Question

The equation of plane which is passing through \((2,1,3)\) and having equal \(X\) and Y-intercept and \(Z\) -intercept 14 is (A) \(11 \mathrm{x}-11 \mathrm{y}+3 \mathrm{z}=42\) (B) \(11 \mathrm{x}+11 \mathrm{y}+3 z=42\) (C) \(11 \mathrm{x}+11 \mathrm{y}-3 \mathrm{z}=42\) (D) \(11 \mathrm{x}+11 \mathrm{y}+3 \mathrm{z}+42=0\)

Short answer

The short answer based on the step-by-step solution is: To find the equation of the plane, calculate the normal vector using the given intercepts, which results in \(\mathbf{n} = \langle 14A, -14A, -A^2 \rangle\). Then, use the point-normal form of the plane equation and plug in the given point \((2,1,3)\) and the normal vector components. Simplify the equation to get the final answer: \(11x - 11y + 3z = 42\). Therefore, the correct choice is (A).

Step-by-step solution

Determine the normal vector

We are given the plane's intercepts, which are the points at which the plane intersects the axes. Based on the information, the X and Y intercepts are equal, meaning the points are \((A,A,0)\) and \((0,A,0)\). Since the Z-intercept is given as 14, the point \((0,0,14)\). As we know, X and Y intercepts are equal, so let it be A. Then, the points are \((A, 0, 0), (0, A, 0), (0, 0, 14)\). Consider the vectors formed by any two points, say by points \((A, 0, 0)\) and \((0, A, 0)\), we get the vector \(\mathbf{u} = \langle -A, A, 0 \rangle\). And, for points \((A, 0, 0)\) and \((0, 0, 14)\), we get the vector \(\mathbf{v} = \langle -A, 0, 14 \rangle\). Calculate the cross product of these vectors to find the normal vector \(\mathbf{n}\).

Calculate the cross product

To find the normal vector, we need to compute the cross product of \(\mathbf{u}\) and \(\mathbf{v}\): \[\mathbf{n} = \mathbf{u} \times \mathbf{v} = \begin{bmatrix} i & j & k \\ -A & A & 0 \\ -A & 0 & 14 \end{bmatrix}\] Expanding the determinant, we get: \[\mathbf{n} = \langle 14A, -14A, -A^2 \rangle\]

Find the equation of the plane

Using the point-normal form of the plane equation: \[A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\] Plug in the point \((2,1,3)\) as \((x_0, y_0, z_0)\) and the components of the normal vector, \(\langle 14A, -14A, -A^2 \rangle\), as \((A, B, C)\): \[14A(x - 2) - 14A (y - 1) - A^2(z - 3) = 0\]

Simplify the equation and identify the answer

Simplify the equation by factoring out common terms to find the answer: \[14A (x - y) - A^2(z - 3) = 0\] Now let's divide the equation by A: \[14(x - y) - Az = 42\] Comparing this equation to the given choices, the correct answer is: (A) \(11x - 11y + 3z = 42\)

Key concepts

Cross Product

When working with vectors in 3-dimensional space, the cross product is a useful tool.
It helps calculate the normal vector to a plane given two directional vectors lying on that plane.
For two vectors, \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), their cross product \( \mathbf{u} \times \mathbf{v} \) is defined as:

• \( (u_2v_3 - u_3v_2)i \)
• \( -(u_1v_3 - u_3v_1)j \)
• \( (u_1v_2 - u_2v_1)k \)

Essentially, this operation computes a vector that is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \), making it invaluable in finding the normal vector to a plane.
Keep in mind, the cross product also follows the right-hand rule, meaning the generated normal vector points in the direction that a screw would move if turned in the direction from \( \mathbf{u} \) to \( \mathbf{v} \).
This is particularly important when dissecting orientations and directions in space.

Normal Vector

A normal vector is a vector that is perpendicular to a plane or surface.
In the context of planes in three-dimensional space, the normal vector plays a crucial role in defining the plane's orientation.
It is derived often via the cross product, as shown in previous sections, which gives a vector perpendicular to two others lying on the plane.

In the plane equation \( Ax + By + Cz = D \), the coefficients \( A \), \( B \), and \( C \) of \( x \), \( y \), and \( z \) respectively, represent components of the plane's normal vector \( \mathbf{n} = \langle A, B, C \rangle \).
This vector remains constant across every point of the plane and helps us understand the tilt or angle of the plane in 3D space.

Intercepts

Intercepts are points where a plane intersects the coordinate axes.
They are a straightforward method to visualize the orientation and position of a plane within the coordinate system.
The X, Y, and Z-intercepts of a plane tell us directly at what points the plane crosses each of the axes.

For example, a point \((A, 0, 0)\) would be the X-intercept, signifying where the plane crosses the X-axis.
Similarly, \((0, A, 0)\) is the Y-intercept, showing the intersection with the Y-axis, and \((0, 0, Z)\) represents the Z-intercept with value \( Z \), showing where it crosses the Z-axis.
These intercepts are particularly helpful when solving for unknown values in equations where planes are involved, as they offer convenient reference points on the axes.

Point-Normal Form

The point-normal form of the equation of a plane is an insightful structure that provides a clear method of determining the plane's equation based on its normal vector.
This form is given by:

• \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]

where \((x_0, y_0, z_0)\) is a point on the plane and \((A, B, C)\) are the components of the plane's normal vector.

The beauty of the point-normal form is that it connects the geometric aspects of the plane (a point and a normal direction) to its algebraic character.
This relationship allows for relatively straightforward manipulation to find the explicit linear equation of the plane.
Furthermore, it serves as a foundation for transitioning to other forms, such as the general form or intercept form by solving and simplifying the equation appropriately.