Math Problem Statement

Example 2.50

Writing an Equation for a Plane Given a Point and a Line

Find an equation of the plane that passes through point (1, 4, 3) and contains the line given by x = (y - 1)/2 = z + 1

Solution

Note

Both V and V are position velfors

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V_{t}

G

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C

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Symmetric equations describe the line that passes through point (0, 1, 1) parallel to vector v_{1} = \langle1, 2, 1\rangle (see the following figure). Use this point and the given point, (1, 4, 3), to identify a second vector parallel to the plane: v_{2} = \langle1 + 0, 4 - 1, 3 - (- 1)\rangle = \langle1, 3, 4\rangle . Use the cross product of these vectors to identify a normal vector for the plane: n = v_{1}*v_{2} = det [[i, j, k], [1, 2, 1], [1, 3, 4]] = (8 - 3) * i - (4 - 1) * j + (3 - 2) * k = 5i - 3j + k

We can also use given point that is (1, 4, 4) Then eq. of plane passing through the given point and containing the line is 5(x - 1) - 3(y - 4) + 1(z - 3) = 0

The scalar equations for the plane are 5x - 3(y - 1) + (z + 1) = 0 and 5x - 3y + z + 4 = 0 eplain where did wwe get the point 0 1 -1

Solution

Let's break down where the point (0,1,1)(0, 1, -1) comes from and how it's used in the solution for finding the equation of the plane.


Step 1: Understanding the Line Equation

The given line is described by the symmetric equations: x=y12=z+1x = \frac{y - 1}{2} = z + 1

This implies the parametric equations of the line:

x=t,y=2t+1,z=t1x = t, \quad y = 2t + 1, \quad z = t - 1

Here, tt is a parameter.


Step 2: Identify a Point on the Line

We can find a specific point on the line by choosing a convenient value for tt. Let’s select t=0t = 0:

x=0,y=2(0)+1=1,z=01=1x = 0, \quad y = 2(0) + 1 = 1, \quad z = 0 - 1 = -1

Therefore, the point (0,1,1)(0, 1, -1) lies on the line.


Step 3: Purpose of the Point in the Solution

The point (0,1,1)(0, 1, -1) is used along with the given point (1,4,3)(1, 4, 3) to define a vector parallel to the plane. This vector is:

v2=(10,41,3(1))=1,3,4\mathbf{v_2} = (1 - 0, 4 - 1, 3 - (-1)) = \langle 1, 3, 4 \rangle


Step 4: Cross Product to Find the Normal Vector

The plane must contain both vectors:

  • v1=1,2,1\mathbf{v_1} = \langle 1, 2, 1 \rangle (the direction vector of the line)
  • v2=1,3,4\mathbf{v_2} = \langle 1, 3, 4 \rangle (the vector between two points)

The cross product of these two vectors gives the normal vector to the plane:

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 1 & 3 & 4 \end{vmatrix}$$ $$= \mathbf{i} (8 - 3) - \mathbf{j} (4 - 1) + \mathbf{k} (3 - 2) = 5\mathbf{i} - 3\mathbf{j} + \mathbf{k}$$ --- ### **Step 5: Equation of the Plane** Now that we have the normal vector $$\mathbf{n} = \langle 5, -3, 1 \rangle$$ and a point on the plane $$(1, 4, 3)$$, the scalar equation for the plane is: $$5(x - 1) - 3(y - 4) + 1(z - 3) = 0$$ Simplifying: $$5x - 3y + z + 4 = 0$$ --- ### **Summary** The point $$(0, 1, -1)$$ was obtained by setting the parameter $$t = 0$$ in the parametric equations of the given line. It served as a convenient point to define the vector $$\mathbf{v_2}$$, which is crucial for finding the normal vector of the plane through the cross product. Would you like a deeper dive into the cross product calculation or any specific step?

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vectors
Planes in 3D
Cross Product
Parametric Equations

Formulas

Cross product of two vectors
Equation of a plane: ax + by + cz + d = 0

Theorems

Cross product to find the normal vector
Equation of a plane using a point and normal vector

Suitable Grade Level

Grades 11-12