Match List-I with List-II
|
List-I |
List-II |
|
(A) Angle between $\hat i-\hat j$ and $\hat j + \hat k$ |
(I) 0 |
|
(B) Angle between $2\hat j-\hat k$ and $\hat j + 2\hat k$ |
(II) $\frac{2\pi}{3}$ |
|
(C) Angle between $\hat i + 2\hat j$ and $5\hat i+ 10\hat j$ |
(III) $\frac{\pi}{6}$ |
|
(D) Angle between $\sqrt{3}\hat i+\hat j$ and $\hat i$ |
(IV) $\frac{\pi}{2}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
|
List-I |
List-II |
|
(A) Angle between $\hat i-\hat j$ and $\hat j + \hat k$ |
(II) $\frac{2\pi}{3}$ |
|
(B) Angle between $2\hat j-\hat k$ and $\hat j + 2\hat k$ |
(IV) $\frac{\pi}{2}$ |
|
(C) Angle between $\hat i + 2\hat j$ and $5\hat i+ 10\hat j$ |
(I) 0 |
|
(D) Angle between $\sqrt{3}\hat i+\hat j$ and $\hat i$ |
(III) $\frac{\pi}{6}$ |
(A) Angle between $\hat i-\hat j$ and $\hat j+\hat k$
Dot product
$(\hat i-\hat j)\cdot(\hat j+\hat k)=0-1+0=-1$
Magnitudes
$|\hat i-\hat j|=\sqrt2,\;|\hat j+\hat k|=\sqrt2$
$\cos\theta=\frac{-1}{2}$
$\theta=\frac{2\pi}{3}$
(A) $\rightarrow$ (II)
(B) Angle between $2\hat j-\hat k$ and $\hat j+2\hat k$
Dot product
$(2\hat j-\hat k)\cdot(\hat j+2\hat k)=2-2=0$
Hence angle $=\frac{\pi}{2}$
(B) $\rightarrow$ (IV)
(C) Angle between $\hat i+2\hat j$ and $5\hat i+10\hat j$
Second vector is scalar multiple of first
Angle $=0$
(C) $\rightarrow$ (I)
(D) Angle between $\sqrt3\hat i+\hat j$ and $\hat i$
Dot product
$(\sqrt3\hat i+\hat j)\cdot\hat i=\sqrt3$
Magnitudes
$|\sqrt3\hat i+\hat j|=2,\;|\hat i|=1$
$\cos\theta=\frac{\sqrt3}{2}$
$\theta=\frac{\pi}{6}$
(D) $\rightarrow$ (III)
Final Matching: (A)-(II), (B)-(IV), (C)-(I), (D)-(III).