Let a= i- k, b=x i+ j+(1-x) k and c=y i+x j+(1+x-y) k.

Question

Let $\vec a=\hat i-\hat k$, $\vec b=x\hat i+\hat j+(1-x)\hat k$ and $\vec c=y\hat i+x\hat j+(1+x-y)\hat k$. Then $[\vec a\ \vec b\ \vec c]$ depends on:

Accepted Answer

$$[\vec a\ \vec b\ \vec c]=\begin{vmatrix}1&0&-1\x&1&1-x\y&x&1+x-y\end{vmatrix} = 1(1+x-y-x+x^2)-0-(x^2-y) = (1-y+x^2)-(x^2-y)=1$$

The result is $1$, independent of both $x$ and $y$.

Solutions · 1