Question

If the projection of the vector $\hat{ i }+2 \hat{ j }+\hat{ k }$ on the sum of the two vectors $2 \hat{ i }+4 \hat{ j }-5 \hat{ k }$ and $-\lambda \hat{ i }+2 \hat{ j }+3 \hat{ k }$ is $1$ , then $\lambda$ is equal to ___

Answer 1

$\vec{ a }=\hat{ i }+2 \hat{ j }+\hat{ k }$
$\vec{ b }=(2-\lambda) \hat{ i }+6 \hat{ j }-2 \hat{ k }$
$\frac{\vec{ a } \cdot \vec{ b }}{|\vec{ b }|}=1, \vec{ a } \cdot \vec{ b }=12-\lambda$
$(\vec{ a } \cdot \vec{ b })=|\vec{ b }|^{2}$
$\lambda^{2}-24 \lambda+144$
$=\lambda^{2}-4 \lambda+4+40$
$20 \lambda=100 $
$\Rightarrow \lambda=5$