Chapter 9, Problem 10T

1. (a) Find a vector perpendicular to the plane that contains the points P(1, 0, 0), Q(2, 0, −1), and R(1, 4, 3).
2. (b) Find an equation for the plane that contains P, Q, and R.
3. (c) Find the area of triangle PQR.

To determine

To find: The vector that is perpendicular to the plane passing through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

Answer to Problem 10T

The vector that is perpendicular to the plane passing through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$ is $4i-3j+4k$.

Explanation of Solution

Given:

The three points are $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

Formula used: The vector passing through two points $A\left(a_1,a_2,a_3\right)$ and $B\left(b_1,b_2,b_3\right)$ is given below,

$\overrightarrow{AB}=\left(b_1-a_1\right)i+\left(b_2-a_2\right)j+\left(b_3-a_3\right)k$ (1)

The Cross Product Formula,

The vectors $u=\langle a_1,a_2,a_3\rangle$ and $v=\langle b_1,b_2,b_3\rangle$ are three-dimensional vectors, then the cross product of u and v is given below,

$u\times v=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$ (2)

The vector $u\times v$ is perpendicular to both vectors u and v.

Calculation:

The given three points are $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

Substitute 1 for $a_{}$, 0 for $a_2$, 0 for $a_3$, 2 for $b_1$, 0 for $b_2$ and $-1$ for $b_3$ in equation (1) to find the vector $\overrightarrow{PQ}$,

$\begin{array}{c}\overrightarrow{PQ}=\left(2-1\right)i+\left(0-0\right)j+\left(-1-0\right)k\\ =i-k\end{array}$

Substitute 1 for $a_{}$, 0 for $a_2$, 0 for $a_3$, 1 for $b_1$, 4 for $b_2$ and 3 for $b_3$ in equation (1) to find the vector $\overrightarrow{PR}$,

$\begin{array}{c}\overrightarrow{PR}=\left(1-1\right)i+\left(4-0\right)j+\left(3-0\right)k\\ =4j+3k\end{array}$

The vector $u\times v$ is perpendicular to both vectors u and v. Similarly $\overrightarrow{PQ}\times \overrightarrow{PR}$ in perpendicular to both vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$. Therefore $\overrightarrow{PQ}\times \overrightarrow{PR}$ is perpendicular to the plane passing through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

Substitute 1 for $a_{}$, 0 for $a_2$, $-1$ for $a_3$, 0 for $b_1$, 4 for $b_2$ and 3 for $b_3$ in equation (2) to find $\overrightarrow{PQ}\times \overrightarrow{PR}$,

$\begin{array}{c}\overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{ccc}i& j& k\\ 1& 0& -1\\ 0& 4& 3\end{array}\right|\\ =\left|\begin{array}{cc}0& -1\\ 4& 3\end{array}\right|i-\left|\begin{array}{cc}1& -1\\ 0& 3\end{array}\right|j+\left|\begin{array}{cc}1& 0\\ 0& 4\end{array}\right|k\\ =\left(0-\left(-4\right)\right)i-\left(3-0\right)j+\left(4-0\right)k\\ =4i-3j+4k\end{array}$

Thus, the vector that is perpendicular to the plane passing through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$ is $4i-3j+4k$.

To determine

To find: The equation of the plane that passes through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

Answer to Problem 10T

The equation of the plane that passes through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$ is $4x-3y+4z=4$.

Explanation of Solution

Given:

The plane passes through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

From part (a), the vector that is perpendicular to the plane passing through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$ is $4i-3j+4k$.

Formula used: The equation of a plane.

The equation of the plane having the normal vector $n=\langle a,b,c\rangle$ and passes through the point $P\left(x_0,y_0,z_0\right)$ is given below,

$a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0$ (3)

Calculation:

The normal vector for the plane is $4i-3j+4k$.

Substitute 4 for $a$, $-3$ for $b$, 4 for $c$, 1 for $x_0$, 0 for $y_0$ and 0 for $z_0$ in equation (3) to find the equation of the plane that has normal vector $4i-3j+4k$ and passes through the point $P\left(1,0,0\right)$,

$\begin{array}{c}\left(4\right)\left(x-1\right)+\left(-3\right)\left(y-0\right)+\left(4\right)\left(z-0\right)=0\\ 4x-4-3y+4z=0\\ 4x-3y+4z=4\end{array}$

Thus, the equation of the plane that passes through the points $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$ is $4x-3y+4z=4$.

To determine

To find: The area of $\Delta PQR$ with vertices $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

Answer to Problem 10T

The area of $\Delta PQR$ with vertices $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$ is $\frac{\sqrt{41}}{2}\text{sq}\text{.}\text{unit}$.

Explanation of Solution

Given:

The vertices of $\Delta PQR$ are $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$.

From part (a), the value of $\overrightarrow{PQ}\times \overrightarrow{PR}$ is $4i-3j+4k$.

Formula used: The Magnitude Formula for the vector $v=\langle a_1,a_2,a_3\rangle$ is given below,

$\left|v\right|=\sqrt{a_1{}^2+a_2{}^2+a_3{}^2}$ (4)

The area of triangle is half the area of parallelogram determined by the vectors u and v,

$A=\frac{1}{2}\left|u\times v\right|$ (5)

Calculation:

Substitute 4 for $a_1$, $-3$ for $a_2$ and 4 for $a_3$ in equation (4) to find the length of the cross product,

$\begin{array}{c}\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|=\sqrt{{\left(4\right)}^2+{\left(-3\right)}^2+{\left(4\right)}^2}\\ =\sqrt{16+9+16}\\ =\sqrt{41}\end{array}$

From equation (5), the area of $\Delta PQR$ is half the area of parallelogram determine by the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$

$\begin{array}{c}A=\frac{1}{2}\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|\\ =\frac{1}{2}\left(\sqrt{41}\right)\\ =\frac{\sqrt{41}}{2}\end{array}$

Thus, the area of $\Delta PQR$ with vertices $P\left(1,0,0\right)$, $Q\left(2,0,-1\right)$ and $R\left(1,4,3\right)$ is $\frac{\sqrt{41}}{2}\text{sq}\text{.}\text{unit}$.