Question

Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are \( -2, 1, -1 \) and \( -3, -4, 1 \).

Hint:

To find the direction ratios of a vector perpendicular to two given lines, compute the cross product of their direction ratios.

Answer 1

Step 1: The direction ratios of the two lines are \( \mathbf{l_1} = (-2, 1, -1) \) and \( \mathbf{l_2} = (-3, -4, 1) \). Step 2: To find the direction ratios of the vector perpendicular to both lines, we take the cross product of \( \mathbf{l_1} \) and \( \mathbf{l_2} \). The cross product is given by: \[ \mathbf{l_1} \times \mathbf{l_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-2 & 1 & -1
-3 & -4 & 1 \end{vmatrix} \] Step 3: Compute the determinant: \[ \mathbf{l_1} \times \mathbf{l_2} = \hat{i} \begin{vmatrix} 1 & -1
-4 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} -2 & -1
-3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} -2 & 1
-3 & -4 \end{vmatrix} \] Step 4: Calculate the 2x2 determinants: \[ \hat{i} \left( (1)(1) - (-1)(-4) \right) = \hat{i} (1 - 4) = -3 \hat{i} \] \[ -\hat{j} \left( (-2)(1) - (-1)(-3) \right) = -\hat{j} (-2 - 3) = 5 \hat{j} \] \[ \hat{k} \left( (-2)(-4) - (1)(-3) \right) = \hat{k} (8 + 3) = 11 \hat{k} \] Step 5: The cross product is: \[ \mathbf{l_1} \times \mathbf{l_2} = -3 \hat{i} + 5 \hat{j} + 11 \hat{k} \] Thus, the direction ratios of the perpendicular vector are: \[ \boxed{-3, 5, 11} \]