Q88 / 180 · COMEDK 2025 (10 May Shift 1) · Mathematics
The value of $\lambda$ for which the angle between lines
$\vec{r}=\hat{\imath}+\hat{\jmath}+\hat{k}+p(2 \hat{\imath}+\hat{\jmath}+2 \hat{k})$ and
$\vec{r}=(1+q) \hat{\imath}+(1+q \lambda) \hat{\jmath}+(1+q) \hat{k}$ is $\dfrac{\pi}{2}$
- A4
- B$-2$
- C$2$
- Correct answerD$-4$
Solution
The direction vector of the first line is $\vec{v}_1 = 2\hat{i} + \hat{j} + 2\hat{k}$.
The second line is given by $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + q(\hat{i} + \lambda\hat{j} + \hat{k})$. Thus, the direction vector of the second line is $\vec{v}_2 = \hat{i} + \lambda\hat{j} + \hat{k}$.
Since the angle between the lines is $\dfrac{\pi}{2}$, the lines are perpendicular, which implies that the dot product of their direction vectors must be zero.
$\vec{v}_1 \cdot \vec{v}_2 = 0$
$(2\hat{i} + \hat{j} + 2\hat{k}) \cdot (\hat{i} + \lambda\hat{j} + \hat{k}) = 0$
$(2)(1) + (1)(\lambda) + (2)(1) = 0$
$2 + \lambda + 2 = 0$
$\lambda + 4 = 0$
$\lambda = -4$
Answer: $-4$
The second line is given by $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + q(\hat{i} + \lambda\hat{j} + \hat{k})$. Thus, the direction vector of the second line is $\vec{v}_2 = \hat{i} + \lambda\hat{j} + \hat{k}$.
Since the angle between the lines is $\dfrac{\pi}{2}$, the lines are perpendicular, which implies that the dot product of their direction vectors must be zero.
$\vec{v}_1 \cdot \vec{v}_2 = 0$
$(2\hat{i} + \hat{j} + 2\hat{k}) \cdot (\hat{i} + \lambda\hat{j} + \hat{k}) = 0$
$(2)(1) + (1)(\lambda) + (2)(1) = 0$
$2 + \lambda + 2 = 0$
$\lambda + 4 = 0$
$\lambda = -4$
Answer: $-4$
Hence, the correct answer is option: 'D'.