Question

Points A, B and C have position vectors \((9,1,1),(8,1,1)\) and \((9,0,2)\). Find (a) the equation of the plane containing A, B and \(\mathrm{C}\) (b) the area of the triangle \(\mathrm{ABC}\).

Short answer

Solution: The equation of the plane containing points A, B, and C is z = 1. The area of triangle ABC is ½ square units.

Step-by-step solution

Find the vectors AB and AC

Let's find the vectors AB and AC by subtracting the position vectors of the corresponding points. $$\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} = (8,1,1) - (9,1,1) = (-1,0,0)$$ $$\overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} = (9,0,2) - (9,1,1) = (0,-1,1)$$

Find the normal vector to the plane

Now, let's find the normal vector to the plane containing points A, B, and C. This can be done by taking the cross product of the vectors AB and AC. $$\overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 0 & 0 \\ 0 & -1 & 1 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}((-1)(-1) - 0) = (0,0,1)$$ The normal vector to the plane is \((0,0,1)\).

Write the equation of the plane

With the normal vector found, we can now write the equation of the plane using the point-normal form of a plane equation. Point-normal form: \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\) Using the coordinates of point A and the components of the normal vector: $$0(x - 9) + 0(y - 1) + 1(z - 1) = 0$$ The equation of the plane is simply \(z - 1 = 0\) or \(z = 1\).

Calculate the area of triangle ABC

To find the area of triangle ABC, we will use the magnitude of the cross product of vectors AB and AC. Area of triangle ABC: $$\frac{1}{2} | \overrightarrow{AB} \times \overrightarrow{AC} |$$ The magnitude of the cross product is: $$| (0,0,1) | = \sqrt{(0)^2 + (0)^2 + (1)^2} = 1$$ Now, we calculate the area of triangle ABC: $$\text{Area} = \frac{1}{2} \times 1 = \frac{1}{2}$$ The area of triangle ABC is \(\frac{1}{2}\) square units.

Key concepts

Understanding Position Vectors

Position vectors are fundamental in the study of geometry and physics, as they provide a means to describe the location of points in space relative to a fixed origin. In essence, a position vector is a directed line segment that starts from the origin of the coordinate system (usually denoted as the point (0,0,0) in three dimensions) and ends at the point in question.

For example, in the exercise provided, point A with a position vector \( (9,1,1) \) effectively communicates that starting from the origin, you move 9 units along the x-axis, 1 unit along the y-axis, and 1 unit along the z-axis to reach point A. Similarly, points B and C have their own position vectors that specify their locations in space. By understanding position vectors, students can easily calculate vectors between points, such as \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \), simply by subtracting the coordinates, which is the first step in solving our original exercise.

It's also crucial to point out that these vectors between points are what we use to define directions and magnitudes of sides within geometries, like triangles in our problem. They provide the base for further calculations such as the cross product, which leads to determining the area of shapes or the equations of planes.

Vector Cross Product

The vector cross product is an operation on two vectors in three-dimensional space that yields a third vector which is perpendicular to the plane containing the original vectors. This outcome, known as the normal vector, is central to defining planes in three dimensions and is calculated using the determinant of a matrix constructed from the components of the two vectors along with unit vectors.

In the context of our exercise, once we have vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \), we can find vector \( \overrightarrow{n} \) by computing the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \). This results in \( \overrightarrow{n}=(0,0,1) \), which is essential for determining the orientation of the plane that contains triangle ABC. The direction of this normal vector is a key piece of information needed to write the equation of the plane.

Importance of the Normal Vector

Not only does the normal vector help in defining the equation of a plane, but it is also instrumental in calculating areas. In three-dimensional space, knowing the direction and magnitude of this perpendicular vector allows for the computation of areas of parallelograms and triangles formed by the original vectors. This leads us to the final concept: calculating the area of a triangle using the vector cross product.

Calculating the Area of a Triangle

When it comes to determining the area of a triangle in three-dimensional space, the vector cross product plays a pivotal role. The magnitude of the cross product between two vectors gives twice the area of the triangle formed by those vectors as sides.

In our given example, we calculated the area of triangle ABC using the magnitude of the cross product \( | \overrightarrow{AB} \times \overrightarrow{AC} | \). The result was \( | (0,0,1) | \) which equals to 1. Consequently, following the formula for the area of a triangle (half the magnitude of the cross product), we get the area as \( \frac{1}{2} \).

Visualizing the Area

The area of a triangle in 3D space can be visualized as the amount of two-dimensional space it covers. It is important for students to connect the abstract calculation involving cross products with the concrete idea of 'space covered'. Through practice, this understanding can deepen, allowing students to grasp more intricate geometric and vector concepts as their studies progress.