If \( \vec{a}=x\hat{\imath}+x\hat{\jmath}+y\hat{k},\ \vec{b}=\hat{\imath}+\hat{k} \) and \( \vec{c}=y\hat{\imath}+y\hat{\jmath}+z\hat{k} \) are three vectors such that \( \vec{a}\times\vec{b}\perp\vec{c} \), then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( y^2=xz \)
Since \( \vec{a}\times\vec{b}\perp\vec{c} \), their dot product is zero.
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\x & x & y\\1 & 0 & 1\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(x-0)-\hat{\jmath}(x-y)+\hat{k}(0-x) \)
\( \displaystyle =x\hat{\imath}-(x-y)\hat{\jmath}-x\hat{k} \)
\( \displaystyle (\vec{a}\times\vec{b})\cdot\vec{c}=xy-(x-y)y-xz \)
\( \displaystyle =xy-xy+y^2-xz \)
\( \displaystyle =y^2-xz \)
Putting this equal to 0 gives \( y^2=xz \).
Shortcut: \( (\vec{a}\times\vec{b})\cdot\vec{c} \) is the scalar triple product, so write it directly as a determinant:
\( \displaystyle \begin{vmatrix}x & x & y\\1 & 0 & 1\\y & y & z\end{vmatrix}=x(0-y)-x(z-y)+y(y-0) \)
\( \displaystyle =y^2-xz \)