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Chapter 12: Problem 5

Find the equation of the plane that passes through the point \((2,-3,1)\) and is perpendicular to the line joining the points \((3,4,-1)\) and \((2,-1,5)\).

Short Answer

Expert verified
Question: Find the equation of the plane that passes through the point (2, -3, 1) and is perpendicular to the line joining the points (3, 4, -1) and (2, -1, 5). Answer: The equation of the plane is given by: x + 5y - 6z + 11 = 0

Step by step solution

01

Find the direction vector of the line.

The direction vector of the given line joining \((3, 4, -1)\) and \((2, -1, 5)\) can be found by subtracting the corresponding coordinates of the given points: Direction vector = \(\begin{bmatrix} 2 - 3 \\ -1 - 4 \\ 5 - (-1) \end{bmatrix} = \begin{bmatrix} -1 \\ -5 \\ 6 \end{bmatrix}\)
02

Use the direction vector as the normal vector of the plane.

Since the plane is perpendicular to the line, the direction vector of the line serves as the normal vector of the plane. Thus, the normal vector is \(\begin{bmatrix} -1 \\ -5 \\ 6 \end{bmatrix}\).
03

Write the general equation of a plane using the normal vector.

The general equation of a plane with normal vector \(\begin{bmatrix} A \\ B \\ C \end{bmatrix}\) and passing through a point \((x_1, y_1, z_1)\) can be written as \(A(x - x_1) + B(y - y_1) + C(z - z_1) = 0\). Plugging in the normal vector and given point \((2, -3, 1)\), we get: \(-1(x - 2) - 5(y + 3) + 6(z - 1) = 0\).
04

Simplify the equation of the plane.

Expanding the equation obtained in the previous step, we have: \(-x + 2 - 5y - 15 + 6z - 6 = 0\). Combining like terms and simplifying, we get the final equation of the plane as: \(x + 5y - 6z + 11 = 0\)

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Key Concepts

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

Equation of a Plane
The equation of a plane in three-dimensional space is a key concept in analytical geometry. Understanding this enables us to describe surfaces and analyze their properties. Generally, a plane is defined using a normal vector and a point that lies on the plane. The normal vector is essential because it is perpendicular to every line that lies on the plane. The standard form of the equation of a plane can be expressed as:
  • If the plane has a normal vector \(\begin{bmatrix} A \ B \ C \end{bmatrix}\) and passes through a point \(x_1, y_1, z_1\), then the equation of the plane is \(A(x - x_1) + B(y - y_1) + C(z - z_1) = 0\).
This equation essentially states that the vector from any point on the plane to the fixed point \(x_1, y_1, z_1\) is perpendicular to the normal vector, maintaining the plane's flat surface without any bend.
Direction Vector
A direction vector is crucial for understanding lines in space. It gives us the orientation or direction that a line takes. To find the direction vector between two points, we simply subtract the coordinates of one point from the other. In our problem, to determine the direction from \(3, 4, -1\) to \(2, -1, 5\), we calculate:
  • Direction vector = \(\begin{bmatrix} 2 - 3 \ -1 - 4 \ 5 - (-1) \end{bmatrix} = \begin{bmatrix} -1 \ -5 \ 6 \end{bmatrix}\).
This vector \(\begin{bmatrix} -1 \ -5 \ 6 \end{bmatrix}\) indicates how the line moves in space, specifically, it shows how much it goes in the negative x-direction, the negative y-direction, and the positive z-direction.
Normal Vector
The normal vector is a fundamental component of a plane's definition. It is a vector that is perpendicular to the surface of the plane. This characteristic makes it indispensable when writing the equation of the plane, as it directly influences the orientation of the plane itself. For the given exercise, because the plane is perpendicular to the line formed by two points, the direction vector of the line serves as the plane's normal vector. Hence, the normal vector of our plane is:
  • Normal vector = \(\begin{bmatrix} -1 \ -5 \ 6 \end{bmatrix}\).
This normal vector ensures that the plane retains its perpendicular alignment to the line, guiding the placement and angle of the plane in space.
Perpendicular Lines
Perpendicularity is an important relationship between two geometrical entities. When we say two lines are perpendicular, they meet at a right angle, precisely 90 degrees. In the context of our problem, the plane is required to be perpendicular to a given line. This implies the normal vector of the plane is the same as the direction vector of that line. Therefore, the perpendicularity ensures that the plane is positioned precisely such that no line on it is parallel to the crossing line. This relationship is crucial in defining the spatial arrangement of the plane relative to the line, maintaining distinct orientations that are key to many geometrical analyses.

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Most popular questions from this chapter

Through the point \(P(a, b, c)\) a plane is drawn at right angles to \(O P\) to meet the axes in \(A, B\) and \(C\). Prove that the area of the triangle \(A B C\) is \(\frac{p^{5}}{2 a b c}\) where \(p\) is the length of \(O P\)

Find the locus of the point whose distance from the origin is 7 times its distance from the plane \(2 x+3 y-6 z=2\).

A plane contains the points \(A(-4,9,-9)\) and \(B(5,-9,6)\) and is perpendicular to the line which joins \(B\) and \(C(4,-6, \mathrm{k})\). Obtain \(k\) and the equation of the plane.

Find the equation of the plane passing through the points: (i) \((8,-2,2),(2,1,-4),(2,4,-6)\) (ii) \((2,2,1),(2,3,2),(-1,3,0)\) (iii) \((2,3,4),(-3,5,1),(4,-1,2)\)

Show that the points \((0,-1,0),(2,1,-1),(1,1,1)\) and \((-3,3,0)\) are coplanar.
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