Question

The plane \(P_{1}\) contains the points \(A, B\) and \(C\), which have position vectors \(\mathbf{a}=-3 \mathbf{i}+2 \mathbf{j}, \mathbf{b}=7 \mathbf{i}+2 \mathbf{j}\) and \(\mathbf{c}=2 \mathbf{i}+3 \mathbf{j}+2 \mathbf{k}\) respectively. Plane \(P_{2}\) passes through \(A\) and is orthogonal to the line \(B C\), whilst plane \(P_{3}\) passes through \(B\) and is orthogonal to the line \(A C\). Find the coordinates of \(\mathbf{r}\), the point of intersection of the three planes.

Short answer

Coordinates \textbf{r} .

Step-by-step solution

- Find Vector BC

First, find the vector \(\textbf{BC}\) by subtracting the position vector of point B from point C: \[\textbf{BC} = \textbf{c} - \textbf{b} = (2 \textbf{i} + 3 \textbf{j} + 2 \textbf{k}) - (7 \textbf{i} + 2 \textbf{j}) = -5 \textbf{i} + \textbf{j} + 2 \textbf{k}\]

- Equation of Plane P2

Plane \(\text{P}_{2}\) contains point A and is orthogonal to \(\textbf{BC}\). The equation of Plane \(\text{P}_{2}\) can be written as: \[ -5(x - (-3)) + 1(y - 2) + 2(z - 0) = 0 \] Simplifying: \[ 5x + y + 2z + 13 = 0.\]

- Find Vector AC

Next, find the vector \(\textbf{AC}\) by subtracting the position vector of A from C: \[\textbf{AC} = \textbf{c} - \textbf{a} = (2 \textbf{i} + 3 \textbf{j} + 2 \textbf{k}) - (-3 \textbf{i} + 2 \textbf{j}) = 5 \textbf{i} + \textbf{j} + 2 \textbf{k}\]

- Equation of Plane P3

Plane \(\text{P}_{3}\) contains point B and is orthogonal to \(\textbf{AC}\). The equation of Plane \(\text{P}_{3}\) can be written as: \[ 5(x - 7) + 1(y - 2) + 2(z - 0) = 0 \] Simplifying: \[ 5x + y + 2z - 37 = 0.\]

- Equation of Plane P1

Plane \(\text{P}_{1}\) contains points A, B, and C. To find the normal vector of \(\text{P}_{1}\), we take the cross product of \(\textbf{AB}\) and \(\textbf{AC}\). Using \(\textbf{a}\) as \(-3 \textbf{i} + 2 \textbf{j}\), \(\textbf{b}\) as \7 \textbf{i} + 2 \textbf{j}\), \(\textbf{c} \textbf{a} - \textbf{b} = \(-3 + 3 \textbf{i} + 2 - 2 \textbf{k}\) = 10 \textbf{i}+0 \textbf{z}\For the cross product: \[| \textbf{i} & \textbf{j} & \textbf{k}| = 10 \textbf{i}\] Given the coefficient Plane \(\text{P}_{1}\):

Key concepts

vector subtraction

Vector subtraction is a fundamental operation in vector algebra. It involves subtracting the corresponding components of two vectors to find the resultant vector. For instance, to find the vector \(\textbf{BC}\) in the exercise, we subtract the components of vector \(\textbf{b}\) from those of vector \(\textbf{c}\). The formula is: \[ \textbf{BC} = \textbf{c} - \textbf{b} \]
This translates to:
\[ \textbf{BC} = (2\textbf{i} + 3\textbf{j} + 2\textbf{k}) - (7\textbf{i} + 2\textbf{j}) = -5\textbf{i} + \textbf{j} + 2\textbf{k} \] This resultant vector \(\textbf{BC}\) helps in defining orthogonality and plane equations.

plane equation

A plane equation in 3-dimensional space defines a flat surface where any point \(\textbf{r} = (x, y, z)\) satisfies the equation. Generally, a plane is represented by:
\[ ax + by + cz + d = 0 \]
Here, \(a, b, \) and \(c\) are the normal vector components, and \(d\) is a constant.
In the exercise, for Plane \(\text{P}_{2}\), passing through point \(A \) and orthogonal to \(\textbf{BC}\), we derive its equation:
\[ -5(x + 3) + 1(y - 2) + 2(z) = 0 \]
Upon simplifying:
\[ 5x + y + 2z + 13 = 0 \]
Similarly, for Plane \(\text{P}_{3}\) passing through point \(B \) and orthogonal to \(\textbf{AC}\):
\[ 5(x - 7) + y + 2(z) = 0 \]
And upon simplifying:
\[ 5x + y + 2z - 37 = 0 \]
The equations help to find the intersection points.

orthogonal planes

Orthogonal planes are planes that intersect at right angles, meaning their normal vectors are orthogonal (perpendicular) to each other. In the exercise, Plane \(\text{P}_{2}\) is orthogonal to the line segment \(\textbf{BC}\) and Plane \(\text{P}_{3}\) is orthogonal to line segment \(\textbf{AC}\).
The orthogonality condition allows the construction of their plane equations. When the planes are orthogonal to vectors \(\textbf{BC}\) and \(\textbf{AC}\), their normal vectors are aligned with these vectors. This ensures that the dot product of these normal vectors will be zero. For example, if \(\textbf{u}\) is normal to Plane \(\text{P}_{2}\) and \(\textbf{v}\) is normal to Plane \(\text{P}_{3}\),
\[ \textbf{u} \cdot \textbf{v} = 0 \]
This idea helps solve the problem by defining the exact orientation of each plane respective to their orthogonal lines.

position vectors

Position vectors are used to pinpoint locations in coordinate space relative to an origin. In this exercise, points \(A, B,\) and \(C\) are given with their position vectors:
\[\textbf{a} = -3 \textbf{i} + 2 \textbf{j}, \textbf{b} = 7 \textbf{i} + 2 \textbf{j}, \textbf{c} = 2 \textbf{i} + 3 \textbf{j} + 2 \textbf{k}\]
These vectors represent coordinates in 3D space - implying their magnitude and direction from the origin.
Using position vectors, we can easily find vector differences like \(\textbf{BC}\) and \(\textbf{AC}\) by subtracting the coordinates of one vector from another. For example, to find vector \(\textbf{AC}\):
\[ \textbf{AC} = \textbf{c} - \textbf{a} = (2 \textbf{i} + 3 \textbf{j} + 2 \textbf{k}) - (-3 \textbf{i} + 2 \textbf{j}) = 5 \textbf{i} + \textbf{j} + 2 \textbf{k}\]
This subtraction helps describe the direction and magnitude of vectors between points, crucial for forming planes.