Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Vectors

Question:

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

Options:

$\frac{1}{24}$

$\frac{5}{72}$

$\frac{7}{72}$

$\frac{7}{48}$

Correct Answer:

$\frac{7}{72}$

Explanation:

We know that

$\left\{\frac{\vec a}{|\vec a|^2}-\frac{\vec b}{|\vec b|^2}\right\}^2=\left\{\frac{\vec a-\vec b}{|\vec a||\vec b|}\right\}$

$∴\left\{\frac{\vec a}{|\vec a|^2}-\frac{\vec b}{|\vec b|^2}\right\}^2=\frac{|\vec a-\vec b|^2}{|\vec a|^2|\vec b|^2}=\frac{1+4+9}{9×16}=\frac{14}{9×16}=\frac{7}{72}$