Question

If the vectors $\vec{ a }=x \hat{ i }+y \hat{ j }+z \hat{ k }$ and such that $\vec{ a }, \vec{ c }$ and $\vec{ b }$ form a right handed system, then $\vec{ c }$ is :

• A. $z \hat{ i }-x \hat{ k }$
• B. $\vec{0}$
• C. $y \hat{ j }$
• D. $-z \hat{ i }+x \hat{ k }$

Answer 1

Answer: (A)

Given that $\vec{a} = x\hat{ i} + y \hat{j} + z \hat{k}$ and $\vec{b} = \hat{j}$ are
such that $\vec{a}. \vec{c}$ and $\vec{b}$ form a right handed system
$\therefore \vec{c} = \vec{b} \times\vec{a} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 0&1&0\\ x&y&z\end{vmatrix}$
$ =\hat{i} z - x \hat{k}$