Chapter 1, Problem 17P

a)

To determine

The distance between the points P and Q.

Explanation of Solution

Given:

$\begin{array}{l}P\left(x_{P,}{y}_P,z_P\right)=\left(-1,4,8\right)\\ Q\left(x_{Q,}{y}_Q,z_Q\right)=\left(2,-1,3\right)\\ R\left(x_{R,}{y}_R,z_R\right)=\left(-1,2,3\right)\end{array}$

Calculation:

Write an expression for the position vector of point P.

  $r_P=x_P{a}_x+y_P{a}_y+z_P{a}_z$

Substitute the respective value in the above equation.

  $r_P=-1a_x+4a_y+8a_z$

Write an expression for the position vector of point Q.

  $r_Q=x_Q{a}_x+y_Q{a}_y+z_Q{a}_z$

Substitute the respective value in the above equation.

  $r_Q=2a_x-1a_y+3a_z$

Write an expression for the distance vector $\left(r_{PQ}\right)$.

  $r_{PQ}=r_Q-r_P$

Substitute the respective values of the variables in above Equation.

  $\begin{array}{c}{r}_{PQ}=\left(2a_x-1a_y+3a_z\right)-\left(\underset{\_}{-1a_x+4a_y+8a_z}.\right)\\ r_{PQ}=\left(2+1\right)a_x+\left(-1-4\right)a_y+\left(3-8\right)a_z\\ r_{PQ}=3a_x-5a_y-5a_z\end{array}$

Calculate the magnitude of the distance vector.

  $\begin{array}{c}\left|r_{PQ}\right|=\sqrt{\left(3^2\right)+\left(-5^2\right)+\left(-5^2\right)}\\ =\sqrt{9+25+25}\\ =\sqrt{59}\\ =7.6811\end{array}$

Thus, the distance between the points P and Q is $\underset{\_}{7.6811}$.

b)

To determine

The distance vector between the points P and R.

Explanation of Solution

Calculation:

Write an expression for the position vector of point R.

  $r_R=x_R{a}_x+y_R{a}_y+z_R{a}_z$

Substitute the respective value in the above equation.

  $r_R=-a_x+2a_y+3a_z$

Write an expression for the distance vector $\left(r_{PR}\right)$.

  $r_{PR}=r_R-r_P$

Substitute the respective values of the variables in above Equation.

  $\begin{array}{c}{r}_{PR}=\left(-a_x+2a_y+3a_z\right)-\left(\underset{\_}{-1a_x+4a_y+8a_z}.\right)\\ r_{PR}=\left(-1+1\right)a_x+\left(2-4\right)a_y+\left(3-8\right)a_z\\ r_{PR}=0a_x-2a_y-5a_z\\ r_{PR}=-2a_y-5a_z\end{array}$

Thus, the distance vector between the points P and R is $\underset{\_}{-2a_y-5a_z}$.

c)

To determine

The angle between QP and QR.

Explanation of Solution

Calculation:

Write an expression for the distance vector $\left(r_{QP}\right)$.

  $r_{QP}=r_P-r_Q$

Substitute the respective values of the variables in above Equation.

  $\begin{array}{c}{r}_{QP}=\left(-1a_x+4a_y+8a_z\right)-\left(2a_x-1a_y+3a_z\right)\\ r_{QP}=\left(-1-2\right)a_x+\left(4+1\right)a_y+\left(8-3\right)a_z\\ r_{QP}=-3a_x+5a_y+5a_z\end{array}$

Calculate the magnitude of the distance vector $r_{PR}$.

  $\begin{array}{c}\left|r_{QP}\right|=\sqrt{\left(-3^2\right)+\left(5^2\right)+\left(5^2\right)}\\ =\sqrt{9+25+25}\\ =\sqrt{59}\\ =7.6811\end{array}$

Write an expression for the distance vector $\left(r_{QP}\right)$.

