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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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 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:
This implies the parametric equations of the line:
Here, 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 . Let’s select :
Therefore, the point lies on the line.
Step 3: Purpose of the Point in the Solution
The point is used along with the given point to define a vector parallel to the plane. This vector is:
Step 4: Cross Product to Find the Normal Vector
The plane must contain both vectors:
- (the direction vector of the line)
- (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?Ask a new question for Free
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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
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