The projection of the vector $\vec a =\hat i+2\hat j-3\hat k$ on the vector $2\hat i +6\hat j+3\hat k$ is

$\frac{5}{7}$

The correct answer is Option (2) → $\frac{5}{7}$

$\vec a=\hat i+2\hat j-3\hat k$

$\vec b=2\hat i+6\hat j+3\hat k$

Projection of $\vec a$ on $\vec b$ is

$\frac{\vec a\cdot\vec b}{|\vec b|^2}\vec b$

$\vec a\cdot\vec b=1\times2+2\times6+(-3)\times3$

$=2+12-9$

$=5$

$|\vec b|^2=2^2+6^2+3^2$

$=4+36+9$

$=49$

Projection $=\frac{5}{49}(2\hat i+6\hat j+3\hat k)$

Projection Length $= \frac57$

$=\frac{10}{49}\hat i+\frac{30}{49}\hat j+\frac{15}{49}\hat k$