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
Answer & explanation
Correct answer: option 2
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 $= \frac57$
$=\frac{10}{49}\hat i+\frac{30}{49}\hat j+\frac{15}{49}\hat k$