Chapter 3: Analytical Solid Geometry

We have learned about points and lines in two-dimensional rectangular coordinate system. In this chapter, we will extend the system to three dimensions.

3.1 Coordinates of a Point in Space

In the plane, each point is associated with an ordered pair of real numbers.

In space, each point is associated with an ordered triple of real numbers.

Through a fixed point, called the origin $O$, draw three mutually perpendicular lines: the x-axis, the y-axis and the z-axis.

A point $P$ in space is determined by an ordered triple $(x,y,z)$ of real numbers as shown in the diagram.

The numbers $x$, $y$, $z$ are called the coordinates of $P$.

The following table:

Plane	Equation	Coordinates
xy-plane	$\ds z=0$	$\ds (x,y,0)$
yz-plane	$\ds x=0$	$\ds (0,y,z)$
xz-plane	$\ds y=0$	$\ds (x,0,z)$

Table 3.1

The following table:

Plane	Equation	Coordinates	Example Equation	Coordinates
Parallel to xy-plane	$\ds z=c$	$\ds (x,y,c)$	$\ds z=1$	$\ds (x,y,1)$
Parallel to yz-plane	$\ds x=a$	$\ds (a,y,z)$	$\ds x=5$	$\ds (5,y,z)$
Parallel to zx-plane	$\ds y=b$	$\ds (x,b,z)$	$\ds y=2$	$\ds (x,2,z)$

Table 3.2

Line perpendicular to the xy-plane

Equation of any line perpendicular to the xy-plane and passing through the point $\ds (a,b,c)$ is $\ds x=a$, $\ds y=b$ and the coordinates of the points on that line are of the form $\ds (a,b,z)$.

Line perpendicular to the yz-plane

Equation of any line perpendicular to the yz-plane and passing through the point $\ds (a,b,c)$ is $\ds y=b$, $\ds z=c$ and the coordinates of the points on the line are of the form $\ds (x,b,c)$.

Line perpendicular to the zx-plane

Equation of any line perpendicular to the zx-plane and passing through the point $\ds (a,b,c)$ is $\ds x=a$, $\ds z=c$ and the coordinates of the points on the line are of the form $\ds (a,y,c)$.

Example 1.

Find the equation of the line through the point $\ds (-3,5,7)$ and perpendicular to
(a) xy-plane
(b) yz-plane
(c) zx-plane.
Find the point of intersection of the line and plane.

Solution

(a) The equation of the line through the point $\ds (-3,5,7)$ and perpendicular to xy-plane is
$\ds x=-3$, $\ds y=5$ or $\ds (-3,5,z)$
The point of intersection of the line and xy-plane is $\ds (-3,5,0)$.

(b) The equation of the line through the point $\ds (-3,5,7)$ and perpendicular to yz-plane is
$\ds y=5$, $\ds z=7$ or $\ds (x,5,7)$
The point of intersection of the line and yz-plane is $\ds (0,5,7)$.

(c) The equation of the line through the point $\ds (-3,5,7)$ and perpendicular to zx-plane is
$\ds z=7$, $\ds x=-3$ or $\ds (-3,y,7)$
The point of intersection of the line and zx-plane is $\ds (-3,0,7)$.

Note that:

Equation of x-axis: $\ds y=0$, $\ds z=0$.
Equation of y-axis: $\ds x=0$, $\ds z=0$.
Equation of z-axis: $\ds x=0$, $\ds y=0$.

Distance between two points

The distance between points $\ds P(x_1,y_1,z_1)$ and $\ds Q(x_2,y_2,z_2)$ is

\displaystyle \begin{aligned} PQ &= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \end{aligned}

Exercise 3.1

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

(a) xy-plane

(b) yz-plane

(c) zx-plane.
Find the equation of the line through the point $\ds (2,3,-4)$ and perpendicular to

(a) xy-plane

(b) yz-plane

(c) zx-plane.

Find the point of intersection of the line and plane.
Find the distance between the points $\ds (2,-3, 5)$ and $\ds (7, 5,-2)$.
Show that the points $\ds (-1, 2, 5)$, $\ds (1, 1, 6)$ and $\ds (0, 5, 6)$ form a right triangle.
Show that the points $\ds (1,-1,2)$, $\ds (3,-2,3)$ and $\ds (5,-3,4)$ are collinear.

