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

Question

Let $\lambda\in\mathbb R$, $\vec a=\lambda\hat i+2\hat j-3\hat k$, $\vec b=\hat i-\lambda\hat j+2\hat k$. If $8(\vec a	imes\vec b) = 8\hat i-40\hat j-24\hat k$, then $|(\vec a+\vec b)	imes(\vec a-\vec b)|^2$ is:

Accepted Answer

$$\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\ \lambda&2&-3\1&-\lambda&2\end{vmatrix} = (4-3\lambda)\hat i-(2\lambda+3)\hat j-(\lambda^2+2)\hat k$$

Matching with $\hat i-5\hat j-3\hat k$ (from $8(\vec a\times\vec b)=8\hat i-40\hat j-24\hat k$): $4-3\lambda=1\Rightarrow\lambda=1$ (checks with the other components).

So $\vec a=\hat i+2\hat j-3\hat k$, $\vec b=\hat i-\hat j+2\hat k$. $\vec a+\vec b=2\hat i+\hat j-\hat k$, $\vec a-\vec b=3\hat j-5\hat k$.

$$(\vec a+\vec b)\times(\vec a-\vec b) = \begin{vmatrix}\hat i&\hat j&\hat k\2&1&-1\0&3&-5\end{vmatrix} = (-5+3)\hat i-(-10)\hat j+6\hat k = -2\hat i+10\hat j+6\hat k$$

$$|\cdots|^2 = 4+100+36 = 140$$

Solutions · 1