If \( \vec{a}=\hat{\imath}+\hat{\jmath}-\hat{k} \) and \( \vec{b}=\hat{\imath}-2\hat{\jmath}+\hat{k} \), then
Match List-I with List-II
| List-I | List-II |
| (A) \( |\vec{a}+\vec{b}| \) | (I) 2 |
| (B) \( |\vec{a}-\vec{b}| \) | (II) \( \sqrt{5} \) |
| (C) \( |\vec{a}.\vec{b}| \) | (III) \( \sqrt{14} \) |
| (D) \( |\vec{a}\times\vec{b}| \) | (IV) \( \sqrt{13} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
(A) \( |\vec{a}+\vec{b}| \)
This matches (II) \( \sqrt{5} \).
\( \vec{a}+\vec{b}=2\hat{\imath}-\hat{\jmath} \)
\( |\vec{a}+\vec{b}|=\sqrt{4+1}=\sqrt{5} \)
(B) \( |\vec{a}-\vec{b}| \)
This matches (IV) \( \sqrt{13} \).
\( \vec{a}-\vec{b}=3\hat{\jmath}-2\hat{k} \)
\( |\vec{a}-\vec{b}|=\sqrt{9+4}=\sqrt{13} \)
(C) \( |\vec{a}.\vec{b}| \)
This matches (I) 2.
\( \vec{a}\cdot\vec{b}=1-2-1=-2 \)
\( |\vec{a}\cdot\vec{b}|=2 \)
(D) \( |\vec{a}\times\vec{b}| \)
This matches (III) \( \sqrt{14} \).
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & 1 & -1\\1 & -2 & 1\end{vmatrix} \)
\( =-\hat{\imath}-2\hat{\jmath}-3\hat{k} \)
\( |\vec{a}\times\vec{b}|=\sqrt{1+4+9}=\sqrt{14} \)
So the matching is (A) - (II), (B) - (IV), (C) - (I), (D) - (III).