If $\displaystyle (x,y,z) = (2,2,-7)$:

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{2+2}{6} = \frac{4}{6} = \frac{2}{3} \\ \frac{y-1}{3} &= \frac{2-1}{3} = \frac{1}{3} \end{aligned}$

Since $\displaystyle \frac{2}{3} \ne \frac{1}{3}$, the point $\displaystyle (2,2,-7)$ is not on the line $\displaystyle PQ$.

(c) If $\displaystyle (x,y,z) = \left(\frac{7}{2},2,2\right)$:

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{\frac{7}{2}+2}{6} = \frac{11}{12} \\ \frac{y-1}{3} &= \frac{2-1}{3} = \frac{1}{3} \end{aligned}$

Since $\displaystyle \frac{11}{12} \ne \frac{1}{3}$, the point $\displaystyle \left(\frac{7}{2},2,2\right)$ is not on the line $\displaystyle PQ$.

(d) If $\displaystyle (x,y,z) = (6,2,10)$:

$\displaystyle \begin{aligned} \frac{x+2}{6} &= \frac{6+2}{6} = \frac{8}{6} = \frac{4}{3} \\ \frac{y-1}{3} &= \frac{2-1}{3} = \frac{1}{3} \end{aligned}$

Since $\displaystyle \frac{4}{3} \ne \frac{1}{3}$, the point $\displaystyle (6,2,10)$ is not on the line $\displaystyle PQ$.

Problem 4

Find the points of intersection of the line joining the two points $\displaystyle (2, 4, 5)$ and $\displaystyle (3, 5, -4)$ with the following planes.

(a) $\displaystyle xy$-plane
(b) $\displaystyle yz$-plane
(c) $\displaystyle zx$-plane
Solution

Let $\displaystyle P(2, 4, 5)$ and $\displaystyle Q(3, 5, -4)$. The directed values are:

$\displaystyle \begin{aligned} \text{directed values of } PQ &= \langle 3-2, 5-4, -4-5 \rangle \\ &= \langle 1, 1, -9 \rangle \end{aligned}$

The coordinates of any point on the line are given by $\displaystyle (x, y, z) = (2+k, 4+k, 5-9k)$.

(a) $\displaystyle xy$-plane ($\displaystyle z = 0$):

$\displaystyle \begin{aligned} 5 - 9k &= 0 \implies k = \frac{5}{9} \\ x &= 2 + \frac{5}{9} = \frac{23}{9}, \quad y = 4 + \frac{5}{9} = \frac{41}{9} \end{aligned}$

The point of intersection is $\displaystyle \left( \frac{23}{9}, \frac{41}{9}, 0 \right)$.

(b) $\displaystyle yz$-plane ($\displaystyle x = 0$):

$\displaystyle \begin{aligned} 2 + k &= 0 \implies k = -2 \\ y &= 4 + (-2) = 2, \quad z = 5 - 9(-2) = 23 \end{aligned}$

The point of intersection is $\displaystyle (0, 2, 23)$.

(c) $\displaystyle zx$-plane ($\displaystyle y = 0$):

$\displaystyle \begin{aligned} 4 + k &= 0 \implies k = -4 \\ x &= 2 + (-4) = -2, \quad z = 5 - 9(-4) = 41 \end{aligned}$

The point of intersection is $\displaystyle (-2, 0, 41)$.

Exercise 3.3

Problem 1

Find $\displaystyle \cos \angle PAQ$ for the followings.

(a) $\displaystyle P(1,2,-1)$, $\displaystyle A(-2,1,5)$, $\displaystyle Q(2,-1,0)$
(b) $\displaystyle P(0,2,-3)$, $\displaystyle A(2,-1,5)$, $\displaystyle Q(-2,3,-1)$
Solution

(a) For $\displaystyle P(1,2,-1)$, $\displaystyle A(-2,1,5)$, $\displaystyle Q(2,-1,0)$:

