Question

Find the projection of $\overrightarrow{A}=10\hat{i}+8\hat{j}-6\hat{k}$ along $\overrightarrow{r}=5\hat{i}+6\hat{j}+9\hat{k}$

• A. $44\sqrt{\frac{2}{71}}$
• B. $22\sqrt{\frac{2}{71}}$
• C. $2.2\sqrt{2}$
• D. $22\sqrt{2}$

Answer 1

The correct option is B $22\sqrt{\frac{2}{71}}$

It is given that, $\overrightarrow{A}=10\hat{i}+8\hat{j}-6\hat{k}$
$\overrightarrow{r}=5\hat{i}+6\hat{j}+9\hat{k}$
projection of $\overline{A}$ along $\overline{r}$ is $A\cos \theta$
$\Rightarrow A\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{r}}{\left|r\right|}=\frac{(10\hat{i}+8\hat{j}-6\hat{k}).(5\hat{i}+6\hat{j}+9\hat{k})}{\sqrt{25+36+81}}$

(Where $\theta$ is the angle between the two vectors)

$=\frac{50+48-54}{\sqrt{142}}=\frac{44}{\sqrt{142}}$

$=\frac{22\times 2}{\sqrt{71\times 2}}=22\sqrt{\frac{2}{71}}\text{units}$

Hence option B is the correct answer