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Chapter 12: Problem 18

$$\text { Find an equation of the following planes.}$$ The plane passing through the points \((-1,1,1),(0,0,2),\) and (3,-1,-2)

Short Answer

Expert verified
Answer: The equation of the plane is 3x - 7y - 6z + 16 = 0.

Step by step solution

01

Calculate two directional vectors

We are given three points on the plane: \(A = (-1, 1, 1)\), \(B = (0, 0, 2)\), and \(C = (3, -1, -2)\). Find two directional vectors using these points: - Vector \(AB\) (from point A to point B): \(\vec{AB} = B-A = (0-(-1), 0-1, 2-1) = (1, -1, 1)\) - Vector \(AC\) (from point A to point C): \(\vec{AC} = C-A = (3-(-1), -1-1, -2-1) = (4, -2, -3)\)
02

Compute the normal vector

Find the normal vector \(\vec{n}\) of the plane using the cross product of vectors \(\vec{AB}\) and \(\vec{AC}\): $$\vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 4 & -2 & -3 \end{vmatrix}$$ First, we compute the cross product using the determinant method: $$\vec{n} = \hat{i}(3) - \hat{j}(7) + \hat{k}(-6)$$ This gives us the normal vector: $$\vec{n} = (3, -7, -6)$$
03

Use the point-normal form of the equation

After finding the normal vector, we can use the point-normal form of the plane's equation: $$A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$$ Here, \((A, B, C) = (3, -7, -6)\) is the normal vector and \((-1, 1, 1)\) is a point on the plane. Plug these values into the equation: $$3(x-(-1)) - 7(y-1) - 6(z-1) = 0$$
04

Simplify the equation

Simplify the plane's equation: $$3(x+1) - 7(y-1) - 6(z-1) = 0$$ $$3x+3 -7y+7 -6z+6 =0$$ From the equation above, we can simplify it to: $$3x - 7y - 6z + 16 = 0$$ So, the equation of the plane passing through the points \((-1,1,1),(0,0,2),\) and (3,-1,-2) is $$3x - 7y - 6z + 16 = 0$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Directional Vectors
To begin solving the problem of finding a plane equation that passes through given points, we need to determine two directional vectors. A directional vector represents the direction from one point to another in a coordinate system.
  • The directional vector \(\vec{AB}\) is calculated from point \(A\) to point \(B\) by subtracting coordinates of \(A\) from those of \(B\): \(B-A = (0-(-1), 0-1, 2-1) = (1, -1, 1)\).
  • Similarly, the directional vector \(\vec{AC}\) is calculated from point \(A\) to point \(C\): \(C-A = (3-(-1), -1-1, -2-1) = (4, -2, -3)\).
These vectors \(\vec{AB} = (1, -1, 1)\) and \(\vec{AC} = (4, -2, -3)\) give us directions that lie within the plane, helping us to further understand the plane's orientation.
Cross Product
The cross product is a mathematical operation that helps find a vector perpendicular to two given vectors in three-dimensional space. For our problem, the cross product will allow us to find the normal vector to the plane.Given directional vectors \(\vec{AB} = (1, -1, 1)\) and \(\vec{AC} = (4, -2, -3)\), the cross product \(\vec{AB} \times \vec{AC}\) is calculated using a determinant:\[\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \1 & -1 & 1 \4 & -2 & -3 \end{vmatrix}\]
  • Calculate \(\hat{i}(3)\) which is 3. It comes from multiplying the cross-products of corresponding components.
  • For \(\hat{j}\), the result is \(-7\), and for \(\hat{k}\), the result is \(-6\).
By solving this determinant, we find the resulting vector to be \((3, -7, -6)\), which we will use as the normal vector of the plane.
Normal Vector
A normal vector is a vector that is perpendicular to a plane. Knowing the normal vector is crucial because it defines the plane's orientation in space.
In our example, the normal vector \(\vec{n} = (3, -7, -6)\) was found by computing the cross product of the directional vectors \(\vec{AB}\) and \(\vec{AC}\). This vector tells us the direction which is orthogonal to the plane, providing stability and direction of the plane's orientation.
  • The components of the normal vector \((3, -7, -6)\) become the coefficients \(A\), \(B\), and \(C\) in the plane equation.
  • By maintaining this orthogonal relationship, it essentially anchors or orients the plane in the three-dimensional space.
Point-Normal Form
The point-normal form of a plane equation makes use of a point on the plane and the normal vector to represent the plane succinctly. The point-normal form is given by:
\[ A(x-x_1) + B(y-y_1) + C(z-z_1) = 0 \]
Where \((A, B, C)\) are components of the normal vector and \((x_1, y_1, z_1)\) is a point on the plane. For this problem:
  • The normal vector is \((3, -7, -6)\), obtained from the cross product.
  • The point on the plane we use is \((-1, 1, 1)\).
Plugging these into the point-normal form, we get:
\[ 3(x+1) - 7(y-1) - 6(z-1) = 0 \]Simplifying this equation leads us to the simple form: \( 3x - 7y - 6z + 16 = 0 \), which is the equation of our plane.

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Most popular questions from this chapter

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