If the volume of a parallelepiped with edges a= i+ j+n k, b=2 i+4 j-n k, c= i+n j+3 k (n0)…

Question

If the volume of a parallelepiped with edges $\vec a=\hat i+\hat j+n\hat k$, $\vec b=2\hat i+4\hat j-n\hat k$, $\vec c=\hat i+n\hat j+3\hat k$ ($n\ge0$) is $158$ cu. units, then:

Accepted Answer

$$[\vec a\ \vec b\ \vec c] = \begin{vmatrix}1&1&n\2&4&-n\1&n&3\end{vmatrix} = 158$$

Expanding gives $3n^2-5n-152=0 \implies (n-8)(3n+19)=0 \implies n=8$ (since $n\ge0$).

With $n=8$: $\vec a=(1,1,8),\vec b=(2,4,-8),\vec c=(1,8,3)$.

$$\vec a\cdot\vec c = 1+8+24=33,\qquad \vec b\cdot\vec c = 2+32-24=10$$

So option B holds.

Solutions · 1