Question

Find a vector of magnitude 11 in the direction opposite to that of ${\displaystyle \overrightarrow{\text{PQ}}}$ where P and Q are the points (1, 3, 2) and (–1, 0, 8), respectively.

Answer 1

The vector with initial point P (1, 3, 2) and terminal point Q (–1, 0, 8) is given by

${\displaystyle \overrightarrow{\text{PQ}}=(-1-1)\hat{\text{i}}+(0-3)\hat{\text{j}}+(8-2)\hat{\text{k}}}$

= ${\displaystyle -2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}$

Thus ${\displaystyle \overrightarrow{\text{OP}}=-\overrightarrow{\text{PQ}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}}$

⇒ ${\displaystyle \left|\overrightarrow{\text{OP}}\right|=\sqrt{2^2+3^2+{(-6)}^2}}$

= ${\displaystyle \sqrt{4+9+36}}$

= ${\displaystyle \sqrt{49}}$

= 7

Therefore, unit vector in the direction of ${\displaystyle \overrightarrow{\text{OP}}}$ is given by

${\displaystyle \widehat{\text{OP}}=\frac{\overrightarrow{\text{OP}}}{\left|\overrightarrow{\text{OP}}\right|}}$

= ${\displaystyle \frac{2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}}{7}}$

Hence, the required vector of magnitude 11 in direction of ${\displaystyle \overrightarrow{\text{OP}}}$ is 11

${\displaystyle \widehat{\text{OP}}=11\left(\frac{2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}}{7}\right)}$

= ${\displaystyle \frac{22}{7}\hat{\text{i}}+\frac{33}{7}\hat{\text{j}}-\frac{66}{7}\hat{\text{k}}}$.