If $\vec{a}$ and $\vec{b}$ are the intersecting face diagonals of a cube of side $x$ in plane $X O Y$ and $Y O Z,$ respectively, with respect to reference frame at the point of intersection of the vectors and sides of cube as the axes, the components of vector $\vec{r} = \vec{a} \times \vec{b}$ are

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$x, - x, x$

$- x^{2}, - x^{2}, x^{2}$

$x^{2}, - x^{2}, x^{2}$

$x, x^{2}, - x$

answer is C.

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Detailed Solution

Here $\vec{a} = x \hat{i} + x \hat{j}$ and $\vec{b} = x \hat{j} + x \hat{k}$ Since $\vec{R}$

$= \vec{a} \times \vec{b}$ we get $\vec{R} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & x & 0 \\ 0 & x & x \end{matrix} |$ $= x^{2} \hat{i} - x^{2} \hat{j} + x^{2} \hat{k}$

Clearly, the components are $R_{x} = x^{2}, R_{y} = - x^{2}, R_{Z}$ $= x^{2}$

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