Match List-I with List-II
| List-I | List-II (value of \(\lambda\)) | ||
| (A) | \(2\hat{\imath}+\lambda\hat{\jmath}+\hat{k}\) is perpendicular to \(\hat{\imath}+\hat{\jmath}-4\hat{k}\) | (I) | 3 |
| (B) | \((\hat{\imath}+2\hat{\jmath}+\lambda\hat{k})\times(2\hat{\imath}+4\hat{\jmath}+6\hat{k}) = \vec{0}\) | (II) | 6 |
| (C) | The area of the parallelogram with adjacent sides \(\hat{\imath}+\hat{\jmath}\) and \(\lambda\hat{k}\) \((\lambda \gt 0)\) is \(4\sqrt{2}\) | (III) | 4 |
| (D) | The projection of \(\lambda\hat{\imath}+\hat{\jmath}+\hat{k}\) on \(2\hat{\imath}+2\hat{\jmath}+\hat{k}\) is 5 | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A) Perpendicular vectors
This matches (IV) 2. The scalar product is zero:
\(2+\lambda-4 = 0, \quad \lambda = 2\)
(B) Zero vector product
This matches (I) 3. A zero vector product means the vectors are parallel, so the components are proportional:
\( \displaystyle \frac{1}{2} = \frac{2}{4} = \frac{\lambda}{6}, \quad \lambda = 3 \)
(C) Area of the parallelogram
This matches (III) 4.
\( \displaystyle (\hat{\imath}+\hat{\jmath})\times\lambda\hat{k} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 0 & \lambda \end{vmatrix} = \lambda\hat{\imath}-\lambda\hat{\jmath} \)
\(\sqrt{2}\,\lambda = 4\sqrt{2}, \quad \lambda = 4\)
(D) Projection
This matches (II) 6.
\( \displaystyle \frac{2\lambda+2+1}{\sqrt{4+4+1}} = 5 \)
\(2\lambda+3 = 15, \quad \lambda = 6\)