Question

The equation of plane which is perpendicular to the planes \(3 \mathrm{x}+\mathrm{y}+\mathrm{z}=0\) and \(\mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=5\) and passing through \((1,3,5)\) is (A) \(x+2 y \bar{z}=0\) (B) \(x-2 y-z=0\) (C) \(x-2 y+z=0\) (D) \(x+2 y-z=0\)

Short answer

The correct answer for the described problem, \(-8x + z + 3 = 0\), is not provided in the given options. The question seems to be invalid.

Step-by-step solution

Find normal vectors for both planes

The normal vector to any plane can be found from the coefficients of x, y, and z in the equation of that plane. For plane \(3x+y+z=0\), the normal vector is \(\textbf{n}_1 = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}\) For plane \(x+2y+3z=5\), the normal vector is \(\textbf{n}_2 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\)

Find the normal vector to our desired plane by taking the cross product of \(\textbf{n}_1\) and \(\textbf{n}_2\)

The cross product of \(\textbf{n}_1\) and \(\textbf{n}_2\) will give the normal vector to the plane that's perpendicular to both planes: \(\textbf{n} = \textbf{n}_1 \times \textbf{n}_2 = \begin{bmatrix} -(1(1)-3(3)) \\ 3(1)-1(3) \\ 1(2)-1(1) \end{bmatrix} = \begin{bmatrix} -8 \\ 0 \\ 1 \end{bmatrix}\)

Plug the normal vector and given point \((1, 3, 5)\) into the equation of a plane

We can now form the general equation of the plane using the normal vector and given point. The equation of a plane in normal form is given by: \(A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\) where (A, B, C) is the normal vector, and \((x_0, y_0,z_0)\) is the given point. By substituting \(\textbf{n} = \begin{bmatrix} -8 \\ 0 \\ 1 \end{bmatrix}\) and \((1, 3, 5)\) into the equation, we obtain: \(-8(x-1) + 0(y-3) + 1(z-5) = 0\) Solving the equation, we get: \(-8x+8+z-5=0\) \(-8x+z = -3\) The final equation can be written as: \(-8x + z + 3 = 0\) Now, let's check the given options. (A) \(x+2 y \bar{z}=0\) (B) \(x-2 y-z=0\) (C) \(x-2 y+z=0\) (D) \(x+2 y-z=0\) None of these options match our result, and the question seems to be invalid. The options given do not contain the correct answer for the described problem.

Key concepts

Normal Vector

Understanding the concept of a normal vector is fundamental when dealing with plane equations in three-dimensional space. A normal vector is a vector that is perpendicular to the surface of a plane. It essentially defines the plane's orientation within the space. For example, if you have a wall, a normal vector would point straight out from it, defining the invisible line at a right angle to the wall's surface.

When we talk about plane equations like \(3x+y+z=0\), the coefficients of \(x\), \(y\), and \(z\) in the equation give us the components of the normal vector. In this specific case, the normal vector is \(\textbf{n}_1 = \begin{bmatrix} 3 \ 1 \ 1 \end{bmatrix}\). This vector is crucial because it not only helps define the position of the plane in space but also, as we'll see in a moment, interacts with other vectors to define new planes.

Cross Product

The cross product of two vectors in three-dimensional space results in another vector that is perpendicular to both of the original vectors. This is a key concept when trying to find a plane that is perpendicular to two given planes. By taking the cross product of the normal vectors of the two planes, we obtain a vector that is normal to both, therefore also normal to the plane which is perpendicular to the original two planes.

The cross product is calculated using the determinant of a matrix composed of the unit vectors \(i, j, k\) along with the components of the two vectors being multiplied. If we have normal vectors \(\textbf{n}_1\) and \(\textbf{n}_2\), their cross product \(\textbf{n}_1 \times \textbf{n}_2\) will be a vector \(\textbf{n}\), which can then be used to determine the equation of the new plane.

Equation of a Plane

The general form of the equation of a plane in three dimensions includes a normal vector and a point through which the plane passes. The formula is expressed as \(A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\), where \(A\), \(B\), and \(C\) are the components of the normal vector, and \((x_0, y_0, z_0)\) represents a point on the plane.

In the exercise example, after finding the normal vector \(\textbf{n}\) using the cross product method described earlier, we plug in the coordinates of the given point through which the plane passes to find its specific equation. This ties together the concepts of normal vector and cross product to produce the equation of a plane. The exercise solution steps show the practical application of this theory to arrive at the plane's equation, which turns out to be different from all provided options, indicating a potential error in the given multiple-choice answers.