Question

Find the value of \( p \) for which the lines \[ \vec{r_1} = \hat{i} + (2\lambda + 1) \hat{j} + (3\lambda + 2) \hat{k} \] and \[ \vec{r_2} = \hat{i} - 3 \hat{j} + (p\mu + 7) \hat{k} \] are perpendicular to each other and also intersect. Also, find the point of intersection of the given lines.

Hint:

To determine if lines are perpendicular, use the dot product of their direction ratios. For intersection, set the position vectors equal and solve component-wise to find the unknowns.

Answer 1

Let us first calculate the direction ratios of the given lines. For the line \( \vec{r_1} \), the direction ratios are: \[ \vec{d_1} = \langle 1, 2, 3 \rangle. \] For the line \( \vec{r_2} \), the direction ratios are: \[ \vec{d_2} = \langle 1, -3, p\mu + 7 \rangle. \] Step 1: Condition for perpendicularity. The two lines are perpendicular if the dot product of their direction ratios is zero: \[ \vec{d_1} \cdot \vec{d_2} = 0. \] So: \[ 1 \cdot 1 + 2 \cdot (-3) + 3 \cdot (p + 7) = 0. \] Simplify: \[ 1 - 6 + 3(p + 7) = 0. \] Expand and solve for \( p \): \[ 1 - 6 + 3p + 21 = 0 \quad \Rightarrow \quad 3p + 16 = 0 \quad \Rightarrow \quad p = -\frac{16}{3}. \] Step 2: Condition for intersection. For the lines to intersect, their position vectors must be equal: \[ \vec{r_1} = \vec{r_2}. \] Equating components: \[ 1 = 1, \quad 2\lambda + 1 = -3, \quad 3\lambda + 2 = p\mu + 7. \] Solve the second equation: \[ 2\lambda + 1 = -3 \quad \Rightarrow \quad \lambda = -2. \] Substitute \( \lambda = -2 \) into the third equation: \[ 3(-2) + 2 = p\mu + 7 \quad \Rightarrow \quad -6 + 2 = p\mu + 7 \quad \Rightarrow \quad p\mu = -11. \] Step 3: Final result. The value of \( p \) is: \[ \boxed{-11}. \] The point of intersection can be determined using \( \lambda = -2 \) and the corresponding values of \( \mu \).