Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | If vector \(\vec{a}\) and \(\vec{b}\) are such that \(\vec{a} = \lambda\vec{b}\) and \(|\vec{a}| = |\vec{b}|\), then | (I) | \(\vec{a}\) and \(\vec{b}\) are orthogonal |
| (B) | Projection vector of \(\vec{a}\) on \(\vec{b}\) | (II) | \([0, 12]\) |
| (C) | \(\vec{a}\) and \(\vec{b}\) are non-zero vectors such that \(|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|\) then | (III) | \(\vec{a} = \pm\vec{b}\) |
| (D) | If \(|\vec{a}| = 4, -3 \le \lambda \le 2\), then the range of \(|\lambda\vec{a}|\) | (IV) | \(\left(\displaystyle\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}\) |
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 vector \(\vec{a}\) and \(\vec{b}\) are such that \(\vec{a} = \lambda\vec{b}\) and \(|\vec{a}| = |\vec{b}|\), then
This matches (III) \(\vec{a} = \pm\vec{b}\)
\( |\vec{a}| = |\lambda||\vec{b}| = |\vec{b}| \)
\( |\lambda| = 1,\quad \lambda = \pm 1 \)
(B) Projection vector of \(\vec{a}\) on \(\vec{b}\)
This matches (IV) \(\left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}\)
It is the scalar projection times the unit vector along \(\vec{b}\):
\( \displaystyle \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|} \cdot \frac{\vec{b}}{|\vec{b}|} = \left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b} \)
(C) \(\vec{a}\) and \(\vec{b}\) are non-zero vectors such that \(|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|\) then
This matches (I) \(\vec{a}\) and \(\vec{b}\) are orthogonal
Squaring both sides:
\( |\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = |\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 \)
\( 4\vec{a}\cdot\vec{b} = 0,\quad \vec{a}\cdot\vec{b} = 0 \)
Non-zero vectors with zero dot product are perpendicular.
(D) If \(|\vec{a}| = 4, -3 \le \lambda \le 2\), then the range of \(|\lambda\vec{a}|\)
This matches (II) \([0, 12]\)
\( |\lambda\vec{a}| = |\lambda||\vec{a}| = 4|\lambda| \)
For \(-3 \le \lambda \le 2\), \(|\lambda|\) is least (0) at \(\lambda = 0\) and largest (3) at \(\lambda = -3\):
\( 0 \le |\lambda| \le 3 \)
\( 0 \le 4|\lambda| \le 12 \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).