Question

The line whose vector equations are \(\vec{r}_1 = 2\hat{i} - 3\hat{j} + 7\hat{k} + \lambda(2\hat{i} + p\hat{j} + 5\hat{k})\) and \(\vec{r}_2 = \hat{i} + 2\hat{j} + 3\hat{k} + \mu(3\hat{i} - p\hat{j} + p\hat{k})\) are perpendicular for all values of \(\lambda\) and \(\mu\). The value of \(p\) is:

• A. \( -1 \)
• B. \( 2 \)
• C. \( 5 \)
• D. \( 6 \)

Hint:

For two lines to be perpendicular, their direction vectors must satisfy a dot product of zero.

Answer 1

The Correct Option is D

Two lines are perpendicular if their direction vectors are orthogonal, meaning their dot product is zero.

Step 1: Identify the direction vectors. The direction vector for \(\vec{r}_1\) is: \[ \vec{d}_1 = 2\hat{i} + p\hat{j} + 5\hat{k}. \] The direction vector for \(\vec{r}_2\) is: \[ \vec{d}_2 = 3\hat{i} - p\hat{j} + p\hat{k}. \]

 Step 2: Compute the dot product. The dot product \(\vec{d}_1 \cdot \vec{d}_2\) is calculated as: \[ \vec{d}_1 \cdot \vec{d}_2 = (2)(3) + (p)(-p) + (5)(p). \] Simplifying yields: \[ \vec{d}_1 \cdot \vec{d}_2 = 6 - p^2 + 5p. \] 

Step 3: Equate the dot product to zero. For perpendicular lines, this condition must hold: \[ 6 - p^2 + 5p = 0. \] 

Step 4: Solve the resulting quadratic equation. Rearrange the equation: \[ p^2 - 5p - 6 = 0. \] Factor the quadratic: \[ (p - 6)(p + 1) = 0. \] The solutions for \( p \) are: \[ p = 6 \quad \text{or} \quad p = -1. \] 

Step 5: Verify the solutions. Both \( p = 6 \) and \( p = -1 \) satisfy the perpendicularity condition. The specific solution required is \( p = 6 \). Final Answer: \[ \boxed{6} \]