3.2 Lines

Directed value of a line segment

For a line segment $\ds PQ$, directed values $\ds \langle l,m,n \rangle$ of $\ds PQ$ where $\ds P$ is $\ds (x_1,y_1,z_1)$ and $\ds Q$ is $\ds (x_2,y_2,z_2)$ is defined by

\displaystyle \begin{aligned} \langle PQ \rangle &= \langle l,m,n \rangle = \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle \end{aligned}

The length of the segment $\ds PQ$ is

\displaystyle \begin{aligned} PQ &= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \\ &= \sqrt{l^2 + m^2 + n^2} \end{aligned}

Real numbers and points on the line

The coordinates of point $\ds R$ on the line $\ds PQ$ with respect to the point $\ds P$ and a real number $\ds k$ are $\ds (x,y,z) = (x_1+kl, y_1+km, z_1+kn)$.

The equation is called coordinate form of the equation of line $\ds PQ$ and $\ds k$ is called a parameter.

In general, if a segment of a line through $\ds (x_1,y_1,z_1)$ has directed values $\ds \langle l,m,n \rangle$ which are not equal to zero, then the equation of the line can be written as

\displaystyle \begin{aligned} \frac{x-x_1}{l} &= \frac{y-y_1}{m} = \frac{z-z_1}{n} \end{aligned}

Since $\ds \langle PR \rangle = \langle x_1+kl-x_1, y_1+km-y_1, z_1+kn-z_1 \rangle = \langle kl, km, kn \rangle$ and $\ds R$ is any point on the line $\ds PQ$ with $\ds \langle PQ \rangle = \langle l,m,n \rangle$, we will defined directed values of a line $\ds PQ$ as $\ds (kl, km, kn)$ where $\ds k$ is a real number.

Example 2.

Given $\ds P(1,2,3)$ and $\ds Q(3,6,5)$, find the coordinates of point $\ds R(x,y,z)$ on the line $\ds PQ$ with respect to the point $\ds P$ and the following parameters.

(a) $\ds k=\frac{1}{2} \quad$ (b) $\ds k=2 \quad$ (c) $\ds k=-2$

Solution

By given $\ds P(1,2,3)$ and $\ds Q(3,6,5)$,

\displaystyle \begin{aligned} \langle PQ \rangle &= \langle l,m,n \rangle \\ &= \langle 3-1, 6-2, 5-3 \rangle \\ &= \langle 2,4,2 \rangle \end{aligned}

(a) For $\ds k=\frac{1}{2}$, $\ds (x,y,z) = \left(1+\frac{1}{2}(2), 2+\frac{1}{2}(4), 3+\frac{1}{2}(2)\right) = (2,4,4)$.

(b) For $\ds k=2$, $\ds (x,y,z) = (1+2(2), 2+2(4), 3+2(2)) = (5,10,7)$.

(c) For $\ds k=-2$, $\ds (x,y,z) = (1+(-2)(2), 2+(-2)(4), 3+(-2)(2)) = (-3,-6,-1)$.

Example 3.

Given $\ds P(-1,2,3)$ and $\ds Q(3,5,-2)$ determine whether or not the following points are on the line $\ds PQ$. If the point is on the line $\ds PQ$, find the corresponding parameter with respect to the point $\ds P$.

(a) $\ds \left(1,\frac{7}{2},\frac{1}{2}\right) \quad$ (b) $\ds (7,8,-7) \quad$ (c) $\ds (-5,-1,8) \quad$ (d) $\ds (7,8,-2)$

Solution

By given $\ds P(-1,2,3)$ and $\ds Q(3,5,-2)$

\displaystyle \begin{aligned} \langle PQ \rangle &= \langle l,m,n \rangle \\ &= \langle 3+1, 5-2, -2-3 \rangle \\ &= \langle 4,3,-5 \rangle \end{aligned}

Thus, the equation of the line $\ds PQ$ is $\ds \frac{x+1}{4} = \frac{y-2}{3} = \frac{z-3}{-5}$

(a) If $\ds (x,y,z) = \left(1,\frac{7}{2},\frac{1}{2}\right)$ then

\displaystyle \begin{aligned} \frac{x+1}{4} &= \frac{1+1}{4} = \frac{1}{2} \\ \frac{y-2}{3} &= \frac{\frac{7}{2}-2}{3} = \frac{1}{2} \\ \frac{z-3}{-5} &= \frac{\frac{1}{2}-3}{-5} = \frac{1}{2} \end{aligned}

So $\ds \frac{x+1}{4} = \frac{y-2}{3} = \frac{z-3}{-5}$ for $\ds \left(1,\frac{7}{2},\frac{1}{2}\right)$.
Hence the point $\ds \left(1,\frac{7}{2},\frac{1}{2}\right)$ is on the line $\ds PQ$ with corresponding parameter is $\ds \frac{1}{2}$.

