Question

Lines \[ \overrightarrow{r} = (i + j - k) + \lambda(2i - 2j + k), \quad \overrightarrow{r} = (4i - 3j + 2k) + \mu(i - 2j + 2k) \] are coplanar. Find the equation of the plane determined by them.

Hint:

To find the equation of a plane determined by two lines, compute the cross product of their direction vectors to get the normal vector.

Answer 1

Step 1: The direction ratios of the two lines are \( \overrightarrow{l_1} = \langle 2, -2, 1 \rangle \) and \( \overrightarrow{l_2} = \langle 1, -2, 2 \rangle \), and the position vector of a point on the first line is \( \overrightarrow{r_1} = \langle 1, 1, -1 \rangle \), and the position vector of a point on the second line is \( \overrightarrow{r_2} = \langle 4, -3, 2 \rangle \). Step 2: To find the equation of the plane, we need the normal vector, which is the cross product of the direction ratios of the two lines: \[ \overrightarrow{n} = \overrightarrow{l_1} \times \overrightarrow{l_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -2 & 1
1 & -2 & 2 \end{vmatrix} \] \[ = \hat{i} \begin{vmatrix} -2 & 1
-2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1
1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2
1 & -2 \end{vmatrix} \] \[ = \hat{i} \left( (-2)(2) - (1)(-2) \right) - \hat{j} \left( (2)(2) - (1)(1) \right) + \hat{k} \left( (2)(-2) - (-2)(1) \right) \] \[ = \hat{i} (-4 + 2) - \hat{j} (4 - 1) + \hat{k} (-4 + 2) \] \[ = -2\hat{i} - 3\hat{j} - 2\hat{k} \] Thus, the normal vector to the plane is \( \overrightarrow{n} = \langle -2, -3, -2 \rangle \). Step 3: The equation of the plane is given by: \[ \overrightarrow{n} \cdot (\overrightarrow{r} - \overrightarrow{r_1}) = 0 \] Substitute the values of \( \overrightarrow{n} = \langle -2, -3, -2 \rangle \) and \( \overrightarrow{r_1} = \langle 1, 1, -1 \rangle \): \[ -2(x - 1) - 3(y - 1) - 2(z + 1) = 0 \] Simplifying: \[ -2x + 2 - 3y + 3 - 2z - 2 = 0 \] \[ -2x - 3y - 2z + 3 = 0 \] Thus, the equation of the plane is: \[ \boxed{2x + 3y + 2z = 3} \]