Question

Let \( L_1 : \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda (\hat{i} - \hat{j} + 2\hat{k}) \), \( \lambda \in \mathbb{R} \)
\( L_2 : \vec{r} = (\hat{j} - \hat{k}) + \mu (3\hat{i} + \hat{j} + p\hat{k}) \), \( \mu \in \mathbb{R} \) and
\( L_3 : \vec{r} = \delta (\ell \hat{i} + m \hat{j} + n \hat{k}) \), \( \delta \in \mathbb{R} \)
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

• A. \((-1, 7, 4)\)
• B. \((-1, -7, 4)\)
• C. \((1, 7, -4)\)
• D. \((1, -7, 4)\)

Answer 1

The Correct Option is A

Identify the direction vectors: \(L_1: \vec{d_1} = \hat{i} - \hat{j} + 2\hat{k}\); \(L_2: \vec{d_2} = 3\hat{i} + \hat{j} + p\hat{k}\); \(L_3: \vec{d_3} = \hat{i} + m\hat{j} - \hat{k}\)

Since \(L_1\) is perpendicular to \(L_2\), we have:

\(\vec{d_1} \times \vec{d_2} = 0\).

\((1)(3) + (-1)(1) + (2)(p) = 0 \Rightarrow 2 + 2p = 0 \Rightarrow p = -1.\)

Since \(L_3\) is perpendicular to both \(L_1\) and \(L_2\): For \(L_3\) perpendicular to \(L_1\):

\(\vec{d_3} \times \vec{d_1} = 0\).

\((1)(1) + (m)(-1) + (-1)(2) = 0 \Rightarrow -m = 1 \Rightarrow m = -1.\)

Substitute \(\delta = -1\) in \(\vec{r} = \delta(\hat{i} - \hat{j} - \hat{k})\) to find the point:

\((-1, 7, 4)\).