$\displaystyle \begin{aligned} \text{directed values of } AP &= \langle 1 - (-2), 2 - 1, -1 - 5 \rangle = \langle 3, 1, -6 \rangle \\ \text{directed values of } AQ &= \langle 2 - (-2), -1 - 1, 0 - 5 \rangle = \langle 4, -2, -5 \rangle \end{aligned}$
$\displaystyle \begin{aligned} \cos \angle PAQ &= \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}} \\ &= \frac{(3)(4) + (1)(-2) + (-6)(-5)}{\sqrt{3^2 + 1^2 + (-6)^2}\sqrt{4^2 + (-2)^2 + (-5)^2}} \\ &= \frac{12 - 2 + 30}{\sqrt{9 + 1 + 36}\sqrt{16 + 4 + 25}} \\ &= \frac{40}{\sqrt{46}\sqrt{45}} = \frac{40}{\sqrt{2070}} \end{aligned}$

(b) For $\displaystyle P(0,2,-3)$, $\displaystyle A(2,-1,5)$, $\displaystyle Q(-2,3,-1)$:

$\displaystyle \begin{aligned} \text{directed values of } AP &= \langle 0 - 2, 2 - (-1), -3 - 5 \rangle = \langle -2, 3, -8 \rangle \\ \text{directed values of } AQ &= \langle -2 - 2, 3 - (-1), -1 - 5 \rangle = \langle -4, 4, -6 \rangle \end{aligned}$
$\displaystyle \begin{aligned} \cos \angle PAQ &= \frac{(-2)(-4) + (3)(4) + (-8)(-6)}{\sqrt{(-2)^2 + 3^2 + (-8)^2}\sqrt{(-4)^2 + 4^2 + (-6)^2}} \\ &= \frac{8 + 12 + 48}{\sqrt{4 + 9 + 64}\sqrt{16 + 16 + 36}} \\ &= \frac{68}{\sqrt{77}\sqrt{68}} = \sqrt{\frac{68}{77}} \end{aligned}$

Problem 2

Determine whether the lines $\displaystyle PQ$ and $\displaystyle RS$ are parallel or skew or intersect. If $\displaystyle PQ$ and $\displaystyle RS$ intersect, are they perpendicular?

(a) $\displaystyle P(1,2,3)$, $\displaystyle Q(4,5,6)$, $\displaystyle R(-2,3,5)$, $\displaystyle S(4, 9, 11)$
(b) $\displaystyle P(3,-1,-3)$, $\displaystyle Q(2,-3,1)$, $\displaystyle R(3,-2,5)$, $\displaystyle S(-1,-2,1)$
(c) $\displaystyle P(4,-2,5)$, $\displaystyle Q(-2,6,1)$, $\displaystyle R(-1,1,4)$, $\displaystyle S(3,3,2)$
(d) $\displaystyle P(-3,-1,6)$, $\displaystyle Q(-1,3,0)$, $\displaystyle R(0, 6, 7)$, $\displaystyle S(-4,-4,-1)$
Solution

(a)

$\displaystyle \begin{aligned} \text{directed values of } PQ &= \langle 3, 3, 3 \rangle \\ \text{directed values of } RS &= \langle 6, 6, 6 \rangle \end{aligned}$

Since their directed values are proportional, the lines $\displaystyle PQ$ and $\displaystyle RS$ are parallel or coincident.
To check if they are coincident, check if point $\displaystyle P(1,2,3)$ lies on line $\displaystyle RS$. The equation of $\displaystyle RS$ is $\displaystyle \frac{x+2}{6} = \frac{y-3}{6} = \frac{z-5}{6}$.
For $\displaystyle P(1,2,3)$, $\displaystyle \frac{1+2}{6} \ne \frac{2-3}{6}$, so $\displaystyle P$ is not on $\displaystyle RS$. Thus, the lines are strictly parallel.

