Let \( \vec{a}=2\hat{\imath}+3\hat{\jmath}+\hat{k} \) and \( \vec{b}=2\hat{\imath}-\hat{\jmath}+2\hat{k} \) be two vectors. Then which of the following are TRUE?
(A) \( \vec{a} \) and \( \vec{b} \) are perpendicular
(B) Projection of \( \vec{a} \) on \( \vec{b} \) is 1
(C) \( |\vec{a}\times\vec{b}|=3\sqrt{13} \)
(D) \( \vec{a}-\vec{b} \) is a unit vector
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) and (C) only
\( \displaystyle \vec{a}\cdot\vec{b}=(2)(2)+(3)(-1)+(1)(2)=3 \)
\( \displaystyle |\vec{b}|=\sqrt{4+1+4}=3 \)
\( \displaystyle \text{Projection of }\vec{a}\text{ on }\vec{b}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}=\frac{3}{3}=1 \)
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & 3 & 1\\2 & -1 & 2\end{vmatrix} \)
\( \displaystyle =(6+1)\hat{\imath}-(4-2)\hat{\jmath}+(-2-6)\hat{k} \)
\( \displaystyle =7\hat{\imath}-2\hat{\jmath}-8\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{49+4+64}=\sqrt{117}=3\sqrt{13} \)
\( \displaystyle \vec{a}-\vec{b}=4\hat{\jmath}-\hat{k} \)
\( \displaystyle |\vec{a}-\vec{b}|=\sqrt{16+1}=\sqrt{17} \)
So \( \vec{a}\cdot\vec{b}=3 \), the projection of \( \vec{a} \) on \( \vec{b} \) is 1, \( |\vec{a}\times\vec{b}|=3\sqrt{13} \) and \( |\vec{a}-\vec{b}|=\sqrt{17} \).
Checking the statements
(A) \( \vec{a} \) and \( \vec{b} \) are perpendicular — Incorrect. \( \vec{a}\cdot\vec{b}=3\ne 0 \).
(B) Projection of \( \vec{a} \) on \( \vec{b} \) is 1 — Correct. \( \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}=\frac{3}{3}=1 \).
(C) \( |\vec{a}\times\vec{b}|=3\sqrt{13} \) — Correct. \( |\vec{a}\times\vec{b}|=\sqrt{117}=3\sqrt{13} \).
(D) \( \vec{a}-\vec{b} \) is a unit vector — Incorrect. Its magnitude is \( \sqrt{17} \), not 1.
Shortcut: For (C), Lagrange’s identity avoids the determinant: \( |\vec{a}|^{2}=14 \), so \( |\vec{a}\times\vec{b}|^{2}=14\times 9-3^{2}=117 \).