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Chapter 12: Problem 11

Find the equation of the plane which passes through the point \((2,-3,4)\) and is parallel to the plane \(2 x-5 y-7 z+15=0\).

Short Answer

Expert verified
Question: Find the equation of the plane that passes through the point (2, -3, 4) and is parallel to the plane 2x - 5y - 7z + 15 = 0. Answer: The equation of the required plane is 2x - 5y - 7z = -9.

Step by step solution

01

Identify the normal vector of the given plane

We can read off the coefficients of the variables \(x\), \(y\), and \(z\) in the equation of the given plane: \(2x - 5y - 7z + 15 = 0\). The normal vector of the given plane is \(\textbf{n} = (2, -5, -7)\).
02

Use the given point and normal vector to find the equation of the required plane

The equation of a plane can be written in the form \(\textbf{n} \cdot \textbf{r} = d\), where \(\textbf{n}\) is the normal vector, \(\textbf{r}\) is the position vector \(\textbf{r} = (x, y, z)\), and \(d\) is a constant. Since the required plane is parallel to the given plane and passes through the point \((2, -3, 4)\), it will have the same normal vector, so we can write the equation of the required plane as: \((2, -5, -7) \cdot (x, y, z) = d\) Expanding the dot product, we have: \(2x - 5y - 7z = d\) Now, we can use the given point \((2, -3, 4)\) to find the value of \(d\): \(2(2) - 5(-3) - 7(4) = d\) Simplify the equation: \(4 + 15 - 28 = d\) \(d = -9\)
03

Write the equation of the required plane

Now that we have the value of \(d\), we can write the equation of the required plane: \(2x - 5y - 7z = -9\) This is the equation of the plane that passes through the point \((2, -3, 4)\) and is parallel to the plane \(2x - 5y - 7z + 15 = 0\).

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Key Concepts

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

Normal Vector
The normal vector is a fundamental concept when dealing with planes in geometry. It is a vector that is perpendicular to every line lying on the plane.
In the context of the equation of a plane, this vector is often derived from the coefficients of the variables in the plane's equation.
  • The equation of the plane might look something like: \( ax + by + cz + d = 0 \), where \( a, b, \) and \( c \) are the coefficients.
  • The normal vector, \( \mathbf{n} \), can be written as \( (a, b, c) \).
Understanding the normal vector is crucial because it simplifies many calculations involving planes, such as finding parallel planes or distances from points to planes.
Parallel Planes
Parallel planes are planes that never intersect; they always maintain the same distance from each other no matter how far they are extended.
For two planes to be parallel, their normal vectors must be the same or scalar multiples of each other.
  • For example, if one plane has a normal vector \( \mathbf{n}_1 = (a, b, c) \) and another plane鈥檚 normal vector is \( \mathbf{n}_2 = k(a, b, c) \), where \( k \) is a scalar, then these planes are parallel.
  • This concept simplifies finding an equation of a new plane that is parallel to a given plane 鈥� you can directly use the normal vector of the given plane.
Knowing about parallel planes assists in tasks like determining the equation of a plane through a specified point that is parallel to another.
Dot Product
The dot product is a mathematical operation that takes two equal-length sequences of numbers and returns a single number.
In three-dimensional space, the dot product is often used with vectors. If you have vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), their dot product is given by: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
This operation is essential in determining relationships between vectors, such as orthogonality.
  • If the dot product is zero, the vectors are orthogonal (perpendicular).
  • If not, it provides insight into the angle or alignment of the vectors.
Within the realm of plane equations, the dot product is used to establish the plane鈥檚 equation with a given normal vector.
Analytical Geometry
Analytical geometry, also known as coordinate or Cartesian geometry, uses algebraic techniques to solve geometric problems.
It allows us to represent geometric objects, like points, lines, and planes, with equations and coordinate systems.
Planes, for instance, can be represented by equations like \( ax + by + cz + d = 0 \).
  • This allows for the analysis of relationships, such as parallelism and perpendicularity, using vector and matrix operations.
  • Furthermore, problems involving distances, midpoints, and angles between geometric objects can be solved more easily.
The power of analytical geometry is immense as it bridges the gap between algebra and geometry, broadening the deductions and conclusions that can be made in geometric spaces.

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Most popular questions from this chapter

Find the equation of the plane which passes through the point \((2,2,4)\) and perpendicular to the planes \(2 x-2 y-4 z+3=0\) and \(3 x+y+6 z-4=0\).

Show that the equations \(b y+c z+d=0, c z+a x+d=0, a x+b y+d=0\) represent planes parallel to \(O X, O Y\) and \(O Z\), respectively.

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.

A plane meets the coordinate axes at \(A, B\) and \(C\) such that the centroid of the triangle is the point \((a, b, c)\). Show that the equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3\)

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