(b)

$\displaystyle \begin{aligned} \text{directed values of } PQ &= \langle -1, -2, 4 \rangle \\ \text{directed values of } RS &= \langle -4, 0, -4 \rangle \end{aligned}$

Directed values are not proportional, so they are not parallel. If they intersect, the equations are:

$\displaystyle \begin{aligned} 3 - t &= 3 - 4s \implies 4s - t = 0 \quad \text{--- (1)} \\ -1 - 2t &= -2 \implies 2t = 1 \implies t = \frac{1}{2} \\ -3 + 4t &= 5 - 4s \quad \text{--- (2)} \end{aligned}$

From (1), if $\displaystyle t = \frac{1}{2}$, then $\displaystyle 4s = \frac{1}{2} \implies s = \frac{1}{8}$. Substitute into (2):

$\displaystyle \begin{aligned} -3 + 4\left(\frac{1}{2}\right) &= -3 + 2 = -1 \\ 5 - 4\left(\frac{1}{8}\right) &= 5 - \frac{1}{2} = \frac{9}{2} \end{aligned}$

Since $\displaystyle -1 \ne \frac{9}{2}$, they do not intersect. The lines are skew.

(c)

$\displaystyle \begin{aligned} \text{directed values of } PQ &= \langle -6, 8, -4 \rangle \\ \text{directed values of } RS &= \langle 4, 2, -2 \rangle \end{aligned}$

Not proportional, so not parallel. If they intersect:

$\displaystyle \begin{aligned} 4 - 6t &= -1 + 4s \implies -6t - 4s = -5 \quad \text{--- (1)} \\ -2 + 8t &= 1 + 2s \implies 8t - 2s = 3 \quad \text{--- (2)} \\ 5 - 4t &= 4 - 2s \implies -4t + 2s = -1 \quad \text{--- (3)} \end{aligned}$

Adding (2) and (3): $\displaystyle 4t = 2 \implies t = \frac{1}{2}$. Then from (2), $\displaystyle 8(1/2) - 2s = 3 \implies 4 - 2s = 3 \implies 2s = 1 \implies s = \frac{1}{2}$.
Check in (1): $\displaystyle -6(1/2) - 4(1/2) = -3 - 2 = -5$.
They satisfy all equations. The lines intersect.
Check for perpendicularity using their directed values $\displaystyle \langle -6, 8, -4 \rangle$ and $\displaystyle \langle 4, 2, -2 \rangle$:

$\displaystyle \begin{aligned} l_1 l_2 + m_1 m_2 + n_1 n_2 &= (-6)(4) + (8)(2) + (-4)(-2) \\ &= -24 + 16 + 8 = 0 \end{aligned}$

Since the sum is $\displaystyle 0$, the intersecting lines are perpendicular.

(d)

$\displaystyle \begin{aligned} \text{directed values of } PQ &= \langle 2, 4, -6 \rangle \\ \text{directed values of } RS &= \langle -4, -10, -8 \rangle \end{aligned}$

Not proportional, so not parallel. If they intersect:

$\displaystyle \begin{aligned} -3 + 2t &= -4s \implies 2t + 4s = 3 \quad \text{--- (1)} \\ -1 + 4t &= 6 - 10s \implies 4t + 10s = 7 \quad \text{--- (2)} \\ 6 - 6t &= 7 - 8s \implies -6t + 8s = 1 \quad \text{--- (3)} \end{aligned}$

Multiply (1) by 2: $\displaystyle 4t + 8s = 6$. Subtract from (2): $\displaystyle 2s = 1 \implies s = \frac{1}{2}$.
Substitute $s=1/2$ into (1): $\displaystyle 2t + 2 = 3 \implies 2t = 1 \implies t = \frac{1}{2}$.
Check in (3): $\displaystyle -6(1/2) + 8(1/2) = -3 + 4 = 1$.
They satisfy all equations. The lines intersect.
Check for perpendicularity:

$\displaystyle \begin{aligned} l_1 l_2 + m_1 m_2 + n_1 n_2 &= (2)(-4) + (4)(-10) + (-6)(-8) \\ &= -8 - 40 + 48 = 0 \end{aligned}$

Since the sum is $\displaystyle 0$, the intersecting lines are perpendicular.

Problem 3

Find the equation of the line passing through the point $\displaystyle (8,-1,-10)$ and perpendicular to the line $\displaystyle (x,y,z)=(1+2k, 2-k, 3-7k)$. Find also the point of intersection of two lines.

