Question

Find the equation of the plane through \((2,-3,2)\) and parallel to the plane of the vectors \(4 \mathbf{i}+3 \mathbf{j}-\mathbf{k}\) and \(2 \mathbf{i}-5 \mathbf{j}+6 \mathbf{k}\).

Short answer

The equation of the plane is \(13x - 22y - 26z - 40 = 0\).

Step-by-step solution

Identify the Normal Vector of the Given Plane

The coefficients of the vectors can help identify a normal vector to the plane. Here, the plane is given indirectly via two vectors that are parallel to it. The cross-product of these vectors will give a vector normal to the plane. Given vectors are \( \mathbf{v_1} = 4 \mathbf{i} + 3 \mathbf{j} - \mathbf{k} \) and \( \mathbf{v_2} = 2 \mathbf{i} - 5 \mathbf{j} + 6 \mathbf{k} \).

Calculate the Cross Product

Calculate the cross product \( \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} \). Using the determinant:\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 3 & -1 \ 2 & -5 & 6 \end{vmatrix} \]Evaluate the determinant, giving:\( \mathbf{n} = (3 \cdot 6 - (-1) \cdot (-5))\mathbf{i} - (4 \cdot 6 - (-1) \cdot 2)\mathbf{j} + (4 \cdot (-5) - 2 \cdot 3)\mathbf{k} \).Simplify to obtain \( \mathbf{n} = (18 - 5)\mathbf{i} - (24 - 2)\mathbf{j} + (-20 - 6)\mathbf{k} \).\( \Rightarrow \mathbf{n} = 13\mathbf{i} - 22\mathbf{j} - 26\mathbf{k} \).

Use Point-Normal Form to Find the Plane Equation

With the normal vector \( \mathbf{n} = 13\mathbf{i} - 22\mathbf{j} - 26\mathbf{k} \) and the point \((2, -3, 2)\), the equation of the plane is given by the point-normal form:\[ 13(x - 2) - 22(y + 3) - 26(z - 2) = 0 \]

Simplify the Equation

Expand the equation:\( 13x - 26 - 22y - 66 - 26z + 52 = 0 \).Combine like terms to obtain the simplified equation:\[ 13x - 22y - 26z - 40 = 0 \].

Key concepts

Cross Product

To solve problems involving planes, especially when given two vectors parallel to the plane, the cross product is a valuable tool. The cross product of two vectors results in a third vector that is perpendicular (or normal) to the originally given vectors.
For example, if you are given two vectors parallel to a plane, like \( \mathbf{v_1} = 4 \mathbf{i} + 3 \mathbf{j} - \mathbf{k}\) and \( \mathbf{v_2} = 2 \mathbf{i} - 5 \mathbf{j} + 6 \mathbf{k}\), calculating the cross product \( \mathbf{v_1} \times \mathbf{v_2} \) gives us a new vector \( \mathbf{n} \), which is normal to the plane.
The cross product is computed using the determinant:

• Arrange the components of \( \mathbf{v_1} \) and \( \mathbf{v_2} \) in a 3x3 matrix.
• Use this matrix to find each component of the resulting vector.

This operation is crucial because the perpendicular vector \( \mathbf{n} \) helps establish the equation of the plane.

Normal Vector

The normal vector is a key element when defining the equation of a plane. It is a vector that is perpendicular to every line on the plane.
In our context, once we obtain the normal vector \( \mathbf{n} \) from the cross product, it provides a direction that is orthogonal to the plane itself. Consider it as a tool showing how the plane is oriented in space.
For our exercise, the computed normal vector is \( 13\mathbf{i} - 22\mathbf{j} - 26\mathbf{k}\). This vector becomes part of the point-normal form of the plane's equation, an essential part of expressing the plane mathematically. Understanding the normal vector's role makes solving for the plane more straightforward and connects the vectors defining the plane to its geometric representation.

Parallel Planes

Understanding parallel planes involves recognizing planes that never intersect and maintain a constant distance from each other.
This occurs in our problem as we seek a plane parallel to another known plane made up of linearly independent vectors.
Here, the given problem specifies vectors \( 4 \mathbf{i} + 3 \mathbf{j} - \mathbf{k} \) and \( 2 \mathbf{i} - 5 \mathbf{j} + 6 \mathbf{k} \). While these vectors define a plane, the problem asks us to find another plane parallel to this existing plane.
To be parallel, two planes must share the same normal vector. Hence, taking the cross product provides this shared vector \( \mathbf{n} \), applying it to the new plane through a different point, ensuring the two planes do not intersect but rather are equidistant apart.

Point-Normal Form

The point-normal form is the go-to equation format when describing a plane using a point on the plane and a normal vector.
Its general formula is
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane and \( a, b, c \) are the components of the normal vector. This format was applied in our solution to find the plane that goes through the point \((2, -3, 2)\) and is parallel to the given vectors.
Our practical application results in the equation:

• Substitute the point and normal vector into the formula.
• Simplify to get \[ 13(x - 2) - 22(y + 3) - 26(z - 2) = 0 \].
• Further simplification sets the final plane equation as \[ 13x - 22y - 26z - 40 = 0 \].

This describes the desired plane in space using the given point and the computed normal vector.