(b) If $\ds (x,y,z) = (7,8,-7)$, then

\displaystyle \begin{aligned} \frac{x+1}{4} &= \frac{7+1}{4} = 2 \\ \frac{y-2}{3} &= \frac{8-2}{3} = 2 \\ \frac{z-3}{-5} &= \frac{-7-3}{-5} = 2 \end{aligned}

So $\ds \frac{x+1}{4} = \frac{y-2}{3} = \frac{z-3}{-5}$ for $\ds (7,8,-7)$.
Hence the point $\ds (7,8,-7)$ is on the line $\ds PQ$ with corresponding parameter is $\ds 2$.

(c) If $\ds (x,y,z) = (-5,-1,8)$, then

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

So $\ds \frac{x+1}{4} = \frac{y-2}{3} = \frac{z-3}{-5}$ for $\ds (-5,-1,8)$.
Hence the point $\ds (-5,-1,8)$ is on the line $\ds PQ$ with corresponding parameter is $\ds -1$.

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

\displaystyle \begin{aligned} \frac{x+1}{4} &= \frac{7+1}{4} = 2 \\ \frac{y-2}{3} &= \frac{8-2}{3} = 2 \\ \frac{z-3}{-5} &= \frac{-2-3}{-5} = 1 \end{aligned}

So $\ds \frac{x+1}{4} = \frac{y-2}{3} \ne \frac{z-3}{-5}$ for $\ds (7,8,-2)$.
Hence the point $\ds (7,8,-2)$ is not on the line $\ds PQ$.

Example 4.

Given $\ds P(2,1,3)$ and $\ds Q(6,-5,3)$, determine whether or not the following points are on the line $\ds PQ$, if the point is on the line $\ds PQ$, find the corresponding parameter with respect to the point $\ds P$.

(a) $\ds (4,-2, 3) \quad$ (b) $\ds (-2, 7, 3) \quad$ (c) $\ds (10, -11, 3) \quad$ (d) $\ds (1, 1, 3)$

Solution

By given $\ds P(2,1,3)$ and $\ds Q(6,-5,3)$,

\displaystyle \begin{aligned} \langle PQ \rangle &= \langle l,m,n \rangle \\ &= \langle 6-2, -5-1, 3-3 \rangle \\ &= \langle 4,-6,0 \rangle \end{aligned}

Then the coordinates of the point $\ds (x,y,z)$ on the line $\ds PQ$ are
$\ds (x,y,z) = (2+4k, 1-6k, 3)$.
This means that the line $\ds PQ$ is on the plane $\ds z=3$.

(a) If $\ds (x,y,z) = (4,-2,3)$, then $\ds (4,-2,3) = (2+4k, 1-6k, 3)$

\displaystyle \begin{aligned} 2+4k &= 4 \\ 1-6k &= -2 \\ k &= \frac{1}{2} \end{aligned}

Thus the point $\ds (4,-2,3)$ is on the line $\ds PQ$ with corresponding parameter $\ds \frac{1}{2}$.

(b) If $\ds (x,y,z) = (-2,7,3)$, then $\ds (-2,7,3) = (2+4k, 1-6k, 3)$

\displaystyle \begin{aligned} 2+4k &= -2 \\ 1-6k &= 7 \\ k &= -1 \end{aligned}

Thus the point $\ds (-2,7,3)$ is on the line $\ds PQ$ with corresponding parameter $\ds -1$.

(c) If $\ds (x,y,z) = (10,-11,3)$, then $\ds (10,-11,3) = (2+4k, 1-6k, 3)$

\displaystyle \begin{aligned} 2+4k &= 10 \\ 1-6k &= -11 \\ k &= 2 \end{aligned}

Thus, the point $\ds (10,-11,3)$ is on the line $\ds PQ$ with corresponding parameter $\ds 2$.

