Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane through \((6,2,-1)\) and perpendicular to the line of intersection of the planes \(4 x-3 y+2 z+5=0\) and \(3 x+2 y-z+11=0\).

Step-by-Step Solution

Answer

The plane's equation is \(-x + 10y + 17z + 3 = 0\).

Step 1: Find the Normal Vector of the Given Planes

The normal vector of the plane \(4x - 3y + 2z + 5 = 0\) is \(\mathbf{n_1} = (4, -3, 2)\), and the normal vector of the plane \(3x + 2y - z + 11 = 0\) is \(\mathbf{n_2} = (3, 2, -1)\).

Step 2: Cross Product for Line of Intersection Normal

The line of intersection of the planes is perpendicular to both normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\). Therefore, find the cross product \(\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2}\).\[\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \4 & -3 & 2 \3 & 2 & -1\end{vmatrix}\]\(\mathbf{n} = ((-3)(-1) - (2)(2)) \mathbf{i} - ((4)(-1) - (2)(3)) \mathbf{j} + ((4)(2) - (-3)(3)) \mathbf{k}\)\(\mathbf{n} = (3 - 4)\mathbf{i} - (-4 - 6)\mathbf{j} + (8 + 9)\mathbf{k}\)\(\mathbf{n} = (-1, 10, 17)\).

Step 3: Equation of the Plane Through Given Point

Now we have a point \((6, 2, -1)\) and a normal vector \((-1, 10, 17)\). The equation of a plane can be written as:\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]Substitute \(a = -1\), \(b = 10\), \(c = 17\), and the point \((x_0, y_0, z_0) = (6, 2, -1)\):\[-1(x - 6) + 10(y - 2) + 17(z + 1) = 0\]Simplifying, this becomes:\[-x + 6 + 10y - 20 + 17z + 17 = 0\]So, the equation is:\[-x + 10y + 17z + 3 = 0\].

Key Concepts

Understanding the Cross Product

The cross product is a mathematical operation that is highly useful when dealing with vectors, especially in three dimensions. It allows you to find a vector that is perpendicular to two given vectors. In this context, the cross product of the normal vectors from the original planes provides a directional vector that is perpendicular to both. The formula for the cross product can be expressed as:

• If you have two vectors, \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), their cross product is given by:\[\mathbf{a} \times \mathbf{b} = (a_2b_3 - a_3b_2) \mathbf{i} - (a_1b_3 - a_3b_1) \mathbf{j} + (a_1b_2 - a_2b_1) \mathbf{k}\]

This operation results in a new vector that is uniquely suited to highlight the line of intersection as described in our problem. Knowing how to compute and interpret the cross product is key to solving many geometric problems involving planes and lines.

What is a Normal Vector?

A normal vector to a plane provides a lot of information about the plane itself. Essentially, it is a vector that is perpendicular to the plane. For the equation of a plane like \( ax + by + cz + d = 0 \), the coefficients \( a, b, \) and \( c \) form the components of a normal vector.

• For the plane \( 4x - 3y + 2z + 5 = 0 \), the normal vector is \( \mathbf{n_1} = (4, -3, 2) \).
• For the plane \( 3x + 2y - z + 11 = 0 \), the normal vector is \( \mathbf{n_2} = (3, 2, -1) \).

Normal vectors are crucial because they help us understand the orientation of the plane. In our problem, the calculated normal vector from the cross product, \((-1, 10, 17)\), becomes crucial as it helps in crafting the equation of the sought-after plane through a specific point.

Exploring the Line of Intersection

The line of intersection between two planes can be somewhat of a puzzle. In geometric space, when two distinct planes meet, they typically form a line at that junction — their line of intersection. To describe this line, you often need a direction vector and a point through which the line passes.

• In our problem, this direction vector is derived from the cross product of the normal vectors of the two planes.
• Thus, the vector \((-1, 10, 17)\) describes the direction of the line where these planes intersect.

Identifying this direction vector is key, as it helps to pinpoint the path along which any plane perpendicular to this line will align itself in space. It plays a foundational role in deriving the equation of our new plane.

Understanding Perpendicular Planes

Perpendicular planes are a special geometric construction where two planes meet at a right angle. For a given line, if a plane is perpendicular to it, the normal vector of the plane and the direction vector of the line are the same.

• In this exercise, we begin with a known point \((6, 2, -1)\) and our calculated direction vector from the cross product, \((-1, 10, 17)\).
• These pieces establish the conditions necessary to construct a plane that is perpendicular to the line of intersection of the given planes.

This principle of aligning the normal vector of the new plane with the direction of the intersection line ensures the plane is oriented perpendicularly. The resulting equation \(-x + 10y + 17z + 3 = 0\) accurately describes this orientation and relation in three-dimensional space.