Question

Find the equation of the plane through \((-1,-2,3)\) and perpendicular to both the planes \(x-3 y+2 z=7\) and \(2 x-2 y-z=-3\).

Short answer

The equation of the plane is \(7x + 5y + 4z = -5\).

Step-by-step solution

Find Normal Vectors

First, identify the normal vectors of the given planes. The normal vector to the plane \(x - 3y + 2z = 7\) is \(\mathbf{n_1} = \langle 1, -3, 2 \rangle\). The normal vector to the plane \(2x - 2y - z = -3\) is \(\mathbf{n_2} = \langle 2, -2, -1 \rangle\).

Compute Cross Product

The direction vector of the desired plane should be perpendicular to both \(\mathbf{n_1}\) and \(\mathbf{n_2}\). Thus, we compute the cross product: \(\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2}\). Calculating the cross product, \(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -3 & 2 \ 2 & -2 & -1 \end{vmatrix}\). Using the determinant, we find \(\mathbf{n} = \langle 7, 5, 4 \rangle\).

Use Point-Normal Form of Plane Equation

Now that we have the normal vector of the desired plane, and a point \((-1, -2, 3)\) through which the plane passes, use the point-normal form equation of the plane: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\). Here, \(a = 7\), \(b = 5\), \(c = 4\), and the point \((x_0, y_0, z_0) = (-1, -2, 3)\).

Write the Plane Equation

Substitute the values into the point-normal form equation: \(7(x + 1) + 5(y + 2) + 4(z - 3) = 0\). Simplifying the equation gives: \(7x + 7 + 5y + 10 + 4z - 12 = 0\). Finally, we combine like terms to get the equation: \(7x + 5y + 4z = -5\).

Key concepts

Normal Vectors

In the realm of plane geometry, a normal vector is a vector that is perpendicular to a plane. When you have the equation of a plane like \( ax + by + cz = d \), the coefficients \(a\), \(b\), and \(c\) form a vector \(\langle a, b, c \rangle\), which is the normal vector of that plane. This vector tells us the orientation of the plane in space. Understanding the normal vector is crucial because it allows us to determine how one plane sits in relation to another in three-dimensional space. If two planes are perpendicular to each other, their normal vectors will also be perpendicular. Conversely, if two normal vectors are parallel, the planes they describe are parallel too. In our problem, we found the normal vectors of two given planes \( \langle 1, -3, 2 \rangle \) and \( \langle 2, -2, -1 \rangle \). These vectors are the stepping stones to defining this new plane that is perpendicular to the planes described by the given equations.

Cross Product

To determine the normal vector of the desired plane, which is perpendicular to two other planes, we employ the cross product. The cross product of two vectors results in a third vector, which is perpendicular to the original two. The formula for the cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is:\[\mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle\]Calculating this for the normal vectors \( \langle 1, -3, 2 \rangle \) and \( \langle 2, -2, -1 \rangle \) gives us the new normal vector \( \langle 7, 5, 4 \rangle \). This is the normal vector of the new plane, ensuring it is perpendicular to the original two planes. The cross product is valuable in contexts where perpendicularity and orientation are crucial, as in the construction of the desired plane.

Point-Normal Form

Point-normal form is a fundamental concept for writing the equation of a plane. When you have a point that lies on the plane and a normal vector, constructing the equation of the plane becomes straightforward.The point-normal form equation is written as:\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]Here, \( (x_0, y_0, z_0) \) is a known point on the plane and \( a, b, c \) are the components of the normal vector of the plane. For our scenario, the point \((-1, -2, 3)\) lies on the plane and the normal vector is \( \langle 7, 5, 4 \rangle \). Plugging these values into the formula helps derive the equation of the plane: \(7(x + 1) + 5(y + 2) + 4(z - 3) = 0\). This form is particularly useful as it gives a direct relationship between the geometry of the plane in space and its algebraic representation.

Plane Geometry

Plane geometry is a branch of mathematics that deals with the shapes and studies within a plane, a flat, two-dimensional surface that extends infinitely. In three-dimensional space, planes play a crucial role in defining and solving spatial problems and relationships.A plane is defined algebraically as \( ax + by + cz = d \), where \((a, b, c)\) is the normal vector to the plane. In real-world applications, planes are used to model flat surfaces, such as ceilings and floors, and are fundamental in fields like computer graphics, architecture, and physics. Understanding the properties of planes, such as how they intersect or align with other planes, underpins many technical fields. In our example, creating a plane that is perpendicular to two other planes means deeply understanding plane geometry. It involves recognizing how the orientations of these planes relate to each other through their normal vectors, ultimately constructing a meaningful geometric picture in the three-dimensional space.