Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three mutually perpendicular vectors of equal magnitudes, then the vector equally inclined to $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ is

• A. $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$
• B. $\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}$
• C. $\frac{\overrightarrow{a}-\overrightarrow{b}-\overrightarrow{c}}{\sqrt{3}}$
• D. $\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{\sqrt{3}}$

Answer 1

Answer: (D)

$\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{\sqrt{3}}$

From the condition given

$\overline{a}-\overline{b}=\overline{b}-\overline{c}=\overline{c}-\overline{a}=0$
$|\overline{a}|=|\overline{b}|=|\overline{c}|$ -----(1)
Let $\overline{V}$ be that Vetor
$\overline{V}-\overline{a}=|\overline{V}||\overline{a}|\cos {\theta }_1$
$\overline{V}-\overline{b}=|\overline{V}||\overline{b}|\cos {\theta }_2$
$\overline{V}-\overline{c}=|\overline{V}||\overline{c}|\cos {\theta }_3$
${\theta }_1={\theta }_2={\theta }_3$ (Vetor equally inlined)
$\overline{V}-\overline{a}=\overline{V}-\overline{b}=\overline{V}-\overline{c}$ -----(2)
So, $\overline{V}=\frac{\overline{a}+\overline{b}+\overline{c}}{\sqrt{3}}$