Question

Find the equation of the plane which passes through the intersecting point of the planes \[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 \, \text{and} \, \vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5, \] and the point \( (1, 1, 1) \).

Hint:

To find the equation of a plane passing through the intersection of two planes, take the cross product of their normal vectors and use a point on the plane to determine the constant \( D \).

Answer 1

The equation of a plane can be written as: \[ \vec{r} \cdot \vec{n} = D, \] where \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) is the position vector, \( \vec{n} \) is the normal vector to the plane, and \( D \) is the constant. We are given two planes: \[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 \text{(Plane 1)}, \] \[ \vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5 \text{(Plane 2)}. \] The point of intersection of the two planes lies on both planes. Hence, the normal vector to the required plane will be the cross product of the normal vectors of Plane 1 and Plane 2.

Step 1: Find the normal vector of the required plane. The normal vector of Plane 1 is \( \vec{n_1} = \hat{i} + \hat{j} + \hat{k} \), and the normal vector of Plane 2 is \( \vec{n_2} = 2\hat{i} + 3\hat{j} + 4\hat{k} \). The normal vector to the required plane is the cross product \( \vec{n_1} \times \vec{n_2} \): \[ \vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 3 & 4 \end{vmatrix}. \] This gives: \[ \vec{n} = \hat{i}(1 \cdot 4 - 1 \cdot 3) - \hat{j}(1 \cdot 4 - 1 \cdot 2) + \hat{k}(1 \cdot 3 - 1 \cdot 2), \] \[ \vec{n} = \hat{i}(4 - 3) - \hat{j}(4 - 2) + \hat{k}(3 - 2), \] \[ \vec{n} = \hat{i} - 2\hat{j} + \hat{k}. \]

Step 2: Find the equation of the plane. The equation of the plane passing through the point \( (1, 1, 1) \) is given by: \[ \vec{r} \cdot \vec{n} = D. \] Substitute \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) and \( \vec{n} = \hat{i} - 2\hat{j} + \hat{k} \) into the equation: \[ (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = D. \] Simplifying the dot product: \[ x - 2y + z = D. \] Now, substitute the point \( (1, 1, 1) \) into the equation to find \( D \): \[ 1 - 2(1) + 1 = D \Rightarrow D = 0. \] Thus, the equation of the plane is: \[ x - 2y + z = 0. \] 
Conclusion: The equation of the plane is: \[ \boxed{x - 2y + z = 0}. \]