Let a= i+3 j- k, b=3 i- j+4 k, c= i+2 j-2 k.

Question

Let $\vec a=\alpha\hat i+3\hat j-\hat k$, $\vec b=3\hat i-\beta\hat j+4\hat k$, $\vec c=\hat i+2\hat j-2\hat k$. If the projection of $\vec a$ on $\vec c$ is $\dfrac{10}3$, and $\vec b	imes\vec c=-6\hat i+10\hat j+7\hat k$, then $\alpha+\beta$ equals:

Accepted Answer

Projection: $\dfrac{\vec a\cdot\vec c}{|\vec c|} = \dfrac{\alpha+6+2}3=\dfrac{10}3 \Rightarrow \alpha=2$.

$$\vec b\times\vec c = \begin{vmatrix}\hat i&\hat j&\hat k\3&-\beta&4\1&2&-2\end{vmatrix} = (2\beta-8)\hat i+10\hat j+(6+\beta)\hat k = (-6,10,7)$$

$2\beta-8=-6 \Rightarrow \beta=1$ (and $6+\beta=7$ confirms).

$$\alpha+\beta = 2+1=3$$

Solutions · 1