Let be three lines such that is perpendicular to and is perpendicular to both and . Then, the point which lies on is [2024]

Question

Let $L_{1} : \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k}), λ \in R$

$L_{2} : \vec{r} = (\hat{j} - \hat{k}) + μ (3 \hat{i} + \hat{j} + p \hat{k}), μ \in R$

$L_{3} : \vec{r} = δ (l \hat{i} + m \hat{j} + n \hat{k}), δ \in R$

be three lines such that $L_{1}$ is perpendicular to $L_{2}$ and $L_{3}$ is perpendicular to both $L_{1}$ and $L_{2}$ . Then, the point which lies on $L_{3}$ is [2024]

Smart Achievers · Accepted Answer

(2)

Given, $L_{1} : \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k}), λ \in R$

$L_{2} : \vec{r} = (\hat{j} - \hat{k}) + μ (3 \hat{i} + \hat{j} + p \hat{k}), μ \in R$

and $L_{3} : \vec{r} = δ (l \hat{i} + m \hat{j} + n \hat{k}), δ \in R$

since, $L_{1}$ is perpendicular to $L_{2}$ .

$∴ (\hat{i} - \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + \hat{j} + p \hat{k}) = 0$

$\Rightarrow 3 - 1 + 2 p = 0 \Rightarrow p = - 1$

Also, $L_{3}$ is perpendicular to both $L_{1}$ and $L_{2}$ .

$∴ L_{3} is parallel to (\hat{i} - \hat{j} + 2 \hat{k}) \times (3 \hat{i} + \hat{j} + p \hat{k})$

Now, $(\hat{i} - \hat{j} + 2 \hat{k}) \times (3 \hat{i} + \hat{j} + p \hat{k})$

$=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 2 \\ 3 & 1 & p \end{matrix} |$ $= \hat{i} (- p - 2) - \hat{j} (p - 6) + \hat{k} (1 + 3)$

$= - \hat{i} + 7 \hat{j} + 4 \hat{k}$ $(∵ p = - 1)$

$∴ l \hat{i} + m \hat{j} + n \hat{k} = - \hat{i} + 7 \hat{j} + 4 \hat{k}$

$\Rightarrow l = - 1, m = 7 and n = 4$

$∴ (- 1, 7, 4) lies on L_{3}$ .

Solution

1 (1, –7, 4)

2 (–1, 7, 4)

3 (–1, –7, 4)

4 (1, 7, –4)

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