(d) If $\ds (x,y,z) = (1,1,3)$, then $\ds (1,1,3) = (2+4k, 1-6k, 3)$

\displaystyle \begin{aligned} 2+4k &= 1 \Rightarrow k = -\frac{1}{4} \\ 1-6k &= 1 \Rightarrow k = 0 \end{aligned}

There is no value $\ds k$ that satisfies this condition. Therefore, the point $\ds (1,1,3)$ is not on the line $\ds PQ$.

Exercise 3.2

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

(a) $\ds k=\frac{1}{2}$

(b) $\ds k=3$

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

(a) $\ds (6,3,-6)$

(b) $\ds (6,5,-5)$

(c) $\ds (-4,0,5)$

(d) $\ds (7,8,-2)$
Given $\ds P(-2,1,3)$ and $\ds Q(4,4,-3)$, determine whether or not the following points are on the line $\ds PQ$. If the point is on the line $\ds PQ$, find the corresponding real number with respect to point $\ds P$.

(a) $\ds (5, 2, 11)$

(b) $\ds (2,2,-7)$

(c) $\ds \left(\frac{7}{2},2,2\right)$

(d) $\ds (6, 2, 10)$
Find the points of intersection of the line joining the two points $\ds (2, 4, 5)$ and $\ds (3, 5, -4)$ with the following planes.

(a) xy-plane

(b) yz-plane

(c) zx-plane

3.3 Parallel, Skew and Perpendicular Lines

Parallel Lines

Two lines are parallel if and only if their directed values are multiples of each other by some real number.

Skew Lines

In space, there are pairs of lines that are neither parallel nor intersect. These pairs of lines are called skew lines.

The Measure of $\ds \angle PAQ$

Consider $\ds P(x_1,y_1,z_1)$, $\ds A(a,b,c)$ and $\ds Q(x_2,y_2,z_2)$.

\displaystyle \begin{aligned} \langle AP \rangle &= \langle l_1,m_1,n_1 \rangle = \langle x_1-a, y_1-b, z_1-c \rangle \\ \langle AQ \rangle &= \langle l_2,m_2,n_2 \rangle = \langle x_2-a, y_2-b, z_2-c \rangle \\ \langle PQ \rangle &= \langle l_3,m_3,n_3 \rangle = \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle \\ &= \langle l_2-l_1, m_2-m_1, n_2-n_1 \rangle \end{aligned}

\displaystyle \begin{aligned} AP^2 + AQ^2 - PQ^2 &= l_1^2 + m_1^2 + n_1^2 + l_2^2 + m_2^2 + n_2^2 - (l_2-l_1)^2 - (m_2-m_1)^2 - (n_2-n_1)^2 \\ &= 2(l_1l_2 + m_1m_2 + n_1n_2) \end{aligned}

By the law of cosines, $\ds \cos \angle PAQ = \frac{AP^2 + AQ^2 - PQ^2}{2AP.AQ}$

\displaystyle \begin{aligned} \cos \angle PAQ &= \frac{2(l_1l_2 + m_1m_2 + n_1n_2)}{2AP.AQ} \\ &= \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}} \end{aligned}

If $\ds l_1l_2 + m_1m_2 + n_1n_2 = 0$ then $\ds \cos \angle PAQ = 0$ and hence $\ds \angle PAQ = 90^\circ$.

Perpendicular Lines

Two lines are perpendicular if and only if they intersect and $\ds l_1l_2 + m_1m_2 + n_1n_2 = 0$ for any directed values $\ds \langle l_1,m_1,n_1 \rangle$ and $\ds \langle l_2,m_2,n_2 \rangle$ of the lines.

Example 5.

Given $\ds P(2,1,3)$, $\ds Q(6,-5,4)$, $\ds R(2,3,4)$ and $\ds S(-1,5,1)$, determine whether the lines $\ds PQ$ and $\ds RS$ are parallel or skew or intersect.

