Match List-I with List-II
| List-I | List-II | ||
| (A) | \(|6\hat{\imath}-2\hat{\jmath}+3\hat{k}|\) | (I) | \(-\displaystyle\frac{1}{3}\) |
| (B) | Maximum value of \(|\lambda\vec{a}|\), where \(|\vec{a}|=4\) and \(-3\le\lambda\le 2\) | (II) | \(\displaystyle\frac{1}{\sqrt{3}}\) |
| (C) | Cosine of the angle between \(\hat{\imath}+\hat{\jmath}-\hat{k}\) and \(\hat{\imath}-\hat{\jmath}+\hat{k}\) | (III) | 12 |
| (D) | Positive value of x for which \(x(\hat{\imath}+\hat{\jmath}+\hat{k})\) is a unit vector | (IV) | 7 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \(|6\hat{\imath}-2\hat{\jmath}+3\hat{k}|\)
This matches (IV) 7.
The magnitude is the square root of the sum of the squares of the components:
\(\sqrt{36+4+9} = \sqrt{49} = 7\)
(B) Maximum value of \(|\lambda\vec{a}|\), where \(|\vec{a}|=4\) and \(-3\le\lambda\le 2\)
This matches (III) 12.
\(|\lambda\vec{a}| = |\lambda|\,|\vec{a}| = 4|\lambda|\)
On \(-3\le\lambda\le 2\) the largest value of \(|\lambda|\) is 3, at \(\lambda = -3\):
\(4\times 3 = 12\)
(C) Cosine of the angle between \(\hat{\imath}+\hat{\jmath}-\hat{k}\) and \(\hat{\imath}-\hat{\jmath}+\hat{k}\)
This matches (I) \(-\frac{1}{3}\).
Using \(\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\):
\(\vec{a}\cdot\vec{b} = 1-1-1 = -1\)
\(|\vec{a}| = |\vec{b}| = \sqrt{3}\)
\( \displaystyle \cos\theta = \frac{-1}{\sqrt{3}\times\sqrt{3}} = -\frac{1}{3} \)
(D) Positive value of x for which \(x(\hat{\imath}+\hat{\jmath}+\hat{k})\) is a unit vector
This matches (II) \(\frac{1}{\sqrt{3}}\).
A unit vector has magnitude 1:
\(|x|\sqrt{1+1+1} = 1\)
\( \displaystyle |x| = \frac{1}{\sqrt{3}} \)
The positive value is \(x = \frac{1}{\sqrt{3}}\).