Solution

Let $\displaystyle P$ be $\displaystyle (8, -1, -10)$. The given line has directed values $\displaystyle \langle 2, -1, -7 \rangle$.
Let $\displaystyle F(1+2k, 2-k, 3-7k)$ be the point of intersection on the given line. The directed values of $\displaystyle PF$ are:

$\displaystyle \begin{aligned} \text{directed values of } PF &= \langle (1+2k) - 8, (2-k) - (-1), (3-7k) - (-10) \rangle \\ &= \langle 2k - 7, 3 - k, 13 - 7k \rangle \end{aligned}$

Since the lines are perpendicular, the sum of products of their directed values is zero:

$\displaystyle \begin{aligned} l_1l_2 + m_1m_2 + n_1n_2 &= 0 \\ 2(2k - 7) - 1(3 - k) - 7(13 - 7k) &= 0 \\ 4k - 14 - 3 + k - 91 + 49k &= 0 \\ 54k - 108 &= 0 \\ k &= 2 \end{aligned}$

The point of intersection $\displaystyle F$ is $\displaystyle (1+2(2), 2-2, 3-7(2)) = (5, 0, -11)$.
The directed values of $\displaystyle PF$ become $\displaystyle \langle 2(2) - 7, 3 - 2, 13 - 7(2) \rangle = \langle -3, 1, -1 \rangle$.
The equation of the required line passing through $\displaystyle (8, -1, -10)$ is:

$\displaystyle \begin{aligned} \frac{x - 8}{-3} &= \frac{y + 1}{1} = \frac{z + 10}{-1} \end{aligned}$

Exercise 3.4

Problem 1

Find the equation of the plane containing

(a) $\displaystyle A(2,-5,4)$, $\displaystyle B(-5,2,4)$ and $\displaystyle C(-2,3,-1)$
(b) $\displaystyle A(4,2,-3)$, $\displaystyle B(1,-2,4)$ and $\displaystyle C(-1,0,3)$.
Solution

(a) $\displaystyle A(2,-5,4)$, $\displaystyle B(-5,2,4)$, $\displaystyle C(-2,3,-1)$.

$\displaystyle \begin{aligned} \text{directed values of } AB &= \langle l_1, m_1, n_1 \rangle = \langle -7, 7, 0 \rangle \\ \text{directed values of } AC &= \langle l_2, m_2, n_2 \rangle = \langle -4, 8, -5 \rangle \end{aligned}$

Let the directed values of the normal to the plane be $\displaystyle \langle a,b,c \rangle$. Since $\displaystyle AB$ and $\displaystyle AC$ lie on the plane, the normal is perpendicular to both. Using analytical geometry formulas:

$\displaystyle \begin{aligned} a &= m_1n_2 - m_2n_1 = (7)(-5) - (8)(0) = -35 \\ b &= n_1l_2 - n_2l_1 = (0)(-4) - (-5)(-7) = -35 \\ c &= l_1m_2 - l_2m_1 = (-7)(8) - (-4)(7) = -56 + 28 = -28 \end{aligned}$

Dividing by $\displaystyle -7$, we get $\displaystyle a = 5$, $\displaystyle b = 5$, and $\displaystyle c = 4$. The directed values of the normal are $\displaystyle \langle 5, 5, 4 \rangle$.
The equation of the plane passing through $\displaystyle A(2, -5, 4)$ is:

$\displaystyle \begin{aligned} ax + by + cz &= ax_1 + by_1 + cz_1 \\ 5x + 5y + 4z &= 5(2) + 5(-5) + 4(4) \\ 5x + 5y + 4z &= 10 - 25 + 16 \\ 5x + 5y + 4z &= 1 \end{aligned}$

(b) $\displaystyle A(4,2,-3)$, $\displaystyle B(1,-2,4)$, $\displaystyle C(-1,0,3)$.

