If \(\vec{a} = 2\hat{\imath} - \hat{\jmath} + 3\hat{k}\) and \(\vec{b} = 2\hat{\imath} + 2\hat{\jmath} + \hat{k}\), then
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Projection of \(\vec{a}\) on \(\vec{b}\) is | (I) | \(-7\hat{\imath} + 4\hat{\jmath} + 6\hat{k}\) |
| (B) | \(\vec{a} \times \vec{b}\) is | (II) | \(\displaystyle\frac{1}{\sqrt{101}}(-7\hat{\imath} + 4\hat{\jmath} + 6\hat{k})\) |
| (C) | unit vector along \(\vec{a} + \vec{b}\) is | (III) | \(\displaystyle\frac{5}{3}\) |
| (D) | Unit vector perpendicular to both \(\vec{a}\) & \(\vec{b}\) is | (IV) | \(\displaystyle\frac{1}{\sqrt{33}}(4\hat{\imath} + \hat{\jmath} + 4\hat{k})\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A) Projection of \(\vec{a}\) on \(\vec{b}\)
This matches (III) \(\frac{5}{3}\). The projection is \(\vec{a}\cdot\vec{b}\) divided by \(|\vec{b}|\):
\(\vec{a}\cdot\vec{b} = 4 - 2 + 3 = 5,\quad |\vec{b}| = \sqrt{4 + 4 + 1} = 3\)
\(\displaystyle \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} = \frac{5}{3}\)
(B) \(\vec{a} \times \vec{b}\)
This matches (I) \(-7\hat{\imath} + 4\hat{\jmath} + 6\hat{k}\):
\(\displaystyle \vec{a} \times \vec{b} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\2&-1&3\\2&2&1\end{vmatrix}\)
\(= \hat{\imath}(-1 - 6) - \hat{\jmath}(2 - 6) + \hat{k}(4 + 2)\)
\(= -7\hat{\imath} + 4\hat{\jmath} + 6\hat{k}\)
(C) unit vector along \(\vec{a} + \vec{b}\)
This matches (IV):
\(\vec{a} + \vec{b} = 4\hat{\imath} + \hat{\jmath} + 4\hat{k},\quad |\vec{a} + \vec{b}| = \sqrt{16 + 1 + 16} = \sqrt{33}\)
\(\displaystyle \text{unit vector} = \frac{1}{\sqrt{33}}(4\hat{\imath} + \hat{\jmath} + 4\hat{k})\)
(D) Unit vector perpendicular to both \(\vec{a}\) & \(\vec{b}\)
This matches (II). The cross product is perpendicular to both vectors, so divide it by its length:
\(|\vec{a} \times \vec{b}| = \sqrt{49 + 16 + 36} = \sqrt{101}\)
\(\displaystyle \text{unit vector} = \frac{1}{\sqrt{101}}(-7\hat{\imath} + 4\hat{\jmath} + 6\hat{k})\)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).