Solution

For $\ds P(2,1,3)$ and $\ds Q(6,-5,4)$, $\ds \langle PQ \rangle = \langle 4,-6,1 \rangle$
For $\ds R(2,3,4)$ and $\ds S(-1,5,1)$, $\ds \langle RS \rangle = \langle -3,2,-3 \rangle$

Since directed values of $\ds PQ$ are not multiple of $\ds RS$,

\displaystyle \begin{aligned} \frac{4}{-3} &\ne \frac{-6}{2} \ne \frac{1}{-3} \end{aligned}

So, two lines are not parallel.

If a point $\ds (x,y,z)$ is on the lines $\ds PQ$ and $\ds RS$, then

\displaystyle \begin{aligned} x &= 2+4s \quad &x &= 2-3t \\ y &= 1-6s \quad &y &= 3+2t \\ z &= 3+s \quad &z &= 4-3t \end{aligned}

for real numbers $\ds s$ and $\ds t$.

Thus,

\displaystyle \begin{aligned} 2+4s &= 2-3t \\ 1-6s &= 3+2t \\ 3+s &= 4-3t \end{aligned}

Solving first two of these equations, we have
$\ds s = -\frac{3}{5}$ and $\ds t = \frac{4}{5}$.

But $\ds 3+\left(-\frac{3}{5}\right) \ne 4-3\left(\frac{4}{5}\right)$ these values of $\ds s$ and $\ds t$ do not satisfy the last equation.
So, the system of equations has no solution and hence the given lines do not intersect.
Therefore, the given lines are skew.

Example 6.

Given $\ds P(0,0,1)$, $\ds Q(3,6,4)$, $\ds R(0,3,1)$ and $\ds S(3,0,4)$, show that the lines $\ds PQ$ and $\ds RS$ are perpendicular.

Solution

$\ds P(0,0,1)$, $\ds Q(3,6,4)$, $\ds R(0,3,1)$, $\ds S(3,0,4)$
$\ds \langle PQ \rangle = \langle l_1,m_1,n_1 \rangle = \langle 3,6,3 \rangle$ and $\ds \langle RS \rangle = \langle l_2,m_2,n_2 \rangle = \langle 3,-3,3 \rangle$.

If a point $\ds (x,y,z)$ is on the lines $\ds PQ$ and $\ds RS$, then

\displaystyle \begin{aligned} x &= 0+3s \quad &x &= 0+3t \\ y &= 0+6s \quad &y &= 3-3t \\ z &= 1+3s \quad &z &= 1+3t \end{aligned}

for real numbers $\ds s$ and $\ds t$.

\displaystyle \begin{aligned} 3s &= 3t \Rightarrow s=t \\ 6s &= 3-3t \Rightarrow 6s = 3-3s \Rightarrow s = \frac{1}{3} \\ 1+3s &= 1+3t \Rightarrow t = \frac{1}{3} \end{aligned}

$\ds x = 3\left(\frac{1}{3}\right) = 1$, $\ds y = 6\left(\frac{1}{3}\right) = 2$, $\ds z = 1+3\left(\frac{1}{3}\right) = 2$.
The point of intersection is $\ds (1, 2, 2)$ and two lines intersect.

\displaystyle \begin{aligned} l_1l_2 + m_1m_2 + n_1n_2 &= (3)(3) + (6)(-3) + (3)(3) = 0 \end{aligned}

Hence $\ds PQ$ and $\ds RS$ are perpendicular.

Example 7.

Find the equation of the line passing through the point $\ds (-4,7,-3)$ and perpendicular to the line $\ds (x,y,z)=(3+2k, -1+3k, 1-k)$. Find also the point of intersection of two lines.

Solution

Directed values of the given line are $\ds \langle 2, 3, -1 \rangle$.
Directed values of the require line are

\displaystyle \begin{aligned} \langle -4-(3+2k), 7-(-1+3k), -3-(1-k) \rangle &= \langle -7-2k, 8-3k, -4+k \rangle \quad \text{for some real number } k \end{aligned}

If two lines are perpendicular, then

\displaystyle \begin{aligned} 2(-7-2k) + 3(8-3k) + (-1)(-4+k) &= 0 \\ -14-4k+24-9k+5-k &= 0 \\ 14k &= 14 \\ k &= 1 \end{aligned}

So, directed values of required line are $\ds \langle -7-2(1), 8-3(1), -4+1 \rangle = \langle -9,5,-3 \rangle$
and the equation of the line is

\displaystyle \begin{aligned} (x,y,z) &= (-4-9t, 7+5t, -3-3t) \end{aligned}

The point of intersection is $\ds (x,y,z) = (5,2,0)$.