$\displaystyle \begin{aligned} \text{directed values of } AB &= \langle l_1, m_1, n_1 \rangle = \langle -3, -4, 7 \rangle \\ \text{directed values of } AC &= \langle l_2, m_2, n_2 \rangle = \langle -5, -2, 6 \rangle \end{aligned}$

Using analytical geometry formulas for the normal $\displaystyle \langle a,b,c \rangle$:

$\displaystyle \begin{aligned} a &= m_1n_2 - m_2n_1 = (-4)(6) - (-2)(7) = -24 + 14 = -10 \\ b &= n_1l_2 - n_2l_1 = (7)(-5) - (6)(-3) = -35 + 18 = -17 \\ c &= l_1m_2 - l_2m_1 = (-3)(-2) - (-5)(-4) = 6 - 20 = -14 \end{aligned}$

Multiplying by $\displaystyle -1$, we get $\displaystyle a=10, b=17, c=14$. The directed values of the normal are $\displaystyle \langle 10, 17, 14 \rangle$.
The equation of the plane passing through $\displaystyle A(4, 2, -3)$ is:

$\displaystyle \begin{aligned} 10x + 17y + 14z &= 10(4) + 17(2) + 14(-3) \\ 10x + 17y + 14z &= 40 + 34 - 42 \\ 10x + 17y + 14z &= 32 \end{aligned}$

Problem 2

Find the equation of the line passing through the point $\displaystyle (3,-2,-2)$ and perpendicular to the plane $\displaystyle -2x+3y-z=4$. Find the point of intersection of the line and the plane.

Solution

The directed values of the normal to the plane $\displaystyle -2x+3y-z=4$ are $\displaystyle \langle -2, 3, -1 \rangle$.
Since the line is perpendicular to the plane, its directed values are also $\displaystyle \langle -2, 3, -1 \rangle$.
The coordinate equations of the line through $\displaystyle (3, -2, -2)$ are:

$\displaystyle \begin{aligned} x &= 3 - 2t \\ y &= -2 + 3t \\ z &= -2 - t \end{aligned}$

To find the point of intersection, substitute these into the plane equation:

$\displaystyle \begin{aligned} -2(3 - 2t) + 3(-2 + 3t) - (-2 - t) &= 4 \\ -6 + 4t - 6 + 9t + 2 + t &= 4 \\ 14t - 10 &= 4 \\ 14t &= 14 \\ t &= 1 \end{aligned}$

Substitute $\displaystyle t = 1$ back to find the coordinates:

$\displaystyle \begin{aligned} x &= 3 - 2(1) = 1 \\ y &= -2 + 3(1) = 1 \\ z &= -2 - (1) = -3 \end{aligned}$

The point of intersection is $\displaystyle (1, 1, -3)$.

Problem 3

Find the equation of the plane containing the point $\displaystyle (2, 3, -1)$ and parallel to the plane $\displaystyle -2x+y+3z=6$.

Solution

Since the required plane is parallel to $\displaystyle -2x+y+3z=6$, its normal has proportional directed values: $\displaystyle \langle -2, 1, 3 \rangle$.
The equation is of the form $\displaystyle -2x + y + 3z = d$.
Since it passes through $\displaystyle (2, 3, -1)$:

$\displaystyle \begin{aligned} d &= -2(2) + 1(3) + 3(-1) \\ d &= -4 + 3 - 3 = -4 \end{aligned}$

The equation of the plane is $\displaystyle -2x + y + 3z = -4$, or equivalently:

$\displaystyle \begin{aligned} 2x - y - 3z &= 4 \end{aligned}$

Exercise 3.5

Problem 1

Find the equation of the sphere with center $\displaystyle C$ and radius $\displaystyle r$.

(a) $\displaystyle C(1,-2, 4)$, $\displaystyle r = 3$
(b) $\displaystyle C(2, 6, -3)$, $\displaystyle r = 2$
(c) $\displaystyle C(2, 3, 5)$, $\displaystyle r = 5$
Solution

The equation of a sphere is $\displaystyle (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 = r^2$.

(a) Center $\displaystyle (1, -2, 4)$, $\displaystyle r = 3$.

$\displaystyle \begin{aligned} (x - 1)^2 + (y + 2)^2 + (z - 4)^2 &= 9 \end{aligned}$

(b) Center $\displaystyle (2, 6, -3)$, $\displaystyle r = 2$.

$\displaystyle \begin{aligned} (x - 2)^2 + (y - 6)^2 + (z + 3)^2 &= 4 \end{aligned}$

(c) Center $\displaystyle (2, 3, 5)$, $\displaystyle r = 5$.

