Match List-I with List-II
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List-I |
List-II |
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(A) If vector $\vec a$ and $\vec b$ are such that $\vec a = λ\vec b$ and $|\vec a| = |\vec b|$, then |
(I) $\vec a$ and $\vec b$ are orthogonal |
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(B) Projection vector of $\vec a$ on $\vec b$ |
(II) $[0, 12]$ |
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(C) $\vec a$ and $\vec b$ are non-zero vectors such that $|\vec a+\vec b|=|\vec a-\vec b|$ then |
(III) $\vec a = ±\vec b$ |
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(D) If $|\vec a| = 4,-3 ≤λ≤ 2$, then the range of $|λ\vec a|$ |
(IV) $\left(\frac{\vec a.\vec b}{|\vec b|^2}\right)\vec b$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) If $\vec{a} = \lambda \vec{b}$ and $|\vec{a}| = |\vec{b}|$
$\Rightarrow |\lambda| = 1 \Rightarrow \lambda = \pm 1 \Rightarrow \vec{a} = \pm \vec{b}$
⟹ (A) → (III)
(B) Projection vector of $\vec{a}$ on $\vec{b}$ is $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$
⟹ (B) → (IV)
(C) $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \Rightarrow \vec{a} \cdot \vec{b} = 0$
⟹ $\vec{a}$ and $\vec{b}$ are orthogonal
⟹ (C) → (I)
(D) $|\lambda \vec{a}| = |\lambda||\vec{a}| = 4|\lambda|$, where $\lambda \in [-3, 2]$
⟹ $|\lambda| \in [0, 3] \Rightarrow |\lambda \vec{a}| \in [0, 12]$
⟹ (D) → (II)