Question

Lines \( \mathbf{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda (2\hat{i} - 2\hat{j} + \hat{k}) \) and \( \mathbf{r} = (4\hat{i} - 3\hat{j} + 2\hat{k}) + \mu (\hat{i} - 2\hat{j} + 2\hat{k}) \) are coplanar. Find the equation of the plane determined by them.

Hint:

For coplanar lines, the plane’s normal is the cross product of direction vectors; use a point on one line.

Answer 1

Lines are coplanar if the vector joining a point on one to a point on the other, and the direction vectors, are coplanar. Points: \( \mathbf{a}_1 = (1, 1, -1) \), \( \mathbf{a}_2 = (4, -3, 2) \). Direction vectors: \( \mathbf{b}_1 = (2, -2, 1) \), \( \mathbf{b}_2 = (1, -2, 2) \).
Coplanarity condition: \[ (\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) = 0. \] \[ \mathbf{a}_2 - \mathbf{a}_1 = (4-1, -3-1, 2-(-1)) = (3, -4, 3). \] \[ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2 \end{vmatrix} = \hat{i}(-4 - (-2)) - \hat{j}(4 - 1) + \hat{k}(-4 - (-2)) = (-2, -3, -2). \] \[ (3, -4, 3) \cdot (-2, -3, -2) = -6 + 12 - 6 = 0. \] Lines are coplanar. Plane equation: Normal is \( \mathbf{b}_1 \times \mathbf{b}_2 = (-2, -3, -2) \). Use point (1, 1, -1): \[ -2(x - 1) - 3(y - 1) - 2(z - (-1)) = 0 \Rightarrow -2x + 2 - 3y + 3 - 2z - 2 = 0 \Rightarrow 2x + 3y + 2z = 3. \] Answer: \( 2x + 3y + 2z = 3 \).