$\displaystyle \begin{aligned} (x - 2)^2 + (y - 3)^2 + (z - 5)^2 &= 25 \end{aligned}$

Problem 2

Check whether the given point $\displaystyle P$ lies inside, outside or on a sphere.

  1. Center $\displaystyle C(0, 0, 0)$, radius $\displaystyle r = 3$ and point $\displaystyle P(1, 1, 1)$.
  2. Center $\displaystyle C(0, 0, 0)$, radius $\displaystyle r = 3$ and point $\displaystyle P(2, 1, 2)$.
  3. Center $\displaystyle C(0, 0, 0)$, radius $\displaystyle r = 3$ and point $\displaystyle P(10, 10, 10)$.
Solution

We calculate the squared distance $\displaystyle CP^2$ and compare it with $\displaystyle r^2 = 9$.

(a) $\displaystyle P(1, 1, 1)$:

$\displaystyle \begin{aligned} CP^2 &= (1-0)^2 + (1-0)^2 + (1-0)^2 = 1 + 1 + 1 = 3 \end{aligned}$

Since $\displaystyle 3 < 9$, point $\displaystyle P$ lies inside the sphere.

(b) $\displaystyle P(2, 1, 2)$:

$\displaystyle \begin{aligned} CP^2 &= (2-0)^2 + (1-0)^2 + (2-0)^2 = 4 + 1 + 4 = 9 \end{aligned}$

Since $\displaystyle 9 = 9$, point $\displaystyle P$ lies on the sphere.

(c) $\displaystyle P(10, 10, 10)$:

$\displaystyle \begin{aligned} CP^2 &= (10-0)^2 + (10-0)^2 + (10-0)^2 = 100 + 100 + 100 = 300 \end{aligned}$

Since $\displaystyle 300 > 9$, point $\displaystyle P$ lies outside the sphere.

Problem 3

Find the equation of the sphere on the join of $\displaystyle (1, -1, 1)$ and $\displaystyle (-3, 4, 5)$ as diameter.

Solution

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

$\displaystyle \begin{aligned} C &= \left(\frac{1 - 3}{2}, \frac{-1 + 4}{2}, \frac{1 + 5}{2}\right) = \left(-1, \frac{3}{2}, 3\right) \end{aligned}$

The radius squared $\displaystyle r^2$ is the squared distance from the center to $\displaystyle (1, -1, 1)$.

$\displaystyle \begin{aligned} r^2 &= (1 - (-1))^2 + \left(-1 - \frac{3}{2}\right)^2 + (1 - 3)^2 \\ &= 2^2 + \left(-\frac{5}{2}\right)^2 + (-2)^2 \\ &= 4 + \frac{25}{4} + 4 = 8 + \frac{25}{4} = \frac{57}{4} \end{aligned}$

The equation of the sphere is:

$\displaystyle \begin{aligned} (x + 1)^2 + \left(y - \frac{3}{2}\right)^2 + (z - 3)^2 &= \frac{57}{4} \end{aligned}$

Problem 4

Find the equation of the plane tangent to the sphere $\displaystyle (x + 2)^2 + (y - 1)^2 + (z + 3)^2 = 27$ at the point $\displaystyle (3, 2, -2)$.

Solution

The center of the sphere is $\displaystyle C(-2, 1, -3)$. The tangent point is $\displaystyle P(3, 2, -2)$.
The directed values of the normal to the tangent plane are along $\displaystyle CP$:

$\displaystyle \begin{aligned} \text{directed values of } CP &= \langle 3 - (-2), 2 - 1, -2 - (-3) \rangle = \langle 5, 1, 1 \rangle \end{aligned}$

The equation of the tangent plane is $\displaystyle 5x + y + z = d$.
Since it passes through $\displaystyle P(3, 2, -2)$:

$\displaystyle \begin{aligned} d &= 5(3) + 1(2) + 1(-2) \\ &= 15 + 2 - 2 = 15 \end{aligned}$

The equation of the tangent plane is:

$\displaystyle \begin{aligned} 5x + y + z &= 15 \end{aligned}$

Problem 5

Find the equation of the sphere with center $\displaystyle (6, -7, -3)$ and touching the plane $\displaystyle 4x - 2y - z = 17$.

