Let a= i+ j- k, b=2 i+ j- k, 0.

Question

Let $\vec a=\alpha\hat i+\hat j-\hat k$, $\vec b=2\hat i+\hat j-\alpha\hat k$, $\alpha>0$. If the projection of $\vec a	imes\vec b$ on $-\hat i+2\hat j-2\hat k$ is 30, then $\alpha$ is:

Accepted Answer

$$\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\ \alpha&1&-1\2&1&-\alpha\end{vmatrix} = (1-\alpha,\ \alpha^2-2,\ \alpha-2)$$

Projection on $(-1,2,-2)$, magnitude 3:
$$\frac{-(1-\alpha)+2(\alpha^2-2)-2(\alpha-2)}3 = 30 \implies 2\alpha^2-\alpha-1=90 \implies 2\alpha^2-\alpha-91=0$$
$$\alpha = \frac{1\pm27}4 \implies \alpha=7\text{ or }-6.5$$

Since $\alpha>0$: $\alpha=7$.

Solutions · 1