Problem 42
Question
Find the equation of the plane through \((1,-1,2)\) and perpendicular to \([-1,1,2]^{\prime}\).
Step-by-Step Solution
Verified Answer
The equation of the plane is \\-x + y + 2z = 2\\.
1Step 1: Understanding the Normal Vector
The problem statement gives the normal vector to the plane as \([-1, 1, 2]\). This means that any plane with this normal must satisfy the equation of the form \(-1x + 1y + 2z = D\).
2Step 2: Finding the Constant D
Substitute the point \(1, -1, 2\) into the plane equation \(-1x + 1y + 2z = D\). The calculation will be: \\(-1(1) + 1(-1) + 2(2) = D\) which simplifies to \(-1 - 1 + 4 = D\). Therefore, \(D = 2\).
3Step 3: Writing the Final Equation
Using the value of \(D\) found in the previous step, the equation of the plane is \(-1x + 1y + 2z = 2\). This can be rewritten more neatly as \-x + y + 2z = 2\.
Key Concepts
Understanding the Normal VectorPerpendicularity in 3D SpaceSubstitution Method in Plane Equations
Understanding the Normal Vector
In geometry, a normal vector is a vector that is perpendicular to a surface. In simpler terms, if you imagine a plane as a sheet of paper, then the normal vector would point straight up from the surface of that sheet at 90 degrees. In the exercise given, the normal vector provided is \([-1, 1, 2]\). This vector is crucial because it helps us to define the orientation of the plane in 3D space.
The equation of a plane can be expressed in the form \(ax + by + cz = D\), where \([a, b, c]\) is the normal vector. The coefficients \(-1, 1,\) and \(2\) in our specific exercise hold positions of \(a, b,\) and \(c\) respectively. This equation generally represents all the points \((x, y, z)\) that lie on the plane. By understanding the role of the normal vector, you can visualize the plane's orientation and how it slices through the three-dimensional space.
The equation of a plane can be expressed in the form \(ax + by + cz = D\), where \([a, b, c]\) is the normal vector. The coefficients \(-1, 1,\) and \(2\) in our specific exercise hold positions of \(a, b,\) and \(c\) respectively. This equation generally represents all the points \((x, y, z)\) that lie on the plane. By understanding the role of the normal vector, you can visualize the plane's orientation and how it slices through the three-dimensional space.
Perpendicularity in 3D Space
Perpendicularity in 3D space is a fascinating aspect of geometry that allows us to determine the relationship between a plane and a vector. Imagine a line shooting straight up from the surface of a plane; this line is perpendicular to the plane and coincides with the normal vector.
In the example exercise, the condition for the plane being perpendicular to the vector \([-1, 1, 2]\) means that this vector is exactly the normal vector of the plane. Visualizing this, imagine the direction given by the normal vector as being orthogonal or at a right angle to any direction within the plane itself.
This notion of perpendicularity is integral when you're working in 3D space because it helps to pinpoint the exact tilt and orientation of the plane. When a plane passes through a particular point and is perpendicular to a given vector, it helps you derive the exact set of points that satisfies the plane equation.
In the example exercise, the condition for the plane being perpendicular to the vector \([-1, 1, 2]\) means that this vector is exactly the normal vector of the plane. Visualizing this, imagine the direction given by the normal vector as being orthogonal or at a right angle to any direction within the plane itself.
This notion of perpendicularity is integral when you're working in 3D space because it helps to pinpoint the exact tilt and orientation of the plane. When a plane passes through a particular point and is perpendicular to a given vector, it helps you derive the exact set of points that satisfies the plane equation.
Substitution Method in Plane Equations
The substitution method is a powerful tool used to find specific components of equations, especially when dealing with equations of planes. The process involves substituting given known values into the general equation to solve for the unknown part. In the exercise, such substitution helped find the constant \(D\) in the equation of the plane.
Since we are given a point, \((1, -1, 2)\), that lies on the plane, we substitute these coordinates into the plane equation \(-1x + 1y + 2z = D\). This results in:
The substitution method is an efficient way to handle such problems, as it breaks the problem into manageable steps and leads directly to the required solution.
Since we are given a point, \((1, -1, 2)\), that lies on the plane, we substitute these coordinates into the plane equation \(-1x + 1y + 2z = D\). This results in:
- \(-1(1)\)
- \(+ 1(-1)\)
- \(+ 2(2) = D\)
The substitution method is an efficient way to handle such problems, as it breaks the problem into manageable steps and leads directly to the required solution.
Other exercises in this chapter
Problem 41
Use a rotation matrix to rotate the vector \(\left[\begin{array}{r}-1 \\\ 2\end{array}\right]\) counterclockwise by the angle \(\pi / 3\).
View solution Problem 41
Write each system in matrix form. (There is no need to solve the systems). $$ \begin{array}{r} 2 x_{1}-x_{2}=4 \\ -x_{1}+2 x_{2}=3 \\ 3 x_{1}=4 \end{array} $$
View solution Problem 42
Use a rotation matrix to rotate the vector \(\left[\begin{array}{r}4 \\\ -1\end{array}\right]\) counterclockwise by the angle \(\pi / 6\).
View solution Problem 42
Write each system in matrix form. (There is no need to solve the systems). $$ \begin{array}{r} x_{1}-3 x_{2}+x_{3}=1 \\ -2 x_{1}+x_{2}-x_{3}=0 \end{array} $$
View solution