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Magazine Chapter 11: Problem 32
Find a plane through the points \(P_{1}(1,2,3), P_{2}(3,2,1)\) and perpendicular to the plane \(4 x-y+2 z=7\)
Short Answer
Expert verified
The equation of the plane is \(2x + 8y + 2z = 28\).
Step by step solution
01
Understand the Problem
We need to find a plane that passes through three points: \( P_1(1,2,3) \), \( P_2(3,2,1) \), and a third point which we'll determine. This plane must also be perpendicular to the given plane with equation \(4x - y + 2z = 7\).
02
Determine Normal Vector of Given Plane
The given plane \(4x - y + 2z = 7\) has a normal vector, \( \mathbf{n} = \langle 4, -1, 2 \rangle \). The plane we are seeking should have a direction vector parallel to this normal vector.
03
Find Another Point that Lies on the Plane to be Found
Since we know two points \( P_1 \) and \( P_2 \) that lie on the plane, we need to find a vector that lies on the plane. We form vector \( \mathbf{v} = P_2 - P_1 = \langle 3-1, 2-2, 1-3 \rangle = \langle 2, 0, -2 \rangle \).
04
Use Cross Product to Find Plane's Normal Vector
To find the normal vector of the sought plane, we take the cross product of \( \mathbf{n} \) and \( \mathbf{v} \). Calculate \( \mathbf{n} \times \mathbf{v} \):\[\mathbf{n} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \4 & -1 & 2 \2 & 0 & -2\end{vmatrix} = \langle 2, 8, 2 \rangle.\]
05
Write the Equation of the Plane
Using the normal vector \( \langle 2, 8, 2 \rangle \) obtained from the cross product and the point \( P_1(1, 2, 3) \), we can write the equation of the plane:\[ 2(x-1) + 8(y-2) + 2(z-3) = 0 \]Simplifying, we get:\[ 2x + 8y + 2z = 28 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Vector
In three-dimensional geometry, understanding the concept of a **normal vector** is crucial when working with planes. A normal vector is a vector that is perpendicular to the plane. It gives information about the orientation of the plane in space. Simply put, it's like a "direction arrow" that points 'straight out' from the plane.
For example, if you have a plane equation of the form:
\[ ax + by + cz = d \]
the normal vector is given by the coefficients \( \langle a, b, c \rangle \). This is because this vector is orthogonal to every vector that lies in the plane.
**Key points about the normal vector:**
For example, if you have a plane equation of the form:
\[ ax + by + cz = d \]
the normal vector is given by the coefficients \( \langle a, b, c \rangle \). This is because this vector is orthogonal to every vector that lies in the plane.
**Key points about the normal vector:**
- It is not necessarily a unit vector, meaning it doesn鈥檛 have to be of length 1.
- When a plane is described by a linear equation, the coefficients of \(x, y,\) and \(z\) in the equation form the components of the plane's normal vector.
- If two planes are parallel, they share the same normal vector direction.
Cross Product
The **cross product** is a way to find a vector that is perpendicular to two given vectors in 3D space. It is particularly helpful when determining a normal vector for a plane that must meet certain conditions, such as being perpendicular to another vector.
To compute the cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the formula is:
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3\end{vmatrix} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle\]
**Important aspects of the cross product:**
To compute the cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the formula is:
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3\end{vmatrix} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle\]
**Important aspects of the cross product:**
- The resulting vector is perpendicular to both input vectors.
- The direction of the resulting vector is determined by the right-hand rule.
- The magnitude of the resulting vector is equal to the area of the parallelogram formed by the two vectors.
Points in 3D Space
Understanding **points in 3D space** is foundational in geometry and vector mathematics. Each point is represented by a coordinate triplet \((x, y, z)\), where each coordinate specifies the point's location along the x, y, and z axes respectively.
When working with planes, it's essential to use points to define the plane's position and orientation in 3D space:
**Working with points:**
When working with planes, it's essential to use points to define the plane's position and orientation in 3D space:
- Two points can be used to form a line.
- Three non-collinear points define a unique plane in space.
**Working with points:**
- You can calculate the distance between points using the Euclidean distance formula.
- We often use points to find direction vectors by subtracting the coordinates of one point from another.
Vector Arithmetic
**Vector arithmetic** involves operations such as addition, subtraction, and scalar multiplication of vectors. These operations are essential in manipulating vectors to solve problems in three-dimensional space.
**Vector Addition and Subtraction:**
**Vector Addition and Subtraction:**
- To add or subtract vectors, you perform component-wise operations. For example, if \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), then:
- Vector addition: \( \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangle \)
- Vector subtraction: \( \mathbf{a} - \mathbf{b} = \langle a_1 - b_1, a_2 - b_2, a_3 - b_3 \rangle \)
- Multiplying a vector by a scalar changes its magnitude but not its direction. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and scalar \( k \), the product \( k\mathbf{a} \) is \( \langle ka_1, ka_2, ka_3 \rangle \).
