Chapter 1, Problem 10P

(a)

To determine

Determine $\left|P+Q-R\right|$.

Explanation of Solution

Given:

$\begin{array}{l}P=2a_x-a_y-2a_z\\ Q=4a_x+3a_y+2a_z\\ R=-a_x+a_y+2a_z\end{array}$

Calculation:

Determine the vector term of $\left(P+Q-R\right)$.

  $\begin{array}{l}P+Q-R=\left(2a_x-a_y-2a_z\right)+\left(4a_x+3a_y+2a_z\right)-\left(-a_x+a_y+2a_z\right)\\ P+Q-R=\left(7a_x+a_y-2a_z\right)\end{array}$

Determine the magnitude of $\left(P+Q-R\right)$.

  $\begin{array}{l}\left|P+Q-R\right|=\sqrt{7^2+1^2+{\left(-2\right)}^2}\\ \left|P+Q-R\right|=\sqrt{54}\\ \left|P+Q-R\right|=7.3485\end{array}$

Thus, the magnitude of $\left(P+Q-R\right)$ is $7.3485$.

(b)

To determine

Determine $P\cdot \left(Q\times R\right)$.

Explanation of Solution

Determine the cross product of Q and R.

  $\begin{array}{l}Q\times R=\left(4a_x+3a_y+2a_z\right)\times \left(-a_x+a_y+2a_z\right)\\ Q\times R=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 4& 3& 2\\ -1& 1& 2\end{array}\right|\\ Q\times R=a_x\left(6-2\right)-a_y\left(8+2\right)+a_z\left(4+3\right)\\ Q\times R=4a_x-10a_y+7a_z\end{array}$

Determine the dot product of P and $\left(Q\times R\right)$.

  $\begin{array}{l}P\cdot \left(Q\times R\right)=\left(2a_x-a_y-2a_z\right)\cdot \left(4a_x-10a_y+7a_z\right)\\ P\cdot \left(Q\times R\right)=8+10-14\\ P\cdot \left(Q\times R\right)=14\end{array}$

Thus, the dot product of P and $\left(Q\times R\right)$ is $14$.

(c)

To determine

Determine $\left(Q\times P\right)\cdot R$.

Explanation of Solution

Determine the cross product of Q and P.

  $\begin{array}{l}Q\times P=\left(4a_x+3a_y+2a_z\right)\times \left(2a_x-a_y-2a_z\right)\\ Q\times P=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 4& 3& 2\\ 2& -1& -2\end{array}\right|\\ Q\times P=a_x\left(-6+2\right)-a_y\left(-8-4\right)+a_z\left(-4-6\right)\\ Q\times P=-4a_x+12a_y-10a_z\end{array}$

Determine the dot product of $Q\times P$ and R.

  $\begin{array}{l}\left(Q\times P\right)\cdot R=\left(-4a_x+12a_y-10a_z\right)\cdot \left(-a_x+a_y+2a_z\right)\\ \left(Q\times P\right)\cdot R=4+12-20\\ \left(Q\times P\right)\cdot R=-4\end{array}$

Thus, the dot product of $\left(Q\times P\right)$ and R is $-4$.

(d)

To determine

Determine $\left(P\times Q\right)\cdot \left(Q\times R\right)$.

Explanation of Solution

Determine the cross product of P and Q.

  $\begin{array}{l}P\times Q=\left(2a_x-a_y-2a_z\right)\times \left(4a_x+3a_y+2a_z\right)\\ P\times Q=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 2& -1& -2\\ 4& 3& 2\end{array}\right|\\ P\times Q=a_x\left(-2+6\right)-a_y\left(4+8\right)+a_z\left(6+4\right)\\ P\times Q=4a_x-12a_y+10a_z\end{array}$

Determine the dot product of $P\times Q$ and $Q\times R$.

  $\begin{array}{l}\left(P\times Q\right)\cdot \left(Q\times R\right)=\left(4a_x-12a_y+10a_z\right)\cdot \left(4a_x-10a_y+7a_z\right)\\ \left(P\times Q\right)\cdot \left(Q\times R\right)=16+120+70\\ \left(P\times Q\right)\cdot \left(Q\times R\right)=206\end{array}$

Thus, the dot product of $P\times Q$ and $Q\times R$ is $206$.

(e)

To determine

Determine $\left(P\times Q\right)\times \left(Q\times R\right)$.

Explanation of Solution

Determine the cross product of $P\times Q$ and $Q\times R$.

  $\begin{array}{l}\left(P\times Q\right)\times \left(Q\times R\right)=\left(4a_x-12a_y+10a_z\right)\cdot \left(4a_x-10a_y+7a_z\right)\\ \left(P\times Q\right)\times \left(Q\times R\right)=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ 4& -12& 10\\ 4& -10& 7\end{array}\right|\\ \left(P\times Q\right)\times \left(Q\times R\right)=a_x\left(-84+100\right)-a_y\left(28-40\right)+a_z\left(-40+48\right)\\ \left(P\times Q\right)\times \left(Q\times R\right)=16a_x+12a_y+8a_z\end{array}$

Thus, the cross product of $P\times Q$ and $Q\times R$ is $16a_x+12a_y+8a_z$.

(f)

To determine

Determine $\cos {\theta }_{PR}$.

Explanation of Solution

Determine the dot product of $P$ and $R$.

  $\begin{array}{l}P\cdot R=\left(2a_x-a_y-2a_z\right)\cdot \left(-a_x+a_y+2a_z\right)\\ P\cdot R=-2-1-4\\ P\cdot R=-7\end{array}$

Determine the magnitude of P.

  $\begin{array}{l}\left|P\right|=\sqrt{2^2+{\left(-1\right)}^2+{\left(-2\right)}^2}\\ \left|P\right|=\sqrt{9}\\ \left|P\right|=3\end{array}$

Determine the magnitude of R.

  $\begin{array}{l}\left|R\right|=\sqrt{{\left(-1\right)}^2+1^2+2^2}\\ \left|R\right|=\sqrt{6}\end{array}$

Write the expression of dot product of P and R.

  $\begin{array}{l}P\cdot R=\left|P\right|\left|R\right|\cos {\theta }_{PR}\\ -7=\left(3\right)\left(\sqrt{6}\right)\cos {\theta }_{PR}\\ \cos {\theta }_{PR}=-0.9526\end{array}$

Thus, the value of $\cos {\theta }_{PR}$ is $-0.9526$.

(g)

To determine

Determine $\sin {\theta }_{PQ}$.

Explanation of Solution

Determine the magnitude of $P\times Q$.

  $\begin{array}{l}\left|P\times Q\right|=\sqrt{4^2+{\left(-12\right)}^2+{10}^2}\\ \left|P\times Q\right|=\sqrt{260}\end{array}$

Determine the magnitude of Q.

  $\begin{array}{l}\left|Q\right|=\sqrt{4^2+3^2+2^2}\\ \left|Q\right|=\sqrt{29}\end{array}$

Write the expression of magnitude of $P\times Q$.

  $\begin{array}{l}\left|P\times Q\right|=\left|P\right|\left|Q\right|\sin {\theta }_{PQ}\\ \sqrt{260}=\left(3\right)\left(\sqrt{29}\right)\sin {\theta }_{PQ}\\ \sin {\theta }_{PQ}=0.9981\end{array}$

Thus, the value of $\sin {\theta }_{PQ}$ is $0.9981$.