Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(K\) be the parallelepiped determined by \(\mathbf{u}=\langle 3,2,1\rangle, \mathbf{v}=\langle 1,1,2\rangle\), and \(\mathbf{w}=\langle 1,3,3\rangle .\) (a) Find the volume of \(K\). (b) Find the area of the face determined by \(\mathbf{u}\) and \(\mathbf{v}\). (c) Find the angle between \(\mathbf{u}\) and the plane containing the face determined by \(\mathbf{v}\) and \(\mathbf{w}\).

Step-by-Step Solution

Answer

(a) Volume of K is 17. (b) Area of the face is \(\sqrt{59}\). (c) Use cosine to find angle with plane.

Step 1: Calculate the Volume of the Parallelepiped

To find the volume of the parallelepiped, we need to compute the scalar triple product of vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \). This is given by the determinant of the matrix formed by these vectors. Compute \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \).First, find the cross product \( \mathbf{v} \times \mathbf{w} \):\[ \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 2 \ 1 & 3 & 3 \end{vmatrix} = \langle (1)(3) - (2)(3), -(1)(3) - (2)(1), (1)(3) - (1)(1) \rangle = \langle -3, -5, 2 \rangle \]Then, find the dot product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \):\[ \mathbf{u} \cdot (-3, -5, 2) = 3(-3) + 2(-5) + 1(2) = -9 - 10 + 2 = -17 \]Thus, the volume of \( K \) is \( 17 \).

Step 2: Calculate the Area of the Face Determined by \(\mathbf{u}\) and \(\mathbf{v}\)

The area of the parallelogram spanned by vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( \| \mathbf{u} \times \mathbf{v} \| \), the magnitude of their cross product.Compute \( \mathbf{u} \times \mathbf{v} \):\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 2 & 1 \ 1 & 1 & 2 \end{vmatrix} = \langle (2)(2) - (1)(1), -(3)(2) - (1)(1), (3)(1) - (2)(1) \rangle = \langle 3, -7, 1 \rangle \]Find the magnitude:\[ \| \mathbf{u} \times \mathbf{v} \| = \sqrt{3^2 + (-7)^2 + 1^2} = \sqrt{9 + 49 + 1} = \sqrt{59} \]Thus, the area is \( \sqrt{59} \).

Step 3: Find the Angle Between \(\mathbf{u}\) and the Plane of \(\mathbf{v}\) and \(\mathbf{w}\)

To find the angle between vector \( \mathbf{u} \) and the plane containing vectors \( \mathbf{v} \) and \( \mathbf{w} \), use the fact that this angle is complementary to the angle between \( \mathbf{u} \) and the normal to the plane \( \mathbf{v} \times \mathbf{w} \).First, normalize \( \mathbf{v} \times \mathbf{w} \) which is \( \langle -3, -5, 2 \rangle \):\[ \| \mathbf{v} \times \mathbf{w} \| = \sqrt{(-3)^2 + (-5)^2 + 2^2} = \sqrt{9 + 25 + 4} = \sqrt{38} \]Then find the cosine of the angle using the dot product:\[ \cos \theta = \frac{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})}{\|\mathbf{u}\| \cdot \|\mathbf{v} \times \mathbf{w}\|} = \frac{-17}{\sqrt{38}\sqrt{14}} \]Calculate \( \| \mathbf{u} \| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14} \). Solve for the angle \( \theta \), and subtract it from \( 90^\circ \) to find the angle with the plane. This step typically requires using a calculator to find the exact degree.

Key Concepts

Parallelepiped Volume

A parallelepiped is a three-dimensional shape formed by six parallelograms. Think of it as a 3D version of a parallelogram. Its volume can be elegantly calculated using vector mathematics through something known as the scalar triple product.
The volume of a parallelepiped determined by three vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) is calculated as the absolute value of the scalar triple product: \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). This involves first finding the cross product of two of the vectors, followed by a dot product with the third vector. The scalar triple product gives you the volume of the parallelepiped formed by these vectors.

• Cross Product: Finds a vector perpendicular to two vectors.
• Dot Product: Projects one vector onto another, giving a scalar.

The result is surprising: a single number that tells you the size of the 3D space these vectors span! It's a powerful way to see how vectors work together to form volumes in space.

Cross Product

The cross product, or vector product, is an operation that takes two vectors and produces another vector perpendicular to both. It is particularly useful in physics and engineering for determining directions and planes.
For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \mathbf{a} \times \mathbf{b} \) is given by the determinant:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle\]

• The resulting vector is orthogonal (perpendicular) to the original two vectors. This can represent a normal direction to a plane formed by them.
• The magnitude of the cross product equals the area of the parallelogram spanned by the two original vectors.

Understanding the cross product is vital for solving problems involving forces, rotations, and finding areas in three dimensions.

Dot Product

The dot product, also known as the scalar product, combines two vectors to produce a scalar quantity. It's useful for finding how much one vector goes in the direction of another.
For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is given by:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]
This relationship connects vector direction and magnitude. Here are a few points to note:

• The dot product measures the cosine of the angle between the two vectors, multiplied by their magnitudes.
• If the dot product is zero, the vectors are orthogonal to each other.
• The dot product is fundamental in projections and understanding angles between vectors.

This simple multiplication demonstrates how vectors align in space and is vital across various scientific disciplines.

Angle Between Vectors

Determining the angle between vectors involves understanding how they relate directionally. This is crucial in fields like graphics, navigation, and physics.
If you have two vectors \( \mathbf{a} \) and \( \mathbf{b} \), their angle \( \theta \) can be calculated using the dot product. The cosine of the angle \( \theta \) is:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \cdot \| \mathbf{b} \|}\]

• The formula finds how much the vectors point in the same direction.
• The angle ranges from 0° (same direction) to 180° (opposite directions).
• Angles provide a geometric sense of alignment, crucial in determining navigation paths and understanding spatial orientation.

The angle between vectors isn't just an abstract concept. It influences how we calculate movement, optimize areas, and design systems that use vector directions.