If \( \vec{a}=3\hat{\imath}+\hat{\jmath}+4\hat{k} \) and \( \vec{b}=\hat{\imath}-\hat{\jmath}+\hat{k} \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \vec{a}.\vec{b} \) | (I) | \( 2\sqrt{3} \) |
| (B) | \( |\vec{a}\times\vec{b}| \) | (II) | 6 |
| (C) | Projection of \( \vec{a} \) on \( \vec{b} \) | (III) | \( \sqrt{42} \) |
| (D) | Projection of \( \vec{b} \) on \( \vec{a} \) | (IV) | \( \displaystyle\frac{3\sqrt{26}}{13} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (II), (B) - (III), (C) - (I), (D) - (IV)
The projections need the magnitudes of both vectors:
\( \displaystyle |\vec{a}|=\sqrt{9+1+16}=\sqrt{26} \)
\( \displaystyle |\vec{b}|=\sqrt{1+1+1}=\sqrt{3} \)
(A) \( \vec{a}\cdot\vec{b} \)
This matches (II) 6.
\( \displaystyle \vec{a}\cdot\vec{b}=3(1)+1(-1)+4(1) \)
\( \displaystyle =3-1+4=6 \)
(B) \( |\vec{a}\times\vec{b}| \)
This matches (III) √42.
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\3 & 1 & 4\\1 & -1 & 1\end{vmatrix} \)
\( \displaystyle =(1+4)\hat{\imath}-(3-4)\hat{\jmath}+(-3-1)\hat{k} \)
\( \displaystyle =5\hat{\imath}+\hat{\jmath}-4\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{25+1+16}=\sqrt{42} \)
(C) Projection of \( \vec{a} \) on \( \vec{b} \)
This matches (I) 2√3.
The projection divides the dot product by the magnitude of the vector projected on:
\( \displaystyle \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}=\frac{6}{\sqrt{3}} \)
\( \displaystyle =2\sqrt{3} \)
(D) Projection of \( \vec{b} \) on \( \vec{a} \)
This matches (IV) 3√26/13.
\( \displaystyle \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}=\frac{6}{\sqrt{26}} \)
\( \displaystyle =\frac{6\sqrt{26}}{26}=\frac{3\sqrt{26}}{13} \)
So the matching is (A)-(II), (B)-(III), (C)-(I), (D)-(IV).
Tip: The projection of a vector on \( \vec{b} \) divides the dot product by \( |\vec{b}| \), the vector it is projected on; swapping the two magnitudes swaps (C) and (D).