Question

lf position vector is given by $\overrightarrow{R}=(-6,-4,-12)$ then the unit vector parallel to $\overrightarrow{R}$ is:

• A. $+\frac{1}{7}(3\hat{i}+2\hat{j}+6\hat{k})$
• B. $-\frac{1}{7}[3\hat{i}+2\hat{j}+6\hat{k}]$
• C. $\pm \frac{1}{7}[3\hat{i}+2\hat{j}-6\hat{k}]$
• D. $\frac{14}{[6\hat{i}+4\hat{j}+12\hat{k}]}$

Answer 1

Answer: (B)

$-\frac{1}{7}[3\hat{i}+2\hat{j}+6\hat{k}]$

A vector divided by its magnitude gives us the unit vector.
Magnitude is given by $\sqrt{a^2+b^2+c^2}$ where the vector is $a\hat{i}+b\hat{j}+c\hat{k}$.
Magnitude of $\overrightarrow{R}=\sqrt{36+16+144}=14$.
Therefore unit vector $\frac{\overrightarrow{R}}{14}=\frac{-(3\hat{i}+2\hat{j}+6\hat{k})}{7}$.