Let a= i+ j+3 k, b=- i- j- k, c= i-2 j- k, with a b=1 and b c=-3.

Question

Let $\vec a=\alpha\hat i+\beta\hat j+3\hat k$, $\vec b=-\beta\hat i-\alpha\hat j-\hat k$, $\vec c=\hat i-2\hat j-\hat k$, with $\vec a\cdot\vec b=1$ and $\vec b\cdot\vec c=-3$. Then $\dfrac13\big[(\vec a	imes\vec b)\cdot\vec c\big]$ is:

Answer: 2

Accepted Answer

$\vec a\cdot\vec b = -\alpha\beta-\alpha\beta-3 = -2\alpha\beta-3=1 \Rightarrow \alpha\beta=-2$.

$\vec b\cdot\vec c = -\beta+2\alpha+1=-3 \Rightarrow \beta-2\alpha=4$. Substituting $\alpha=-2/\beta$: $\beta+\dfrac4\beta=4 \Rightarrow \beta^2-4\beta+4=0 \Rightarrow \beta=2,\ \alpha=-1$.

$\vec a=(-1,2,3)$, $\vec b=(-2,1,-1)$, $\vec c=(1,-2,-1)$.

$$\vec a\times\vec b = \begin{vmatrix}\hat i&\hat j&\hat k\-1&2&3\-2&1&-1\end{vmatrix} = (-5,-7,3)$$
$$(\vec a\times\vec b)\cdot\vec c = -5+14-3=6 \implies \frac13(6)=2$$

Solutions · 1