  $r_{QR}=r_R-r_Q$

Substitute the respective values of the variables in above Equation.

  $\begin{array}{c}{r}_{QR}=\left(-a_x+2a_y+3a_z\right)-\left(2a_x-1a_y+3a_z\right)\\ r_{QR}=\left(-1-2\right)a_x+\left(2+1\right)a_y+\left(3-3\right)a_z\\ r_{QR}=-3a_x+3a_y+0a_z\end{array}$

Calculate the magnitude of the distance vector $r_{PR}$.

  $\begin{array}{c}\left|r_{QR}\right|=\sqrt{\left(-3^2\right)+\left(3^2\right)+\left(0^2\right)}\\ =\sqrt{9+9+0}\\ =\sqrt{18}\\ =\text{4}\text{.2426}\end{array}$

Calculate the dot product of $\left(r_{QP}\cdot r_{QR}\right)$.

  $\begin{array}{c}{r}_{QP}\cdot r_{QR}=\left(-3a_x+5a_y+5a_z\right)\cdot \left(-3a_x+3a_y+0a_z\right)\\ =\left(9+15+0\right)\\ =24\end{array}$

Write the expression for the dot product of $r_{QP}\cdot r_{QR}$.

  $r_{QP}\cdot r_{QR}=\left|r_{QP}\right|\times \left|r_{QP}\right|\times \cos {\theta }_{AB}$ (I).

Rearrange the equation (I).

  ${\theta }_{AB}={\cos }^{-1}\left[\frac{r_{QP}\cdot r_{QR}}{\left|r_{QP}\right|\times \left|r_{QP}\right|}\right]$

Substitute the respective values of the variables in above Equation.

  $\begin{array}{c}{\theta }_{AB}={\cos }^{-1}\left[\frac{24}{\sqrt{59}\times \sqrt{18}}\right]\\ ={\cos }^{-1}\left(0.73645\right)\\ =42.57°\end{array}$

Thus, the angle between vector QP and QR is $\left({\theta }_{AB}\right)$ is $\underset{\_}{42.57°}$.

d)

To determine

The area of triangle PQR.

Explanation of Solution

Calculation:

Write the expression for the area of the triangle PQR.

$\begin{array}{c}{A}_{PQR}=\left|\frac{1}{2}{r}_{PQ}\times r_{QR}\right|\\ =\left|\frac{1}{2}\left(3a_x-5a_y-5a_z\right)\times \left(-3a_x+3a_y+0a_z\right)\right|\\ =\frac{1}{2}\times \sqrt{{15}^2+{15}^2+{\left(-6\right)}^2}\\ =11.023\end{array}$

Thus, the area of triangle PQR is $\underset{\_}{11.023}$.

e)

To determine

The perimeter of triangle PQR.

Explanation of Solution

Calculation:

Write an expression for the distance vector $\left(r_{RP}\right)$.

  $r_{RP}=r_P-r_R$

Substitute the respective values of the variables in above Equation.

  $\begin{array}{c}{r}_{RP}=\left(-1a_x+4a_y+8a_z\right)-\left(-1a_x+2a_y+3a_z\right)\\ r_{RP}=\left(-1+1\right)a_x+\left(4-2\right)a_y+\left(8-3\right)a_z\\ r_{RP}=0a_x+2a_y+5a_z\\ r_{RP}=2a_y+5a_z\end{array}$

Calculate the magnitude of the vector $r_{RP}$.

  $\begin{array}{c}\left|r_{RP}\right|=\sqrt{0^2+2^2+5^2}\\ =\sqrt{29}\\ =5.3852\end{array}$

Perimeter of triangle is the sum of the sides of the triangle.

  $P=r_{PQ}+r_{QR}+r_{RP}$

Substitute the respective values in the above equation.

  $\begin{array}{c}P=7.6811+4.2426+5.3852\\ =17.309\end{array}$

Thus, the perimeter of triangle PQR is $\underset{\_}{17.309}$.