Chapter 2, Problem 2.1P

Given P(4, 2, 1) and ${\text{A}}_{\text{PQ}}{\text{=2a}}_{\text{x}}{\text{+4a}}_{\text{y}}{\text{+6a}}_{\text{z}}$, find the point Q.

To determine

The coordinates ofpoint $Q$ .

Answer to Problem 2.1P

The coordinates ofpoint $Q$ is $\left(6,6,7\right)$ .

Explanation of Solution

Given:

The coordinates of point $P$ is $\left(4,2,1\right)$ .

Vector $A_{\text{PQ}}$ is $2a_x+4a_y+6a_z$ .

Calculation:

The point $Q$ is $\left(Q_x,Q_y,Q_z\right)$ .

The vector $A_{\text{PQ}}$ can be written as,

  $A_{\text{PQ}}=\left(Q_x-P_x\right)a_x+\left(Q_y-P_y\right)a_y+\left(Q_z-P_z\right)a_z$

Compare the given vector $A_{\text{PQ}}$ with the above vector.

  $\begin{array}{c}{Q}_x-P_x=2\\ Q_x-4=2\\ Q_x=6\end{array}$

The value of $Q_y$ is calculated as,

  $\begin{array}{c}{Q}_y-P_y=4\\ Q_y-2=4\\ Q_y=6\end{array}$

The value of $Q_z$ is calculated as,

  $\begin{array}{c}{Q}_z-P_z=6\\ Q_z-1=6\\ Q_z=7\end{array}$

Therefore, the point $Q$ is $\left(6,6,7\right)$ .