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:

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

Answer 2

To solve this problem, we need to determine a point lying on the line \( L_3 \), given the conditions on \( L_1, L_2, \) and \( L_3 \).

1. Understand the given lines:
   • Line \( L_1 \): Given by the vector equation \( \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda (\hat{i} - \hat{j} + 2\hat{k}) \). The direction vector is \( \vec{a_1} = \hat{i} - \hat{j} + 2\hat{k} \).
   • Line \( L_2 \): Given by \( \vec{r} = (\hat{j} - \hat{k}) + \mu (3\hat{i} + \hat{j} + p\hat{k}) \). The direction vector is \( \vec{a_2} = 3\hat{i} + \hat{j} + p\hat{k} \).
   • Line \( L_3 \): Given by \( \vec{r} = \delta (\ell \hat{i} + m \hat{j} + n \hat{k}) \). The direction vector is \( \vec{a_3} = \ell \hat{i} + m \hat{j} + n \hat{k} \).
2. Perpendicularity conditions:
   • Line \( L_1 \) is perpendicular to \( L_2 \) implies \( \vec{a_1} \cdot \vec{a_2} = 0 \). \((\hat{i} - \hat{j} + 2\hat{k}) \cdot (3\hat{i} + \hat{j} + p\hat{k})\)
     • Calculate the dot product: \(1 \times 3 + (-1) \times 1 + 2 \times p = 0\)
     • Simplifying gives: \(2p + 2 = 0 \Rightarrow p = -1\)
   • Line \( L_3 \) is perpendicular to both \( L_1 \) and \( L_2 \).
     • Perpendicular to \( L_1 \): \((\ell, m, n) \cdot (1, -1, 2) = 0 \Rightarrow \ell - m + 2n = 0\)
     • Perpendicular to \( L_2 \): \((\ell, m, n) \cdot (3, 1, -1) = 0 \Rightarrow 3\ell + m - n = 0\)
3. Solve the equations:
   • From \( \ell - m + 2n = 0 \) and \( 3\ell + m - n = 0 \), solve:
     • First Equation: \(m = \ell + 2n\)
     • Substitute in Second Equation: \(3\ell + (\ell + 2n) - n = 0 \Rightarrow 4\ell + n = 0 \Rightarrow n = -4\ell\)
   • Substitute \( n = -4\ell \) into \( m = \ell + 2n \): \(m = \ell + 2(-4\ell) = \ell - 8\ell = -7\ell\)
   • Thus, \((\ell, m, n) = (\ell, -7\ell, -4\ell) = \ell(1, -7, -4)\)
4. Choose a point on \( L_3 \):
   • A point on \( L_3 \) for \(\ell = -1\) is \((-1, 7, 4)\), matches with option \((-1, 7, 4)\).

Therefore, the point which lies on \( L_3 \) is \((-1, 7, 4)\).