If \( \vec{a}=4\hat{\imath}+5\hat{\jmath}-\hat{k},\vec{b}=\hat{\imath}-4\hat{\jmath}+5\hat{k} \) and \( \vec{c}=3\hat{\imath}+\hat{\jmath}-\hat{k} \) . If a vector \( \vec{d} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and \( \vec{c}.\vec{d}=21 \), then unit vector along \( \vec{d} \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{\sqrt{3}}\left(\hat{\imath}-\hat{\jmath}-\hat{k}\right) \)
A vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) is along \( \vec{a}\times\vec{b} \).
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\4 & 5 & -1\\1 & -4 & 5\end{vmatrix} \)
\( \displaystyle =(25-4)\hat{\imath}-(20+1)\hat{\jmath}+(-16-5)\hat{k} \)
\( \displaystyle =21\hat{\imath}-21\hat{\jmath}-21\hat{k} \)
Let \( \vec{d}=t\,(\hat{\imath}-\hat{\jmath}-\hat{k}) \). Then
\( \displaystyle \vec{c}\cdot\vec{d}=t\,(3-1+1)=3t \)
\( \displaystyle 3t=21\;\Rightarrow\;t=7 \)
\( \displaystyle \vec{d}=7\hat{\imath}-7\hat{\jmath}-7\hat{k} \)
\( \displaystyle \hat{d}=\frac{\vec{d}}{|\vec{d}|} \)
\( \displaystyle =\frac{7(\hat{\imath}-\hat{\jmath}-\hat{k})}{7\sqrt{3}} \)
\( \displaystyle =\frac{1}{\sqrt{3}}(\hat{\imath}-\hat{\jmath}-\hat{k}) \)
Shortcut: The unit vector must be perpendicular to \( \vec{a}=4\hat{\imath}+5\hat{\jmath}-\hat{k} \). The dot product of \( \vec{a} \) with the vector in brackets in each option:
Option (1): (2)(4) + (2)(5) + (2)(−1) = 16
Option (2): (1)(4) + (−1)(5) + (−1)(−1) = 0
Option (3): (1)(4) + (1)(5) + (−1)(−1) = 10
Option (4): (1)(4) + (1)(5) + (1)(−1) = 8
Only Option (2) gives 0, so it is the only possible answer.