Q158 / 240 · KCET 2026 · Mathematics
The value of $\lambda$ for which the vectors $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ are orthogonal is
- A$\dfrac{5}{2}$
- Correct answerB$\dfrac{-5}{2}$
- C$\dfrac{2}{5}$
- D$\dfrac{-2}{5}$
Solution
For two vectors $\vec{a}$ and $\vec{b}$ to be orthogonal, their dot product must be zero.
$\vec{a} \cdot \vec{b} = 0$
$(2\hat{i} + \lambda\hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 0$
$(2)(1) + (\lambda)(2) + (1)(3) = 0$
$2 + 2\lambda + 3 = 0$
$2\lambda + 5 = 0$
$\lambda = \dfrac{-5}{2}$
Answer: $\dfrac{-5}{2}$
$\vec{a} \cdot \vec{b} = 0$
$(2\hat{i} + \lambda\hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 0$
$(2)(1) + (\lambda)(2) + (1)(3) = 0$
$2 + 2\lambda + 3 = 0$
$2\lambda + 5 = 0$
$\lambda = \dfrac{-5}{2}$
Answer: $\dfrac{-5}{2}$
Hence, the correct answer is option: 'B'.