Let a be coplanar with b=2 i+ j+ k, c= i- j+ k, perpendicular to d=3 i+2 j+6 k, with | a|=…

Question

Let $\vec a$ be coplanar with $\vec b=2\hat i+\hat j+\hat k$, $\vec c=\hat i-\hat j+\hat k$, perpendicular to $\vec d=3\hat i+2\hat j+6\hat k$, with $|\vec a|=\sqrt{10}$. A possible value of $[\vec a\ \vec b\ \vec c]+[\vec a\ \vec b\ \vec d]+[\vec a\ \vec c\ \vec d]$ is:

Accepted Answer

$\vec a=\lambda\vec b+\mu\vec c = (2\lambda+\mu)\hat i+(\lambda-\mu)\hat j+(\lambda+\mu)\hat k$.

$\vec a\cdot\vec d=0$: $3(2\lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)=0 \implies 14\lambda+7\mu=0 \implies \mu=-2\lambda$

So $\vec a=(0,-3\lambda,-\lambda)$. $|\vec a|=\sqrt{10}\lambda=\sqrt{10}\Rightarrow\lambda=\pm1$.

Since $\vec a$ is coplanar with $\vec b,\vec c$: $[\vec a\ \vec b\ \vec c]=0$, so the sum reduces to $[\vec a\ \vec b\ \vec d]+[\vec a\ \vec c\ \vec d] = [\vec a\ (\vec b+\vec c)\ \vec d]$:
$$\begin{vmatrix}0&-3\lambda&-\lambda\3&0&2\3&2&6\end{vmatrix} = 3\lambda(12)+\lambda(6) = 42\lambda$$

Taking $\lambda=-1$: $42(-1)=-42$.

Solutions · 1