Exercise 3.3

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

(a) $\ds P(1,2,-1)$, $\ds A(-2,1,5)$, $\ds Q(2,-1,0)$

(b) $\ds P(0,2,-3)$, $\ds A(2,-1,5)$, $\ds Q(-2,3,-1)$
Determine whether the lines $\ds PQ$ and $\ds RS$ are parallel or skew or intersect. If $\ds PQ$ and $\ds RS$ intersect, are they perpendicular?

(a) $\ds P(1,2,3)$, $\ds Q(4,5,6)$, $\ds R(-2,3,5)$, $\ds S(4, 9, 11)$

(b) $\ds P(3,-1,-3)$, $\ds Q(2,-3,1)$, $\ds R(3,-2,5)$, $\ds S(-1,-2,1)$

(c) $\ds P(4,-2,5)$, $\ds Q(-2,6,1)$, $\ds R(-1,1,4)$, $\ds S(3,3,2)$

(d) $\ds P(-3,-1,6)$, $\ds Q(-1,3,0)$, $\ds R(0, 6, 7)$, $\ds S(-4,-4,-1)$
Find the equation of the line passing through the point $\ds (8,-1,-10)$ and perpendicular to the line $\ds (x,y,z)=(1+2k, 2-k, 3-7k)$. Find also the point of intersection of two lines.

3.4 Planes

A plane is determined by three points which are not on the same line.

Let $\ds P(x,y,z)$ be any point on the plane through $\ds A$, $\ds B$, $\ds C$. Since $\ds A$, $\ds B$, $\ds C$ are not on the same line, line segment joining any two points will intersect each other.

Let $\ds AB$ intersect $\ds AC$ at $\ds A$. Draw a line through $\ds P$ parallel to $\ds AC$.

This line will meet $\ds AB$ at $\ds R(x_1+sl_1, y_1+sm_1, z_1+sn_1)$ for some parameter $\ds s$.

As shown in the given figure, coordinates of any point $\ds (x,y,z)$ on the plane are

\displaystyle \begin{aligned} x &= x_1+sl_1+tl_2 \\ y &= y_1+sm_1+tm_2 \\ z &= z_1+sn_1+tn_2 \end{aligned}

for some parameter $\ds t$.

Let

\displaystyle \begin{aligned} a &= m_1n_2 - m_2n_1 \\ b &= n_1l_2 - n_2l_1 \\ c &= l_1m_2 - l_2m_1 \end{aligned}

Then $\ds al_1 + bm_1 + cn_1 = 0$ and $\ds al_2 + bm_2 + cn_2 = 0$.
Thus, the plane equation is

\displaystyle \begin{aligned} ax+by+cz &= ax_1+by_1+cz_1 \end{aligned}

Cartesian form: $\ds ax+by+cz=d$ where $\ds d = ax_1+by_1+cz_1$.

The line $\ds l$ with equation $\ds \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ is perpendicular to both of the lines $\ds AB$ and $\ds AC$, so the line $\ds l$ is perpendicular to the plane $\ds ABC$.

Hence any line with directed values $\ds \langle ka, kb, kc \rangle$, for some parameter $\ds k$, is perpendicular to the plane $\ds ABC$.

Example 8.

Find the equation of the plane containing $\ds A(1,0,1)$, $\ds B(3,6,4)$ and $\ds C(-2,3,1)$.

Solution

$\ds A(1,0,1)$, $\ds B(3,6,4)$, $\ds C(-2,3,1)$

\displaystyle \begin{aligned} \langle AB \rangle &= \langle 2,6,3 \rangle \\ \langle AC \rangle &= \langle -3,3,0 \rangle \end{aligned}

\displaystyle \begin{aligned} 2a + 6b + 3c &= 0 \\ -3a + 3b + 0c &= 0 \end{aligned}

Using cross multiplication method:

\displaystyle \begin{aligned} a &= 6(0) - 3(3) = -9 \\ b &= 3(-3) - 0(2) = -9 \\ c &= 2(3) - (-3)(6) = 24 \\ d &= ax_1+by_1+cz_1 \\ &= -9(1) + (-9)(0) + 24(1) \\ &= 15 \end{aligned}

The equation of the plane

\displaystyle \begin{aligned} -9x - 9y + 24z &= 15 \\ 3x + 3y - 8z &= -5 \end{aligned}

Example 9.

