Match List-I with List-II.
| List-I | List-II | ||
| (A) | Angle between $\hat{i}-2\hat{j}+3k$ and $2\hat{i}+\hat{j}$ is | (I) | $cos^{-1}\frac{2}{\sqrt{18}}$ |
| (B) | Angle between $\hat{i}+\hat{j}+2\hat{k}$ and $2\hat{i} + 2\hat{j} +4\hat{k}$ is | (II) | 0 |
| (C) | Angle between $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i} + \hat{j} +\hat{k}$ is | (III) | 90° |
| (D) | Angle between $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i} + \hat{j} +\hat{k}$ is | (IV) | $cos^{-1}\frac{1}{3}$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
for two vector $\vec a, \vec b$
angle θ between them
$⇒\vec a.\vec b=|\vec a||\vec b|\cos θ⇒θ=\cos^{-1}(\frac{\vec a.\vec b}{|\vec a||\vec b|})$
(A) $\vec a=\hat i-2\hat j+3\hat k,\vec b=2\hat i+\hat j$
$\vec a.\vec b=2-2=0⇒θ=90°$ (III)
(B) $\vec a=\hat{i}+\hat{j}+2\hat{k},\vec b=2\hat{i} + 2\hat{j} +4\hat{k}$
$\vec a=2\vec b⇒θ=0°$ (II)
(C) $\vec a=2\hat{i}-\hat{j}+\hat{k},\vec b=\hat{i} + \hat{j} +\hat{k}$
$\vec a.\vec b=2-1+1=2,|\vec a|=\sqrt{6},|\vec b|=\sqrt{3}$
so $θ=\cos^{-1}(\frac{2}{\sqrt{18}})$ (I)
(D) $\vec a=\hat i+\hat j-\hat k,\vec b=\hat i+\hat j+\hat k$
$\vec a.\vec b=1+1-1=1,|\vec a|=\sqrt{3}=|\vec b|$
$θ=\cos^{-1}(\frac{1}{3})$ (IV)