Problem 17 - Thomas Calculus

Question

In Exercises \(15-18\) , a. Find the area of the triangle determined by the points \(P, Q\) ,and \(R .\) b. Find a unit vector perpendicular to plane \(P Q R\) . $$ P(2,-2,1), \quad Q(3,-1,2), \quad R(3,-1,1) $$

Step-by-Step Solution

Answer

Area: \(\frac{1}{2}\sqrt{2}\); Unit vector: \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)\).

Step 1: Determine Vectors PQ and PR

First, we calculate the vectors \(\vec{PQ}\) and \(\vec{PR}\) from the given points. The vector \(\vec{PQ}\) is found by subtracting the coordinates of \(P\) from that of \(Q\):\[\vec{PQ} = (3 - 2, -1 + 2, 2 - 1) = (1, 1, 1)\]Similarly, \(\vec{PR}\) is calculated by subtracting the coordinates of \(P\) from those of \(R\):\[\vec{PR} = (3 - 2, -1 + 2, 1 - 1) = (1, 1, 0)\]

Step 2: Compute the Cross Product of Vectors PQ and PR

Next, we find the cross product \(\vec{PQ} \times \vec{PR}\). This gives us a vector normal to the plane containing points \(P, Q, R\):\[\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix}\]Calculating the determinant, we obtain:\[= \mathbf{i}(1 \cdot 0 - 1 \cdot 1) - \mathbf{j}(1 \cdot 0 - 1 \cdot 1) + \mathbf{k}(1 \cdot 1 - 1 \cdot 1)\]\[= -\mathbf{i} + \mathbf{j} + 0\mathbf{k} = (-1, 1, 0)\]

Step 3: Compute the Area of the Triangle PQR

The area of the triangle is half the magnitude of the cross product of \(\vec{PQ}\) and \(\vec{PR}\). Calculate the magnitude of \(\vec{PQ} \times \vec{PR}\):\[\|\vec{PQ} \times \vec{PR}\| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2}\]Thus, the area of the triangle is:\[\text{Area} = \frac{1}{2} \sqrt{2}\]

Step 4: Find a Unit Vector Perpendicular to Plane PQR

To find a unit vector perpendicular to the plane, we normalize the cross product vector \((-1, 1, 0)\). The magnitude of this vector is \(\sqrt{2}\), so the unit vector is:\[\text{Unit vector} = \frac{1}{\sqrt{2}}(-1, 1, 0)\]Simplified, the unit vector is:\[\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)\]

Key Concepts

Cross Product

The cross product is an essential concept in vector calculus. In vector algebra, the cross product of two vectors results in a third vector that is perpendicular to both of the original vectors. This operation only applies in three-dimensional space, making it particularly useful for 3D problems.

To calculate the cross product between two vectors \( \vec{a} \) and \( \vec{b} \), you can use a determinant involving the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):

• Arrange a matrix with \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row.
• Place components of \( \vec{a} \) in the second row.
• Place components of \( \vec{b} \) in the third row.

For the given exercise in the solution, vectors \( \vec{PQ} \) and \( \vec{PR} \) are derived from points \( P, Q, \) and \( R \). Their cross product gives a vector that is orthogonal to the plane containing these points, which is calculated as \((-1, 1, 0)\).

A use for the cross product includes finding perpendicular vectors, which is crucial for understanding the geometry of three-dimensional shapes.

Area of a Triangle

Finding the area of a triangle in a 3D coordinate system can be achieved using the cross product of two vectors that define the triangle. Imagine a triangle formed by three points \( P, Q, \) and \( R \). You can construct vectors \( \vec{PQ} \) and \( \vec{PR} \) from these points.

The area of the triangle is given by half the magnitude of the cross product of \( \vec{PQ} \) and \( \vec{PR} \). Mathematically, \[\text{Area} = \frac{1}{2} \|\vec{PQ} \times \vec{PR}\|\] In our exercise, after finding the cross product of vectors \( (1, 1, 1) \) and \( (1, 1, 0) \), we get \((-1, 1, 0)\). Then, the magnitude of this vector, \( \sqrt{2} \), helps compute the area of the triangle as \( \frac{1}{2} \sqrt{2} \).

Using the cross product simplifies calculations without having to delve into complex geometry equations, providing an efficient tool for area computation.

Unit Vector

A unit vector is a vector that has a magnitude of one. It is typically used to indicate direction only without regard to magnitude. To create a unit vector from a given vector, you divide each component of the vector by the vector's magnitude.

In our problem, once we have the normal vector \((-1, 1, 0)\) which is perpendicular to the plane \(PQR\), we can convert it to a unit vector. Here's how:

• Calculate the magnitude of the vector using the formula:\[|\vec{v}| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{2}\]
• Divide each component of \((-1, 1, 0)\) by \(\sqrt{2}\).

This results in the unit vector \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right) \).

Unit vectors are particularly important in physics and engineering as they describe directions in a standardized way. They help simplify solving vector equations by separating direction from magnitude.