Chapter 19: Problem 20
The point \((1,2,4)\) lies in plane \(P\) which is perpendicular to the vector \(\mathbf{a}=2 \mathbf{i}-3 \mathbf{j}+\mathbf{k}\). Given \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) is the position vector of another point in the plane find the equation of the plane.
Short Answer
Expert verified
The equation of the plane P is: \(2x - 3y + z = 0\).
Step by step solution
01
Write the point-normal form of the equation of a plane
The point-normal form of the equation of a plane is given by:
\[(\mathbf{r} - \mathbf{r}_0) \cdot \mathbf{a} = 0\]
where \(\mathbf{r}\) is the position vector of any point in the plane, \(\mathbf{r}_0\) is the position vector of a known point in the plane, and \(\mathbf{a}\) is the normal vector to the plane.
02
Substitute the known values
We are given that \((1,2,4)\) lies in plane P, so the position vector of this point is \(\mathbf{r}_0 = \mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\). We are also given the normal vector to the plane: \(\mathbf{a} = 2\mathbf{i}-3\mathbf{j}+\mathbf{k}\).
Substitute these values into the point-normal form equation:
\[(\mathbf{r} - (\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})) \cdot (2\mathbf{i}-3\mathbf{j}+\mathbf{k}) = 0\]
03
Expand the dot product and simplify the equation
Now, expand the dot product in the equation:
\[((x - 1)\mathbf{i} + (y-2)\mathbf{j} + (z-4)\mathbf{k}) \cdot (2\mathbf{i}-3\mathbf{j}+\mathbf{k}) = 0\]
\((x-1)(2) + (y-2)(-3) + (z-4)(1) = 0\)
\(2(x-1) - 3(y-2) + (z-4) = 0\)
Now, distribute and simplify:
\(2x - 2 - 3y + 6 + z - 4 = 0\)
\(2x - 3y + z + 0 = 0\)
04
Write the final equation of the plane
The equation of the plane P is:
\(2x - 3y + z = 0\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Point-Normal Form of a Plane
The point-normal form is a fundamental way to describe a plane in three-dimensional space. This equation is particularly useful because it directly incorporates a normal vector, which is key in defining the plane's orientation.
The form is written as \[(\mathbf{r} - \mathbf{r}_0) \cdot \mathbf{a} = 0\]where:
The form is written as \[(\mathbf{r} - \mathbf{r}_0) \cdot \mathbf{a} = 0\]where:
- \(\mathbf{r}\) is the position vector of any arbitrary point \((x, y, z)\) on the plane.
- \(\mathbf{r}_0\) is the position vector of a specific point known to lie on the plane, given as \((x_0, y_0, z_0)\).
- \(\mathbf{a}\) is a normal vector perpendicular to the plane.
Position Vectors
Position vectors are a simple yet crucial concept in vector mathematics, providing a blueprint for identifying the location of a point in space. A position vector points from the origin to a specific point \((x, y, z)\).
In general, the position vector is expressed as \[\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}\]where:
In general, the position vector is expressed as \[\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}\]where:
- \(x\), \(y\), and \(z\) are the coordinates of the point.
- \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) are the unit vectors along the x, y, and z axes, respectively.
Dot Product
The dot product, also known as the scalar product, is a fundamental operation in vector mathematics, determining how much one vector extends in the direction of another. For two vectors \(\mathbf{u}\) and \(\mathbf{v}\), the dot product is given by:\[\mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y + u_z v_z\]This results in a scalar (a single number) rather than a vector.
The dot product satisfies the following:
The dot product satisfies the following:
- If \(\mathbf{u} \cdot \mathbf{v} = 0\), the vectors are perpendicular to each other.
- The dot product provides a way to measure the angle between two vectors, crucial for understanding their relative orientation.
Normal Vector
A normal vector is essential for defining the orientation of a plane. This vector is orthogonal, or at right angles, to every vector lying on the plane's surface.
When given in the form of \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\), the normal vector helps in establishing the plane's equation by ensuring all directional components are accounted for.
When given in the form of \(\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}\), the normal vector helps in establishing the plane's equation by ensuring all directional components are accounted for.
- A plane can only be changed by altering its normal vector, making it pivot or rotate about points on the plane.
- The magnitude of the normal vector doesn't affect the plane's equation since direction, not length, is crucial.