Let \(\vec{a}=\hat{\imath}+\hat{\jmath}\), \(\vec{b}=\hat{\jmath}+\hat{k}\) and \(\vec{c}=2\hat{\imath}+\hat{\jmath}+3\hat{k}\). The vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and satisfies \(\vec{c}\cdot\vec{d} = 12\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3\hat{\imath}-3\hat{\jmath}+3\hat{k}\)
A vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is parallel to their cross product:
\( \displaystyle \vec{a}\times\vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} \)
\( = \hat{\imath}(1-0)-\hat{\jmath}(1-0)+\hat{k}(1-0)\)
\( = \hat{\imath}-\hat{\jmath}+\hat{k}\)
So \(\vec{d} = t(\hat{\imath}-\hat{\jmath}+\hat{k})\) for some scalar \(t\). The condition on the scalar product fixes \(t\):
\(\vec{c}\cdot\vec{d} = t(2-1+3) = 4t\)
\(4t = 12\)
\(t = 3\)
\(\vec{d} = 3\hat{\imath}-3\hat{\jmath}+3\hat{k}\)
Why the other options are incorrect
Option (1) \(-3\hat{\imath}+3\hat{\jmath}-3\hat{k}\) — Incorrect. It is perpendicular to both, but its scalar product with \(\vec{c}\) is \(-12\).
Option (2) \(\hat{\imath}-\hat{\jmath}+\hat{k}\) — Incorrect. It is \(\vec{a}\times\vec{b}\) itself; its scalar product with \(\vec{c}\) is 4.
Option (4) \(4\hat{\imath}-4\hat{\jmath}+4\hat{k}\) — Incorrect. Its scalar product with \(\vec{c}\) is 16.