Problem 18 - University Calculus: Early Transcendentals

Question

a. Find the area of the triangle determined by the points \(P, Q\) and \(R\). b. Find a unit vector perpendicular to plane \(P Q R\). $$P(-2,2,0), \quad Q(0,1,-1), \quad R(-1,2,-2)$$

Step-by-Step Solution

Answer

The area is \(\frac{\sqrt{14}}{2}\). The unit perpendicular vector is \(\left(\frac{-2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\right)\).

Step 1: Find Vectors for Two Sides of the Triangle

To find the area of the triangle and a perpendicular vector, we need vectors for two sides of the triangle. Let's calculate the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\):\[\overrightarrow{PQ} = Q - P = (0 - (-2), 1 - 2, -1 - 0) = (2, -1, -1)\]\[\overrightarrow{PR} = R - P = (-1 - (-2), 2 - 2, -2 - 0) = (1, 0, -2)\]

Step 2: Compute the Cross Product of the Vectors

The area of the triangle is half the magnitude of the cross product of \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\). Compute the cross product:\[\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & -1 \ 1 & 0 & -2 \end{vmatrix} = (-2)\mathbf{i} - (-1)(-2)\mathbf{j} + (0)(-1)\mathbf{k} = (-2, 3, 1)\]

Step 3: Calculate the Magnitude of the Cross Product

Find the magnitude of the vector \((-2, 3, 1)\) to use in calculating the triangle area:\[\|(-2, 3, 1)\| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}\]

Step 4: Determine the Area of the Triangle

The area of the triangle \(PQR\) is half the magnitude of the cross product:\[\text{Area} = \frac{1}{2} \times \sqrt{14} = \frac{\sqrt{14}}{2}\]

Step 5: Find Unit Vector Perpendicular to the Plane

To find a unit vector perpendicular to the triangle's plane, normalize the cross product \((-2, 3, 1)\):\[\text{Unit vector} = \frac{1}{\sqrt{14}}(-2, 3, 1) = \left(\frac{-2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}\right)\]

Key Concepts

Cross Product

When you need to find a vector that is perpendicular to two given vectors, the cross product is your best tool. This operation is essential in vector calculus, especially when dealing with problems in 3D space. For any two vectors \(\mathbf{A} = (a_1, a_2, a_3)\) and \(\mathbf{B} = (b_1, b_2, b_3)\), the cross product \(\mathbf{A} \times \mathbf{B}\) results in a new vector that is orthogonal to both \(\mathbf{A}\) and \(\mathbf{B}\). The formula for this is:

• \(\mathbf{C} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\),

which computes as \((a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\). This vector is not only perpendicular but also intrinsic to calculating the area of a parallelogram formed by \(\mathbf{A}\) and \(\mathbf{B}\). Additionally, the magnitude of this cross product gives you twice the area of the triangle created by the vectors. Remember, the cross product is different from the dot product, as it results in a vector, not a scalar.

Triangle Area in 3D Space

Calculating the area of a triangle in 3D space can seem complex, but it's simplified by using the cross product. When the vertices of the triangle are known, such as points \(P, Q,\) and \(R\), we first determine vectors along two sides of the triangle, such as \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\). Once we have these, we use the cross product to find a vector perpendicular to the plane containing the triangle. The magnitude of this cross product vector provides twice the area of the triangle:

• \(\text{Area} = \frac{1}{2} \times \| \overrightarrow{PQ} \times \overrightarrow{PR} \|\)

In simpler terms, you find the cross product, calculate its magnitude, and then take half of that value to get your triangle's area. This method is efficient since it directly accounts for the spatial orientation of the triangle's vertices.

Unit Vector

A unit vector is a vector that has a length of one and indicates direction. In tasks involving vectors in 3D space, such as defining perpendicularity to a plane, unit vectors are fundamental. After obtaining a vector, like the cross product from our previous steps, you can convert it into a unit vector by normalizing it. This is done by dividing each component of the vector by its magnitude.

• Given a vector \(\mathbf{A} = (x, y, z)\), the unit vector \(\mathbf{u}\) in the direction of \(\mathbf{A}\) is \(\mathbf{u} = \frac{1}{\|\mathbf{A}\|}(x, y, z)\).

This effectively scales the vector \(\mathbf{A}\) to have a magnitude of one, ensuring it purely represents direction, not magnitude. Unit vectors are extremely valuable in defining directions of lines and orientations in space, serving as a standard for vector operations.

Plane Perpendicularity

When tasked with finding a vector perpendicular to a plane defined by points, such as \(P, Q,\) and \(R\), understanding plane perpendicularity is crucial. A vector perpendicular to a plane is orthogonal to any vector lying on that plane. This is where the cross product comes into play again. By calculating the cross product of vectors along two edges of the triangle, typically \(\overrightarrow{PQ}\) and \(\overrightarrow{PR}\), you obtain a vector that is perpendicular to the plane. This vector's components directly relate to the orientations of each axis:

• If it's part of coordinate geometry, this perpendicular vector can help in forming equations of planes or identifying normal directions.

Once a perpendicular vector is identified, it usually needs to be converted into a unit vector to establish a standard directional reference. This capability of finding perpendicular directions is crucial in 3D modeling, computer graphics, and physics simulations.