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: Equation of the first line. 
The vector equation of the line passing through \( A(1, 2, 3) \) and \( B(6, 8, 11) \) is: \[ \vec{r}_1 = \vec{A} + t(\vec{B} - \vec{A}) \] where \( \vec{A} = (1, 2, 3) \) and \( \vec{B} = (6, 8, 11) \). The direction vector \( \vec{B} - \vec{A} \) is: \[ \vec{B} - \vec{A} = (6 - 1, 8 - 2, 11 - 3) = (5, 6, 8) \] So the parametric equation of the first line is: \[ \vec{r}_1 = (1, 2, 3) + t(5, 6, 8) \] or in component form: \[ x_1 = 1 + 5t, \quad y_1 = 2 + 6t, \quad z_1 = 3 + 8t \] 

Step 2: Equation of the second line. 
The parametric form of the second line is given as: \[ \vec{r}_2 = (4, 1, 0) + \lambda(6, 2, 1) \] So the parametric equations for the second line are: \[ x_2 = 4 + 6\lambda, \quad y_2 = 1 + 2\lambda, \quad z_2 = \lambda \] 

Step 3: Find the point of intersection. 
At the point of intersection, the coordinates of the two lines must be equal. Therefore: \[ x_1 = x_2, \quad y_1 = y_2, \quad z_1 = z_2 \] From \( x_1 = x_2 \): \[ 1 + 5t = 4 + 6\lambda \quad \Rightarrow \quad 5t - 6\lambda = 3 \quad \text{(Equation 1)} \] From \( y_1 = y_2 \): \[ 2 + 6t = 1 + 2\lambda \quad \Rightarrow \quad 6t - 2\lambda = -1 \quad \text{(Equation 2)} \] From \( z_1 = z_2 \): \[ 3 + 8t = \lambda \quad \Rightarrow \quad 8t - \lambda = -3 \quad \text{(Equation 3)} \] We now solve the system of three equations: - Equation 1: \( 5t - 6\lambda = 3 \) - Equation 2: \( 6t - 2\lambda = -1 \) - Equation 3: \( 8t - \lambda = -3 \) 

Step 4: Solving the system of equations. From Equation 3: \[ \lambda = 8t + 3 \] Substitute \( \lambda = 8t + 3 \) into Equation 2: \[ 6t - 2(8t + 3) = -1 \] \[ 6t - 16t - 6 = -1 \] \[ -10t = 5 \quad \Rightarrow \quad t = -\frac{1}{2} \] Now substitute \( t = -\frac{1}{2} \) into \( \lambda = 8t + 3 \): \[ \lambda = 8\left(-\frac{1}{2}\right) + 3 = -4 + 3 = -1 \] 

Step 5: Find the point of intersection. Substitute \( t = -\frac{1}{2} \) into the parametric equations of the first line: \[ x_1 = 1 + 5\left(-\frac{1}{2}\right) = 1 - \frac{5}{2} = -\frac{3}{2} \] \[ y_1 = 2 + 6\left(-\frac{1}{2}\right) = 2 - 3 = -1 \] \[ z_1 = 3 + 8\left(-\frac{1}{2}\right) = 3 - 4 = -1 \] Hence, the coordinates of the point of intersection are: \[ \left(-\frac{3}{2}, -1, -1\right) \] 

Step 6: Find the equation of the line passing through the point of intersection and perpendicular to both lines. The direction vector of the line passing through the point of intersection and perpendicular to both lines is given by the cross product of the direction vectors of the two lines. The direction vector of the first line is \( \vec{d_1} = (5, 6, 8) \) and the direction vector of the second line is \( \vec{d_2} = (6, 2, 1) \). The cross product \( \vec{d_1} \times \vec{d_2} \) is: \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 6 & 8 \\ 6 & 2 & 1 \end{vmatrix} \] \[ = \hat{i}(6 \cdot 1 - 8 \cdot 2) - \hat{j}(5 \cdot 1 - 8 \cdot 6) + \hat{k}(5 \cdot 2 - 6 \cdot 6) \] \[ = \hat{i}(6 - 16) - \hat{j}(5 - 48) + \hat{k}(10 - 36) \] \[ = \hat{i}(-10) - \hat{j}(-43) + \hat{k}(-26) \] \[ = (-10, 43, -26) \] Therefore, the direction vector of the perpendicular line is \( (-10, 43, -26) \). The parametric equations of the line passing through the point of intersection \( \left(-\frac{3}{2}, -1, -1\right) \) and having direction vector \( (-10, 43, -26) \) are: \[ x = -\frac{3}{2} - 10t, \quad y = -1 + 43t, \quad z = -1 - 26t \] 

Final Answer: 
The coordinates of the point of intersection are \( \left(-\frac{3}{2}, -1, -1\right) \). The equation of the line passing through this point and perpendicular to both lines is: \[ x = -\frac{3}{2} - 10t, \quad y = -1 + 43t, \quad z = -1 - 26t \]