Problem 30 - Essential Calculus Early Transcendentals

Question

(a) Find a nonzero vector orthogonal to the plane through the points \(P, Q,\) and \(R,\) and (b) find the area of triangle \(P Q R .\) $$P(0,0,-3), \quad Q(4,2,0), \quad R(3,3,1)$$

Step-by-Step Solution

Answer

A nonzero vector orthogonal to the plane is \((-1, -7, 6)\), and the area of triangle \(PQR\) is \(\frac{1}{2} \times \sqrt{86}\).

Step 1: Determine Vectors in the Plane

First, find two vectors that lie in the plane. These vectors can be obtained by subtracting the coordinates of point \(P\) from those of points \(Q\) and \(R\) respectively.- Vector \(\vec{PQ} = Q - P = (4 - 0, 2 - 0, 0 - (-3)) = (4, 2, 3)\).- Vector \(\vec{PR} = R - P = (3 - 0, 3 - 0, 1 - (-3)) = (3, 3, 4)\).

Step 2: Find the Cross Product

The cross product of \(\vec{PQ}\) and \(\vec{PR}\) gives a vector that is orthogonal to the plane. Compute the cross product:\[\vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 2 & 3 \ 3 & 3 & 4 \end{vmatrix}\]Expanding the determinant:\[= \mathbf{i}(2 \cdot 4 - 3 \cdot 3) - \mathbf{j}(4 \cdot 4 - 3 \cdot 3) + \mathbf{k}(4 \cdot 3 - 2 \cdot 3)\]\[= \mathbf{i}(-1) - \mathbf{j}(7) + \mathbf{k}(6)\]\[= (-1, -7, 6)\]

Step 3: Calculate the Area of Triangle PQR

The area of triangle \(PQR\) can be found using the magnitude of the cross product from the previous step. The area is half the magnitude of \(\vec{PQ} \times \vec{PR}\):First, find the magnitude:\[\|\vec{PQ} \times \vec{PR}\| = \sqrt{(-1)^2 + (-7)^2 + 6^2}\]\[= \sqrt{1 + 49 + 36} = \sqrt{86}\]Thus, the area of the triangle is:\[\text{Area} = \frac{1}{2} \times \sqrt{86}\]

Key Concepts

Cross Product

The cross product is a fascinating operation in vector calculus that involves two vectors in three-dimensional space. Given two vectors \( \vec{a} \) and \( \vec{b} \), their cross product, denoted as \( \vec{a} \times \vec{b} \), results in another vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \). This orthogonal vector plays a crucial role when dealing with planes and areas.
To compute the cross product, we use the determinant method. For vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), the cross product is given by:

• \( \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)
• Calculate the determinant by expanding along the first row.
• The resulting vector \( (c_1, c_2, c_3) \) where each component is found using this expansion.

This process yields a new vector that not only tells us the direction perpendicular to the original vectors but also provides insight into the rotational direction given by the right-hand rule.

Orthogonal Vectors

Orthogonal vectors are a central concept in vector calculus and occur when two vectors are perpendicular to each other. When two vectors \( \vec{a} \) and \( \vec{b} \) are orthogonal, their dot product is zero: \( \vec{a} \cdot \vec{b} = 0 \). This principle is often employed in determining the normal vector to a plane, as seen with the cross product.
To find a vector orthogonal to a plane defined by two vectors within that plane, such as \( \vec{PQ} \) and \( \vec{PR} \), you calculate their cross product. The result is a vector at a right angle to both, meaning it is orthogonal to the plane itself.
Orthogonality is key in numerous applications, such as finding planes of projection and simplifying complex vector equations. In the context of geometry, an orthogonal vector is itself indicative of symmetry and balance, making it a powerful concept in mathematical modeling and physics.

Magnitude of a Vector

The magnitude of a vector gives the "length" or "size" of the vector. For a vector \( \vec{v} = (v_1, v_2, v_3) \), its magnitude \( \|\vec{v}\| \) is calculated using the formula:\[ \|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
This calculation is fundamental when interpreting vectors in a spatial context, as it tells you the actual distance the vector points from the origin to the point \((v_1, v_2, v_3)\).
The magnitude plays a pivotal role when you want to determine physical quantities like force or velocity, where direction and intensity matter. Additionally, in geometry, when assessing areas of shapes like triangles formed by vectors, the magnitude of the cross product directly relates to the area calculation. For instance, the area of a triangle is half the magnitude of the vector resulting from the cross product of two sides. This connection between magnitude and geometric properties highlights the intertwined nature of algebraic operations and geometric interpretations.