Find the equation of the line that passes through the point $\ds (-1,3,2)$ and perpendicular to the plane $\ds 3x-2y-z=3$. Find the point of intersection of the line and the given plane.

Solution

$\ds \langle a,b,c \rangle = \langle 3,-2,-1 \rangle$.
The equation of the line is

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

\displaystyle \begin{aligned} \frac{x+1}{3} &= \frac{y-3}{-2} = \frac{z-2}{-1} = k \quad \text{(say)} \end{aligned}

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

\displaystyle \begin{aligned} 3x-2y-z &= 3 \\ 3(3k-1) - 2(-2k+3) - (-k+2) &= 3 \\ 9k-3 + 4k-6 + k-2 &= 3 \\ 14k &= 14 \\ k &= 1 \end{aligned}

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

Hence the point of intersection is $\ds (2, 1, 1)$.

Example 10.

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

Solution

$\ds \langle a,b,c \rangle = \langle 3,-2,-3 \rangle$
Thus the equation of the required plane is

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

The point $\ds (-1,3,2)$ lies on the required plane.

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

The equation of the required plane is

\displaystyle \begin{aligned} 3x-2y-3z &= -15 \end{aligned}

Exercise 3.4

Find the equation of the plane containing

(a) $\ds A(2,-5,4)$, $\ds B(-5,2,4)$ and $\ds C(-2,3,-1)$

(b) $\ds A(4,2,-3)$, $\ds B(1,-2,4)$ and $\ds C(-1,0,3)$.
Find the equation of the line passing through the point $\ds (3,-2,-2)$ and perpendicular to the plane $\ds -2x+3y-z=4$. Find the point of intersection of the line and the plane.
Find the equation of the plane containing the point $\ds (2, 3, -1)$ and parallel to the plane $\ds -2x+y+3z=6$.

3.5 Sphere

The distance between center $\ds (x_1,y_1,z_1)$ and any point $\ds (x,y,z)$ of a sphere is radius $\ds r$.

\displaystyle \begin{aligned} r &= \sqrt{(x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2} \end{aligned}

The equation of the sphere with center $\ds (x_1,y_1,z_1)$ and radius $\ds r$ is

\displaystyle \begin{aligned} (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 &= r^2 \end{aligned}

Example 11

Find the equation of the plane tangent to the sphere $\ds (x-2)^2+(y-1)^2+(z+1)^2=14$ at the point $\ds (3,4,1)$.

Solution

$\ds \langle CP \rangle = \langle 1,3,2 \rangle$

The equation of the plane is

$\ds x+3y+2z = d$

Since $\ds P(3,4,1)$ is on the plane, so we get

\displaystyle \begin{aligned} 3+3(4)+2(1) &= d \\ d &= 17 \end{aligned}

The equation of the plane is $\ds x+3y+2z=17$.

Example 12

Find the equation of the sphere with center $\ds (0, 1, 0)$ and touching the plane $\ds x-2y+2z+5=0$.

Solution

The equation of the line that passes through the center $\ds C(0,1,0)$ and perpendicular to the plane $\ds x-2y+2z+5=0$ is

\displaystyle \begin{aligned} \frac{x-0}{1} &= \frac{y-1}{-2} = \frac{z-0}{2} \end{aligned}

\displaystyle \begin{aligned} \frac{x}{1} &= \frac{y-1}{-2} = \frac{z}{2} = s \quad \text{(say)} \end{aligned}

\displaystyle \begin{aligned} x &= s \\ y &= -2s+1 \\ z &= 2s \end{aligned}

If one of these points $\ds P$ is on the plane, then

\displaystyle \begin{aligned} s - 2(-2s+1) + 2(2s) + 5 &= 0 \\ s - 2 + 4s + 4s + 5 &= 0 \\ 9s &= -3 \\ s &= -\frac{1}{3} \end{aligned}

\displaystyle \begin{aligned} x &= -\frac{1}{3} \\ y &= -2\left(-\frac{1}{3}\right)+1 = \frac{5}{3} \\ z &= 2\left(-\frac{1}{3}\right) = -\frac{2}{3} \end{aligned}

The point of intersection $\ds = \left(-\frac{1}{3},\frac{5}{3},-\frac{2}{3}\right)$.