Solution

The radius $\displaystyle r$ is the perpendicular distance from the center $\displaystyle C(6, -7, -3)$ to the plane $\displaystyle 4x - 2y - z - 17 = 0$.

$\displaystyle \begin{aligned} r &= \frac{|4(6) - 2(-7) - 1(-3) - 17|}{\sqrt{4^2 + (-2)^2 + (-1)^2}} \\ &= \frac{|24 + 14 + 3 - 17|}{\sqrt{16 + 4 + 1}} \\ &= \frac{24}{\sqrt{21}} \end{aligned}$

Thus, $\displaystyle r^2 = \frac{576}{21} = \frac{192}{7}$.
The equation of the sphere is:

$\displaystyle \begin{aligned} (x - 6)^2 + (y + 7)^2 + (z + 3)^2 &= \frac{192}{7} \end{aligned}$

Problem 6

What is the equation of the sphere which passes through the points $\displaystyle (3, 0, 2)$, $\displaystyle (-1, 1, 1)$ and $\displaystyle (2, -5, 4)$ and whose center lies on the plane $\displaystyle 2x + 3y + 4z = 6$?

Solution

Let the center be $\displaystyle C(a, b, c)$. The distance from $\displaystyle C$ to each point is the radius $\displaystyle r$, so their squares are equal.

$\displaystyle \begin{aligned} r^2 &= (3-a)^2 + (0-b)^2 + (2-c)^2 = a^2+b^2+c^2 - 6a - 4c + 13 \quad \text{--- (1)} \\ r^2 &= (-1-a)^2 + (1-b)^2 + (1-c)^2 = a^2+b^2+c^2 + 2a - 2b - 2c + 3 \quad \text{--- (2)} \\ r^2 &= (2-a)^2 + (-5-b)^2 + (4-c)^2 = a^2+b^2+c^2 - 4a + 10b - 8c + 45 \quad \text{--- (3)} \end{aligned}$

Equating (1) and (2):

$\displaystyle \begin{aligned} -6a - 4c + 13 &= 2a - 2b - 2c + 3 \\ 8a - 2b + 2c &= 10 \\ 4a - b + c &= 5 \quad \text{--- (4)} \end{aligned}$

Equating (2) and (3):

$\displaystyle \begin{aligned} 2a - 2b - 2c + 3 &= -4a + 10b - 8c + 45 \\ 6a - 12b + 6c &= 42 \\ a - 2b + c &= 7 \quad \text{--- (5)} \end{aligned}$

Subtracting (5) from (4):

$\displaystyle \begin{aligned} 3a + b &= -2 \implies b = -3a - 2 \end{aligned}$

Substitute $\displaystyle b$ into (5):

$\displaystyle \begin{aligned} a - 2(-3a - 2) + c &= 7 \\ a + 6a + 4 + c &= 7 \\ 7a + c &= 3 \implies c = 3 - 7a \end{aligned}$

The center lies on $\displaystyle 2x + 3y + 4z = 6$, so $\displaystyle 2a + 3b + 4c = 6$. Substitute $\displaystyle b$ and $\displaystyle c$:

$\displaystyle \begin{aligned} 2a + 3(-3a - 2) + 4(3 - 7a) &= 6 \\ 2a - 9a - 6 + 12 - 28a &= 6 \\ -35a + 6 &= 6 \\ -35a &= 0 \implies a = 0 \end{aligned}$

Then $\displaystyle b = -3(0) - 2 = -2$ and $\displaystyle c = 3 - 7(0) = 3$. The center is $\displaystyle (0, -2, 3)$.
Substitute the center into (1) to find $\displaystyle r^2$:

$\displaystyle \begin{aligned} r^2 &= (3 - 0)^2 + (0 - (-2))^2 + (2 - 3)^2 \\ &= 9 + 4 + 1 = 14 \end{aligned}$

The equation of the sphere is:

$\displaystyle \begin{aligned} x^2 + (y + 2)^2 + (z - 3)^2 &= 14 \end{aligned}$