Match List-I with List-II
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List-I |
List-II |
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(A) The vectors $λ\hat i+\hat j+2\hat k$ and $\hat i+\hat j+\hat k$ are perpendicular if $λ$ is equal to |
(I) 1 |
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(B) The vectors $3\hat i+6\hat j -\hat k$ and $2\hat i+4\hat j-λ\hat k$ are collinear if $λ$ is equal to |
(II) -3 |
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(C) The number of vectors of unit-length which are perpendicular to both the vectors $\vec a=\hat i+\hat j+2\hat k$ and $\vec b = 3\hat i-\hat j+5\hat k$ is |
(III) 2/3 |
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(D) If $|\vec a| = 1$ and $\vec a+\vec b=\vec 0$, then $|\vec b|$ is equal to |
(IV) 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)-(III), (C)-(IV), (D)-(I)
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List-I |
List-II |
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(A) The vectors $λ\hat i+\hat j+2\hat k$ and $\hat i+\hat j+\hat k$ are perpendicular if $λ$ is equal to |
(II) -3 |
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(B) The vectors $3\hat i+6\hat j -\hat k$ and $2\hat i+4\hat j-λ\hat k$ are collinear if $λ$ is equal to |
(III) 2/3 |
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(C) The number of vectors of unit-length which are perpendicular to both the vectors $\vec a=\hat i+\hat j+2\hat k$ and $\vec b = 3\hat i-\hat j+5\hat k$ is |
(IV) 2 |
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(D) If $|\vec a| = 1$ and $\vec a+\vec b=\vec 0$, then $|\vec b|$ is equal to |
(I) 1 |
(A) For perpendicularity: $(\lambda,1,2)\cdot(1,1,1)=0\Rightarrow \lambda+1+2=0\Rightarrow \lambda=-3$
(B) Collinear: $\frac{3}{2}=\frac{6}{4}=\frac{-1}{-\lambda}\Rightarrow \frac{1}{\lambda}=\frac{3}{2}\Rightarrow \lambda=\frac{2}{3}$
(C) Vectors perpendicular to both $\vec a$ and $\vec b$ are along $\vec a\times\vec b\neq\vec 0$, so two unit vectors ($\pm$ direction) ⇒ $2$
(D) $|{\vec a}|=1,\ \vec a+\vec b=\vec 0\Rightarrow \vec b=-\vec a\Rightarrow |\vec b|=1$
The correct matching is A–II, B–III, C–IV, D–I.