If $\vec a$ and $\vec b$ are to non-collinear vectors such that $|\vec a|=3,|\vec b|=4$ and $\vec a-\vec b=\hat i+2\hat j+3\hat k$, then the value of $\left\{\frac{\vec a}{|\vec a|^2}-\frac{\vec b}{|\vec b|^2}\right\}^2$ is equal to
Answer & explanation
Correct answer: option 3
We know that
$\left\{\frac{\vec a}{|\vec a|^2}-\frac{\vec b}{|\vec b|^2}\right\}^2=\left\{\frac{\vec a-\vec b}{|\vec a||\vec b|}\right\}^2$
$∴\left\{\frac{\vec a}{|\vec a|^2}-\frac{\vec b}{|\vec b|^2}\right\}^2=\frac{|\vec a-\vec b|^2}{|\vec a|^2|\vec b|^2}=\frac{1+4+9}{9×16}=\frac{14}{9×16}=\frac{7}{72}$