Radius

\displaystyle \begin{aligned} CP &= \sqrt{\left(-\frac{1}{3}-0\right)^2 + \left(\frac{5}{3}-1\right)^2 + \left(-\frac{2}{3}-0\right)^2} \\ &= \sqrt{\frac{1}{9}+\frac{4}{9}+\frac{4}{9}} \\ &= 1 \end{aligned}

The equation of the sphere is

\displaystyle \begin{aligned} (x-0)^2 + (y-1)^2 + (z-0)^2 &= 1 \\ x^2 + (y-1)^2 + z^2 &= 1 \end{aligned}

Example 13

Find the equation of a sphere that passes through the points $\ds (9, 0, 0)$, $\ds (3, 13, 5)$ and $\ds (11,0, 10)$, given that its center lies on the yz-plane.

Solution

The equation of the sphere with center $\ds (x_1,y_1,z_1)$ and radius $\ds r$ is

\displaystyle \begin{aligned} (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 &= r^2 \end{aligned}

$\ds (x-0)^2 + (y-y_1)^2 + (z-z_1)^2 = r^2$, since its center lies on the yz-plane.

At $\ds (9,0,0)$,

\displaystyle \begin{aligned} (9-0)^2 + (0-y_1)^2 + (0-z_1)^2 &= r^2 \\ 81 + y_1^2 + z_1^2 &= r^2 \quad \text{(1)} \end{aligned}

At $\ds (3, 13, 5)$,

\displaystyle \begin{aligned} (3-0)^2 + (13-y_1)^2 + (5-z_1)^2 &= r^2 \\ 9 + (13-y_1)^2 + (5-z_1)^2 &= r^2 \quad \text{(2)} \end{aligned}

At $\ds (11, 0, 10)$,

\displaystyle \begin{aligned} (11-0)^2 + (0-y_1)^2 + (10-z_1)^2 &= r^2 \\ 121 + y_1^2 + (10-z_1)^2 &= r^2 \quad \text{(3)} \end{aligned}

From (1) and (3),

\displaystyle \begin{aligned} 81 + y_1^2 + z_1^2 &= 121 + y_1^2 + (10-z_1)^2 \\ 81 + z_1^2 &= 121 + 100 - 20z_1 + z_1^2 \\ 20z_1 &= 140 \\ z_1 &= 7 \end{aligned}

From (1) and (2),

\displaystyle \begin{aligned} 81 + y_1^2 + z_1^2 &= 9 + (13-y_1)^2 + (5-z_1)^2 \\ 81 + y_1^2 + 7^2 &= 9 + 169 - 26y_1 + y_1^2 + (5-7)^2 \\ 81 + 49 &= 178 - 26y_1 + 4 \\ 130 &= 182 - 26y_1 \\ 26y_1 &= 52 \\ y_1 &= 2 \end{aligned}

Center $\ds = (0,2,7)$

Substitute $\ds x_1=0$, $\ds y_1=2$ and $\ds z_1=7$ in equation (1),

\displaystyle \begin{aligned} 81 + (2)^2 + 7^2 &= r^2 \\ 81 + 4 + 49 &= r^2 \\ r^2 &= 134 \end{aligned}

Hence the equation of the sphere is

\displaystyle \begin{aligned} x^2 + (y-2)^2 + (z-7)^2 &= 134 \end{aligned}

Exercise 3.5

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) $\displaystyle C(2, 3, 5)$, $\displaystyle r = 5$
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 $\displaystyle P(10, 10, 10)$.
Find the equation of the sphere on the join of $\displaystyle (1, -1, 1)$ and $\displaystyle (-3, 4, 5)$ as diameter.
Find the equation of the plane tangent to the sphere $\displaystyle (x + 2)^2 + (y - 1)^2 + (z + 3)^2 = 27$ at the point $\displaystyle (3, 2, -2)$.
Find the equation of the sphere with center $\displaystyle (6, -7, -3)$ and touching the plane $\displaystyle 4x - 2y - z = 17$.
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 $\displaystyle 2x + 3y + 4z = 6$?