Suppose that \( \vec{a}=\hat{\imath}+\hat{\jmath},\ \vec{b}=\hat{\imath}-\hat{\jmath} \) and \( \vec{c}=\hat{\imath}+\hat{\jmath}+\hat{k} \). If \( \hat{n} \) is a unit vector such that \( \vec{a}.\hat{n}=0 \) and \( \vec{b}.\hat{n}=0 \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( |\vec{a}| \) | (I) | 1 |
| (B) | \( \vec{a}.\vec{b} \) | (II) | \( \sqrt{2} \) |
| (C) | \( \left|\vec{a}+\vec{b}+\vec{c}\right| \) | (III) | 0 |
| (D) | \( |\vec{c}\cdot\hat{n}| \) | (IV) | \( \sqrt{11} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
(A) \( |\vec{a}| \)
This matches (II) \( \sqrt{2} \).
\( \displaystyle |\vec{a}|=\sqrt{1^2+1^2}=\sqrt{2} \)
(B) \( \vec{a}.\vec{b} \)
This matches (III) 0.
\( \displaystyle \vec{a}\cdot\vec{b}=(1)(1)+(1)(-1)=0 \)
(C) \( \left|\vec{a}+\vec{b}+\vec{c}\right| \)
This matches (IV) \( \sqrt{11} \).
\( \displaystyle \vec{a}+\vec{b}+\vec{c}=3\hat{\imath}+\hat{\jmath}+\hat{k} \)
\( \displaystyle \left|\vec{a}+\vec{b}+\vec{c}\right|=\sqrt{9+1+1}=\sqrt{11} \)
(D) \( |\vec{c}\cdot\hat{n}| \)
This matches (I) 1. \( \vec{a} \) and \( \vec{b} \) both lie in the xy-plane, so a unit vector perpendicular to both is \( \hat{n}=\pm\hat{k} \) (check: \( \vec{a}\times\vec{b}=-2\hat{k} \)). The modulus makes the answer the same for either sign.
\( \displaystyle \vec{c}\cdot\hat{n}=\pm 1 \)
\( \displaystyle |\vec{c}\cdot\hat{n}|=1 \)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).