If \( \vec{a}=2\hat{\imath}+\hat{\jmath} \) and \( \vec{b}=\hat{\jmath}-\hat{k} \), then Match List-I with List-II
| LIST-I | LIST-II | ||
| (A) | Unit vector along \( \vec{a}+\vec{b} \) | (I) | \( \displaystyle\frac{1}{3}\left(-\hat{\imath}+2\hat{\jmath}+2\hat{k}\right) \) |
| (B) | Unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) | (II) | 1 |
| (C) | \( |\vec{a}+\vec{b}| \) | (III) | \( \displaystyle\frac{1}{3}\left(2\hat{\imath}+2\hat{\jmath}-\hat{k}\right) \) |
| (D) | \( \vec{a}\cdot\vec{b} \) | (IV) | 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
\( \displaystyle \vec{a}+\vec{b}=2\hat{\imath}+2\hat{\jmath}-\hat{k} \)
\( \displaystyle |\vec{a}+\vec{b}|=\sqrt{4+4+1}=3 \)
(A) Unit vector along \( \vec{a}+\vec{b} \)
This matches (III) \( \frac{1}{3}\left(2\hat{\imath}+2\hat{\jmath}-\hat{k}\right) \).
A unit vector is the vector divided by its magnitude:
\( \displaystyle \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{1}{3}\left(2\hat{\imath}+2\hat{\jmath}-\hat{k}\right) \)
(B) Unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \)
This matches (I) \( \frac{1}{3}\left(-\hat{\imath}+2\hat{\jmath}+2\hat{k}\right) \).
The cross product is perpendicular to both:
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & 1 & 0\\0 & 1 & -1\end{vmatrix} \)
\( \displaystyle =(-1-0)\hat{\imath}-(-2-0)\hat{\jmath}+(2-0)\hat{k} \)
\( \displaystyle =-\hat{\imath}+2\hat{\jmath}+2\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{1+4+4}=3 \)
So the unit vector is \( \frac{1}{3}\left(-\hat{\imath}+2\hat{\jmath}+2\hat{k}\right) \) (its negative is the other such unit vector).
(C) \( |\vec{a}+\vec{b}| \)
This matches (IV) 3, since \( |\vec{a}+\vec{b}|=\sqrt{4+4+1}=3 \).
(D) \( \vec{a}\cdot\vec{b} \)
This matches (II) 1.
\( \displaystyle \vec{a}\cdot\vec{b}=(2)(0)+(1)(1)+(0)(-1)=1 \)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
Shortcut: (C) = 3 and (D) = 1 need only one line each, and Option (4) is the only option with (C)-(IV) and (D)-(II).