Question

A line passing through the points \(A(1,2,3)\) and \(B(6,8,11)\) intersects the line \[ \vec r = 4\hat i + \hat j + \lambda(6\hat i + 2\hat j + \hat k) \] Find the coordinates of the point of intersection. Hence write the equation of a line passing through the point of intersection and perpendicular to both the lines.

Hint:

To find a line perpendicular to two given lines in 3D, take the cross product of their direction vectors. This gives a direction vector perpendicular to both.

Answer 1

Step 1: Write the equation of line AB.
Direction vector of line \(AB\): \[ \vec{AB} = (6-1,\,8-2,\,11-3) \] \[ = (5,6,8) \] Thus parametric form: \[ x = 1 + 5t \] \[ y = 2 + 6t \] \[ z = 3 + 8t \] Step 2: Write the second line in parametric form.
Given \[ \vec r = (4,1,0) + \lambda(6,2,1) \] Thus \[ x = 4 + 6\lambda \] \[ y = 1 + 2\lambda \] \[ z = \lambda \] Step 3: Find intersection point.
At intersection both coordinates must match. From \(z\): \[ 3 + 8t = \lambda \] Substitute into \(y\): \[ 2 + 6t = 1 + 2(3 + 8t) \] \[ 2 + 6t = 1 + 6 + 16t \] \[ 2 + 6t = 7 + 16t \] \[ -10t = 5 \] \[ t = -\frac12 \] Now \[ \lambda = 3 + 8(-1/2) \] \[ \lambda = -1 \] Step 4: Coordinates of intersection.
Substitute \(t=-\frac12\): \[ x = 1 + 5(-1/2) = -\frac{3}{2} \] \[ y = 2 + 6(-1/2) = -1 \] \[ z = 3 + 8(-1/2) = -1 \] Thus point of intersection is \[ \left(-\frac{3}{2},-1,-1\right) \] Step 5: Find a line perpendicular to both lines.
Direction vectors: First line: \[ \vec d_1 = (5,6,8) \] Second line: \[ \vec d_2 = (6,2,1) \] Required direction is \[ \vec d = \vec d_1 \times \vec d_2 \] \[ = \begin{vmatrix} \hat i & \hat j & \hat k
5 & 6 & 8
6 & 2 & 1 \end{vmatrix} \] \[ = \hat i(6\cdot1-8\cdot2) - \hat j(5\cdot1-8\cdot6) + \hat k(5\cdot2-6\cdot6) \] \[ = \hat i(-10) + \hat j(43) + \hat k(-26) \] Thus direction vector \[ (-10,43,-26) \] Step 6: Write equation of required line.
Passing through \[ \left(-\frac{3}{2},-1,-1\right) \] Equation of required line: \[ \frac{x+\frac{3}{2}}{-10}=\frac{y+1}{43}=\frac{z+1}{-26} \] Final Answer:
Point of intersection: \[ \left(-\frac{3}{2},-1,-1\right) \] Equation of required line: \[ \frac{x+\frac{3}{2}}{-10}=\frac{y+1}{43}=\frac{z+1}{-26} \]