Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | The vectors \(\lambda\hat{\imath}+\hat{\jmath}+2\hat{k}\) and \(\hat{\imath}+\hat{\jmath}+\hat{k}\) are perpendicular if \(\lambda\) is equal to | (I) | 1 |
| (B) | The vectors \(3\hat{\imath}+6\hat{\jmath}-\hat{k}\) and \(2\hat{\imath}+4\hat{\jmath}-\lambda\hat{k}\) are collinear if \(\lambda\) is equal to | (II) | \(-3\) |
| (C) | The number of vectors of unit-length which are perpendicular to both the vectors \(\vec{a}=\hat{\imath}+\hat{\jmath}+2\hat{k}\) and \(\vec{b}=3\hat{\imath}-\hat{\jmath}+5\hat{k}\) is | (III) | 2/3 |
| (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)
(A) The vectors \(\lambda\hat{\imath}+\hat{\jmath}+2\hat{k}\) and \(\hat{\imath}+\hat{\jmath}+\hat{k}\) are perpendicular if \(\lambda\) is equal to
This matches (II) −3. Perpendicular vectors have zero dot product:
\(\lambda(1) + 1(1) + 2(1) = 0\)
\(\lambda = -3\)
(B) The vectors \(3\hat{\imath}+6\hat{\jmath}-\hat{k}\) and \(2\hat{\imath}+4\hat{\jmath}-\lambda\hat{k}\) are collinear if \(\lambda\) is equal to
This matches (III) 2/3. Collinear vectors have proportional components:
\(\displaystyle \frac{3}{2} = \frac{6}{4} = \frac{-1}{-\lambda}\)
\(\displaystyle \frac{1}{\lambda} = \frac{3}{2}\)
\(\displaystyle \lambda = \frac{2}{3}\)
(C) The number of vectors of unit-length which are perpendicular to both the vectors \(\vec{a}=\hat{\imath}+\hat{\jmath}+2\hat{k}\) and \(\vec{b}=3\hat{\imath}-\hat{\jmath}+5\hat{k}\) is
This matches (IV) 2. A vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is along \(\vec{a}\times\vec{b}\):
\(\displaystyle \vec{a}\times\vec{b} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & 1 & 2\\ 3 & -1 & 5\end{vmatrix}\)
\(= (5+2)\hat{\imath} - (5-6)\hat{\jmath} + (-1-3)\hat{k} = 7\hat{\imath}+\hat{\jmath}-4\hat{k}\)
This is a non-zero vector, so there are exactly two unit vectors along this line, one in each direction:
\(\displaystyle \pm\frac{7\hat{\imath}+\hat{\jmath}-4\hat{k}}{\sqrt{66}}\)
(D) If \(|\vec{a}| = 1\) and \(\vec{a}+\vec{b}=\vec{0}\), then \(|\vec{b}|\) is equal to
This matches (I) 1:
\(\vec{b} = -\vec{a}\)
\(|\vec{b}| = |\vec{a}| = 1\)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).