Q71 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
Let $\vec{a} = \lambda\hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}, \vec{c} = \hat{i} + \mu\hat{j} + \hat{k}$
If $\vec{a}$ is parallel to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$ then $\lambda - \mu = \ldots$
If $\vec{a}$ is parallel to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$ then $\lambda - \mu = \ldots$
- A$-2$
- B$-1$
- Correct answerC$2$
- D$1$
Solution
Given $\vec{a} = \lambda\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.
Since $\vec{a}$ is parallel to $\vec{b}$, their corresponding components are proportional:
$\dfrac{\lambda}{2} = \dfrac{1}{4} = \dfrac{1}{4}$
$\Rightarrow \lambda = \dfrac{1}{2}$
Given $\vec{c} = \hat{i} + \mu\hat{j} + \hat{k}$.
Since $\vec{b}$ is perpendicular to $\vec{c}$, their dot product is zero:
$\vec{b} \cdot \vec{c} = 0$
$(2)(1) + (4)(\mu) + (4)(1) = 0$
$2 + 4\mu + 4 = 0$
$4\mu = -6$
$\Rightarrow \mu = -\dfrac{3}{2}$
Calculating $\lambda - \mu$:
$\lambda - \mu = \dfrac{1}{2} - \left(-\dfrac{3}{2}\right) = \dfrac{1}{2} + \dfrac{3}{2} = 2$
Answer: $2$
Since $\vec{a}$ is parallel to $\vec{b}$, their corresponding components are proportional:
$\dfrac{\lambda}{2} = \dfrac{1}{4} = \dfrac{1}{4}$
$\Rightarrow \lambda = \dfrac{1}{2}$
Given $\vec{c} = \hat{i} + \mu\hat{j} + \hat{k}$.
Since $\vec{b}$ is perpendicular to $\vec{c}$, their dot product is zero:
$\vec{b} \cdot \vec{c} = 0$
$(2)(1) + (4)(\mu) + (4)(1) = 0$
$2 + 4\mu + 4 = 0$
$4\mu = -6$
$\Rightarrow \mu = -\dfrac{3}{2}$
Calculating $\lambda - \mu$:
$\lambda - \mu = \dfrac{1}{2} - \left(-\dfrac{3}{2}\right) = \dfrac{1}{2} + \dfrac{3}{2} = 2$
Answer: $2$
Hence, the correct answer is option: 'C'.