Question

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^\circ$ about its line of intersection with the plane $x + y + z = 1$. The equation of the plane in the new position is:

• A. 3x+9y +z+17=0
• B. 3x+9y +z =17
• C. 3x−9y −z =17
• D. 3x+9y −z =17

Hint:

When rotating a plane about the line of intersection with another plane, use the rotation matrix for 3D space to apply the transformation and obtain the new plane equation.

Answer 1

The Correct Option is B

Step 1: To find the equation of the plane after a 90° rotation about the line of intersection, we first need to find the equation of the line of intersection between the two given planes.

Step 2: The planes are \(2x - y + 3z + 5 = 0\) and \(x + y + z = 1\). The line of intersection can be found by solving these two plane equations simultaneously.

Step 3: Solve the system of equations for two of the variables in terms of the third. After substitution and solving, we find the parametric equations for the line of intersection.

Step 4: Next, we apply the 90° rotation using the rotation matrix for 3D space. The rotation matrix for rotating around the line of intersection is derived from the axis of the line and the angle of rotation.

Step 5: Apply the rotation transformation to the plane equation, and simplify to get the new equation \(3x + 9y + z = 17\).

Answer 2

Equation of a Plane Rotated Through 90 Degrees

We are given two planes: $P_1: 2x - y + 3z + 5 = 0$ and $P_2: x + y + z - 1 = 0$. The plane $P_1$ is rotated through $90^\circ$ about its line of intersection with the plane $x + y + z = 1$. We need to find the equation of the plane in its new position.

Step 1: Equation of a plane passing through the line of intersection

The equation of a plane passing through the line of intersection of $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$:

$(2x - y + 3z + 5) + \lambda (x + y + z - 1) = 0$

$(2 + \lambda)x + (-1 + \lambda)y + (3 + \lambda)z + (5 - \lambda) = 0$

Step 2: The new plane is perpendicular to the original plane

The plane is rotated through $90^\circ$ about its line of intersection with $x + y + z = 1$. This implies the new position of the plane is perpendicular to the original plane $2x - y + 3z + 5 = 0$.

Step 3: Find the normal vectors of the planes

The normal vector to the original plane $2x - y + 3z + 5 = 0$ is $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$.

The normal vector to the new plane $(2 + \lambda)x + (-1 + \lambda)y + (3 + \lambda)z + (5 - \lambda) = 0$ is $\mathbf{n}_2 = \begin{pmatrix} 2 + \lambda \\ -1 + \lambda \\ 3 + \lambda \end{pmatrix}$.

Step 4: Use the condition for perpendicularity of planes

Since the planes are perpendicular, their normal vectors are perpendicular, and their dot product is zero:

$\mathbf{n}_1 \cdot \mathbf{n}_2 = (2)(2 + \lambda) + (-1)(-1 + \lambda) + (3)(3 + \lambda) = 0$

$4 + 2\lambda + 1 - \lambda + 9 + 3\lambda = 0$

$14 + 4\lambda = 0$

$4\lambda = -14$

$\lambda = -\frac{7}{2}$

Step 5: Substitute the value of $\lambda$ into the equation of the new plane

$(2 - \frac{7}{2})x + (-1 - \frac{7}{2})y + (3 - \frac{7}{2})z + (5 - (-\frac{7}{2})) = 0$

$(\frac{4 - 7}{2})x + (\frac{-2 - 7}{2})y + (\frac{6 - 7}{2})z + (\frac{10 + 7}{2}) = 0$

$-\frac{3}{2}x - \frac{9}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$

Step 6: Simplify the equation

Multiply the equation by $-2$ to eliminate fractions and the negative sign for the $x$ term:

$3x + 9y + z - 17 = 0$

$3x + 9y + z = 17$

The equation of the plane in its new position is $3x + 9y + z = 17$.

Final Answer: (B) $3x + 9y + z = 17$