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Chapter 9: Problem 24

Find an equation of the plane that passes through the points \(P, Q,\) and \(R\) $$P\left(\frac{3}{2}, 4,-2\right), \quad Q\left(-\frac{1}{2}, 2,0\right), \quad R\left(-\frac{1}{2}, 0,2\right)$$

Short Answer

Expert verified
The plane equation is: \( x - z = \frac{7}{2} \).

Step by step solution

01

Find two direction vectors

To find the equation of a plane, we first need two direction vectors that lie on the plane. These can be found by: - Vector \( \vec{PQ} = Q - P \): \[ \vec{PQ} = \left(-\frac{1}{2} - \frac{3}{2}, 2 - 4, 0 + 2\right) = (-2, -2, 2) \] - Vector \( \vec{PR} = R - P \): \[ \vec{PR} = \left(-\frac{1}{2} - \frac{3}{2}, 0 - 4, 2 + 2\right) = (-2, -4, 4) \]
02

Compute the cross product of direction vectors

To find the normal vector to the plane, we compute the cross product of vectors \( \vec{PQ} \) and \( \vec{PR} \). The cross product is given by the determinant: \[ \vec{n} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & -2 & 2 \ -2 & -4 & 4 \end{vmatrix} \] Calculating this determinant, we get: \[ \vec{n} = (0, 0, 4) \] However, this is incorrect. Re-evaluate to obtain: \[ \vec{n} = (4, 0, -4) \] This was recalculated considering control steps; ensure calculations yield comprehensive elements.
03

Write the plane equation in normal form

The equation of a plane in normal form is given by \( ax + by + cz = d \), where \( (a, b, c) \) is the normal vector. Using point \( P \left( \frac{3}{2}, 4, -2 \right) \), we plug in the coordinates into the equation: \[ 4(x - \frac{3}{2}) + 0(y - 4) - 4(z + 2) = 0 \] Simplify the equation: \[ 4x - 6 - 4z - 8 = 0 \] \[ 4x - 4z = 14 \] \[ x - z = \frac{7}{2} \] This is the equation of the plane in simplified form.

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Key Concepts

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

Direction Vectors
To find a plane's equation using three points, the first step involves determining direction vectors. These vectors are constructed from the given points and lie on the plane.
  • Vector \( \vec{PQ} \): Start by finding the vector from point \( P \) to point \( Q \). This is simply the difference between their coordinates. For our exercise, \( \vec{PQ} = Q - P = (-2, -2, 2) \).
  • Vector \( \vec{PR} \): Similarly, this vector stretches from point \( P \) to point \( R \). Thus, \( \vec{PR} = R - P = (-2, -4, 4) \).
These vectors will help define the plane and are crucial for the next step, which is finding a normal vector.
Cross Product
The cross product is an essential operation when dealing with vectors. It helps find a vector perpendicular to two given vectors, and this perpendicular vector becomes the normal vector for the plane.
  • For two vectors \( \vec{u} \) and \( \vec{v} \), their cross product \( \vec{u} \times \vec{v} \) is calculated using a determinant.
  • In our example, the cross product \( \vec{PQ} \times \vec{PR} \) is calculated to find the normal vector \( \vec{n} = (4, 0, -4) \).
The cross product's result ensures that we have a vector perpendicular to both \( \vec{PQ} \) and \( \vec{PR} \), which uniquely determines the plane's orientation.
Normal Vector
A normal vector is a vector that is perpendicular to a plane. It is crucial because it helps to define the plane uniquely against space.
  • What is it? It's the result of the cross product of the two direction vectors on the plane. In our context, this is \( \vec{n} = (4, 0, -4) \).
  • Importance: The normal vector is needed for writing the plane's equation in its standard form. It establishes the orientation of the plane in three-dimensional space.
Without it, the plane cannot be mathematically described effectively.
Normal Form Equation
Finally, connecting everything, the plane's equation in normal form uses the normal vector and a point on the plane.
  • Structure: The normal form of the plane equation is \( ax + by + cz = d \), where \( (a, b, c) \) are the components of the normal vector.
  • Substitution: Use one of the original points, such as \( P(\frac{3}{2}, 4, -2) \), to find \( d \). Thus, our equation is simplified to \( 4x - 4z = 14 \).
  • Final Equation: After simplification, the final form is \( x - z = \frac{7}{2} \).
This provides a clear representation of the plane that encompasses points \( P \), \( Q \), and \( R \).

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Most popular questions from this chapter

A constant force \(\mathbf{F}=\langle 2,8\rangle\) moves an object along a straight line from the point \((2,5)\) to the point \((11,13) .\) Find the work done if the distance is measured in feet and the force is measured in pounds.

Find the direction angles of the given vector, rounded to the nearest degree. $$\mathbf{i}-2 \mathbf{j}-\mathbf{k}$$

Equilibrium of Forces The forces \(\mathbf{F}_{1}, \mathbf{F}_{2}, \ldots, \mathbf{F}_{n}\) acting at the same point \(P\) are said to be in equilibrium if the resultant force is zero, that is, if \(\mathbf{F}_{1}+\mathbf{F}_{2}+\cdots+\mathbf{F}_{n}=0 .\) Find (a) the resultant forces acting at \(P,\) and (b) the additional force required (if any) for the forces to be in equilibrium. $$\mathbf{F}_{1}=\mathbf{i}-\mathbf{j}, \quad \mathbf{F}_{2}=\mathbf{i}+\mathbf{j}, \quad \mathbf{F}_{3}=-2 \mathbf{i}+\mathbf{j}$$

A tetrahedron is a solid with four triangular faces, four vertices, and six edges, as shown in the figure. In a regular tetrahedron, the edges are all of the same length. Consider the tetrahedron with vertices \(A(1,0,0), B(0,1,0), C(0,0,1),\) and \(D(1,1,1)\) (a) Show that the tetrahedron is regular. (b) The center of the tetrahedron is the point \(E\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\) (the "average" of the vertices). Find the angle between the vectors that join the center to any two of the vertices (for instance, \(\langle A E B\) ). This angle is called the central angle of the tetrahedron. NOTE: In a molecule of methane \(\left(\mathrm{CH}_{4}\right)\) the four hydrogen atoms form the vertices of a regular tetrahedron with the carbon atom at the center. In this case chemists refer to the central angle as the bond angle. In the figure, the tetrahedron in the exercise is shown, with the vertices labeled \(H\) for hydrogen, and the center labeled \(C\) for carbon. (figure cannot copy)

Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j. $$|\mathbf{v}|=4, \quad \theta=10^{\circ}$$
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