Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The equation of the plane is \(7x + 5y + 4z + 5 = 0\).

Step 1: Finding the Normal Vectors

For the given planes, identify the normal vectors. The normal vector of the plane \(x-3y+2z=7\) is \(\mathbf{n_1} = \langle 1, -3, 2 \rangle\) and the normal vector of \(2x-2y-z=-3\) is \(\mathbf{n_2} = \langle 2, -2, -1 \rangle\). A plane perpendicular to both will have a normal vector in the direction of the cross product of \(\mathbf{n_1}\) and \(\mathbf{n_2}\).

Step 2: Calculating the Cross Product

To find a vector perpendicular to both \(\mathbf{n_1}\) and \(\mathbf{n_2}\), compute the cross product \(\mathbf{n_1} \times \mathbf{n_2}\):\[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & -3 & 2 \2 & -2 & -1 \\end{vmatrix} = \mathbf{i}(3 \cdot (-1) - 2 \cdot (-2)) - \mathbf{j}(1 \cdot (-1) - 2 \cdot 2) + \mathbf{k}(1 \cdot (-2) - (-3) \cdot 2)\]\[= \mathbf{i}(3 + 4) - \mathbf{j}(-1 - 4) + \mathbf{k}(-2 + 6) = \langle 7, 5, 4 \rangle.\]

Step 3: Using the Normal Vector and Point to Find the Plane Equation

The equation of a plane 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 a point on the plane. Substitute \((a, b, c) = (7, 5, 4)\) and the point \((-1, -2, 3)\) into the equation:\[ 7(x + 1) + 5(y + 2) + 4(z - 3) = 0 \]Simplifying the equation gives:\[ 7x + 7 + 5y + 10 + 4z - 12 = 0 \]\[ 7x + 5y + 4z + 5 = 0 \].

Step 4: Final Equation of the Plane

The final equation of the plane that passes through \((-1,-2,3)\) and is perpendicular to both given planes is \(7x + 5y + 4z + 5 = 0\). Simplify if necessary to standardize, but the equation is already in its simplest form.

Key Concepts

Cross Product

The cross product is a mathematical operation between two vectors in three-dimensional space. It results in a third vector that is perpendicular to both of the original vectors. This is crucial in determining the direction of a normal vector for a plane.

• The cross product is only defined in three-dimensional space, where the vectors can be represented by three components.
• It is not commutative. This means \(\mathbf{A} \times \mathbf{B} \) will have the opposite direction to \(\mathbf{B} \times \mathbf{A} \).
• The magnitude of the resulting vector is proportional to the sine of the angle between the original vectors and is also equivalent to the area of the parallelogram that the vectors span.

To calculate the cross product, we use the determinant of a matrix containing the unit vectors and the components of each vector:\[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix}\]where \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the unit vectors and \(a_1, b_1, c_1\) and \(a_2, b_2, c_2\) are the components of the given vectors.

Normal Vector

In geometry, a normal vector to a surface is one that is perpendicular to the surface at a given point. For planes, the normal vector plays a critical role in defining the plane's equation.

• Every plane has exactly one normal vector at any position on the plane, but that vector can point in either direction.
• The coefficients \(a, b, c\) in the plane equation \(ax + by + cz + d = 0\) are the components of the normal vector.
• The normal vector is essential for determining the orientation of the plane in three-dimensional space.

By using normal vectors of planes, we can understand the relationships between different planes, such as perpendicularity or parallelism.

Perpendicular Planes

Two planes are perpendicular if their normal vectors are perpendicular. The cross product of two normal vectors gives a new vector that is perpendicular to both.

• To check if two planes are perpendicular, we can calculate the dot product of their normal vectors. If the dot product equals zero, the planes are perpendicular.
• For the given exercise, we used the cross product to find a normal vector for a new plane that is perpendicular to both original planes.
• This property is useful in many applications, such as constructing perpendicular planes in computer graphics or physics problems.

The determination of a plane perpendicular to other planes serves both geometric and practical applications, making cross product derived normal vectors invaluable.

Three-Dimensional Geometry

Three-dimensional geometry deals with shapes and figures that have width, depth, and height. It's a vast subject involving many fundamental concepts including points, lines, planes, and shapes like cubes and spheres.

• In 3D geometry, we often use a coordinate system that defines every point in space using three numbers: \(x, y, z \).
• Understanding the spatial relationships between different geometric entities is crucial for solving complex geometric problems.
• Key operations include calculating distances, angles, and intersections between different elements, all requiring a solid grasp of three-dimensional spatial reasoning.

The relationship between planes, especially how they intersect or align can be fully explored using these concepts, essential for many scientific and engineering disciplines.