Question

If the lines \( \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2} \) and \( \frac{x - 1}{3k} = \frac{y - 1}{1} = \frac{z - 6}{-7} \) are perpendicular to each other, find the value of \( k \) and hence write the vector equation of a line perpendicular to these two lines passing through the point \( (3, -4, 7) \).

Hint:

For perpendicular lines, use the dot product of their direction vectors to solve for unknown parameters.

Answer 1

Step 1: Determine the value of \( k \)
The direction ratios for the first line are \( \langle -3, 2k, 2 \rangle \) and for the second line are \( \langle 3k, 1, -7 \rangle \). Since the lines are perpendicular, their dot product is zero:\[(-3)(3k) + (2k)(1) + (2)(-7) = 0\]Solving for \( k \):\[-9k + 2k - 14 = 0 \implies -7k = 14 \implies k = -2\]Step 2: Calculate the vector equation of the perpendicular line
With \( k = -2 \), the direction vectors are:\[\vec{b}_1 = \langle -3, -4, 2 \rangle, \quad \vec{b}_2 = \langle -6, 1, -7 \rangle\]The vector perpendicular to both \( \vec{b}_1 \) and \( \vec{b}_2 \) is found using the cross product:\[\vec{b} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-3 & -4 & 2
-6 & 1 & -7 \end{vmatrix} = 26\hat{i} - 33\hat{j} - 27\hat{k}\]The equation of the required line is:\[\vec{r} = \langle 3, -4, 7 \rangle + \lambda(26\hat{i} - 33\hat{j} - 27\hat{k})\]