Match List-I with List-II:
|
List-I |
List-II |
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(A) $4\hat i-2\hat j-4\hat k$ |
(I) A vector perpendicular to both $\hat i+2\hat j+\hat k$ and $2\hat i+2\hat j+3\hat k$ |
|
(B) $-4\hat i-4\hat j + 2\hat k$ |
(II) Direction ratios are $-2, 1, 2$ |
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(C) $2\hat i-4\hat j + 4\hat k$ |
(III) Angle with the vector $\hat i-2\hat j-\hat k$ is $\cos^{-1}(\frac{1}{\sqrt{6}})$ |
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(D) $4\hat i-\hat j-2\hat k$ |
(IV) Dot product with $-2\hat i+\hat j+ 3\hat k$ is 10 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
(A) $4\hat i-2\hat j-4\hat k$ have d.c. as -
$\frac{4}{\sqrt{36}}\hat i-\frac{2}{\sqrt{36}}\hat j-\frac{4}{\sqrt{36}}\hat k$
$\frac{2}{3}\hat i-\frac{1}{3}\hat j-\frac{2}{3}\hat k$
∴ D.r. is → 2, -1, 2 or -2, 1, 2
(B) $(-4\hat i-4\hat j + 2\hat k).(-2\hat i+\hat j+3\hat k)=8-4+6$
$=10$
(C) $\cos θ=\frac{2(1)+(-4)(-2)+(4)(-1)}{\sqrt{36}\sqrt{6}}=\frac{2+8-4}{6\sqrt{6}}=\frac{1}{\sqrt{6}}$
(D) $\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&1\\2&2&3\end{vmatrix}=4\hat i-\hat j-2\hat k$
∴ This is perpendicular to both of them.