Match List - I with List - II.
| List - I | List - II | ||
| Two given vectors | Projection of \(\vec{a}\) on \(\vec{b}\) | ||
| (A) | \(\vec{a}=\hat{\imath}-\hat{\jmath},\ \vec{b}=\hat{\imath}+\hat{\jmath}\) | (I) | \(\displaystyle\frac{2}{\sqrt{5}}\) |
| (B) | \(\vec{a}=\hat{\imath}+\hat{\jmath},\ \vec{b}=2\hat{\imath}-\hat{k}\) | (II) | 0 |
| (C) | \(\vec{a}=\hat{\jmath}+\hat{k},\ \vec{b}=\hat{\imath}+\hat{k}\) | (III) | \(\sqrt{2}\) |
| (D) | \(\vec{a}=2\hat{\imath}+3\hat{\jmath},\ \vec{b}=\hat{\imath}-\hat{k}\) | (IV) | \(\displaystyle\frac{1}{\sqrt{2}}\) |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
The projection of \(\vec{a}\) on \(\vec{b}\) is \(\displaystyle\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\).
(A) \(\vec{a}\cdot\vec{b} = 1 - 1 = 0\), so the projection is 0 → (II).
(B) \(\vec{a}\cdot\vec{b} = 2\), \(|\vec{b}| = \sqrt{4+1} = \sqrt{5}\):
\( \displaystyle \frac{2}{\sqrt{5}} \rightarrow \text{(I)} \)
(C) \(\vec{a}\cdot\vec{b} = 1\), \(|\vec{b}| = \sqrt{2}\):
\( \displaystyle \frac{1}{\sqrt{2}} \rightarrow \text{(IV)} \)
(D) \(\vec{a}\cdot\vec{b} = 2\), \(|\vec{b}| = \sqrt{2}\):
\( \displaystyle \frac{2}{\sqrt{2}} = \sqrt{2} \rightarrow \text{(III)} \)
So (A) - (II), (B) - (I), (C) - (IV), (D) - (III).