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 also perpendicular. This means that the dot product of their direction vectors must be zero. 

Step 1: Determine the direction vectors. The direction vector of $\vec{r}_1$ is: $$\vec{d}_1 = 2\hat{i} + p\hat{j} + 5\hat{k}.$$ The direction vector of $\vec{r}_2$ is: $$\vec{d}_2 = 3\hat{i} - p\hat{j} + p\hat{k}.$$

 Step 2: Calculate the dot product. The dot product of $\vec{d}_1$ and $\vec{d}_2$ is: $$\vec{d}_1 \cdot \vec{d}_2 = (2)(3) + (p)(-p) + (5)(p).$$ Simplifying the expression: $$\vec{d}_1 \cdot \vec{d}_2 = 6 - p^2 + 5p.$$ 

Step 3: Set the dot product equal to zero. Since the lines are perpendicular, the dot product must be zero: $$6 - p^2 + 5p = 0.$$ 

Step 4: Solve the quadratic equation. Rearrange the terms: $$p^2 - 5p - 6 = 0.$$ Factor the quadratic expression: $$(p - 6)(p + 1) = 0.$$ Thus, the possible solutions for $p$ are: $$p = 6 \quad \text{or} \quad p = -1.$$ 

Step 5: Confirm the solution. Both $p = 6$ and $p = -1$ satisfy the condition of perpendicularity. In this case, the solution required is $p = 6$. Final Answer: $$\boxed{6}$$