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 is 9. (b) Area is \(\sqrt{35}\). (c) Angle: \(90^\circ - \phi\), where \(\phi = \cos^{-1}\left(\frac{-9}{14}\right)\).

Step 1: Finding the Volume of the Parallelepiped

The volume of a parallelepiped determined by vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) is given by the absolute value of the scalar triple product \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). First, calculate the cross product \( \mathbf{v} \times \mathbf{w} \):\[\mathbf{v} \times \mathbf{w} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 1 & 2 \1 & 3 & 3 \\end{vmatrix} = \mathbf{i}(1 \cdot 3 - 2 \cdot 3) - \mathbf{j}(1 \cdot 3 - 2 \cdot 1) + \mathbf{k}(1 \cdot 3 - 1 \cdot 1)\]Simplify the expression:\[\mathbf{v} \times \mathbf{w} = \langle -3, -1, 2 \rangle\]Now compute the dot product:\[\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \langle 3, 2, 1 \rangle \cdot \langle -3, -1, 2 \rangle = 3(-3) + 2(-1) + 1(2)\]Calculate the result:\[= -9 - 2 + 2 = -9\]Take the absolute value:\[| -9 | = 9\]Thus, the volume of the parallelepiped \( K \) is 9.

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

The area of a parallelogram determined by vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by the magnitude of their cross product \( \| \mathbf{u} \times \mathbf{v} \| \). Compute the cross product:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \3 & 2 & 1 \1 & 1 & 2 \\end{vmatrix} = \mathbf{i}(2 \cdot 2 - 1 \cdot 1) - \mathbf{j}(3 \cdot 2 - 1 \cdot 1) + \mathbf{k}(3 \cdot 1 - 2 \cdot 1)\]Simplify the expression:\[\mathbf{u} \times \mathbf{v} = \langle 3, -5, 1 \rangle\]Now, compute the magnitude:\[\| \mathbf{u} \times \mathbf{v} \| = \sqrt{3^2 + (-5)^2 + 1^2} = \sqrt{9 + 25 + 1} = \sqrt{35}\]Thus, the area of the face is \( \sqrt{35} \).

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

To find the angle between \( \mathbf{u} \) and the plane containing \( \mathbf{v} \) and \( \mathbf{w} \), first find a normal vector to the plane which is \( \mathbf{v} \times \mathbf{w} = \langle -3, -1, 2 \rangle \) as calculated earlier. The angle \( \theta \) between \( \mathbf{u} \) and the plane is given by:\[\theta = 90^\circ - \phi\]where \( \phi \) is the angle between \( \mathbf{u} \) and the normal vector, and is found using:\[\cos \phi = \frac{\mathbf{u} \cdot \mathbf{n}}{\| \mathbf{u} \| \| \mathbf{n} \|}\]First, calculate \( \mathbf{u} \cdot \mathbf{n} \) where \( \mathbf{n} = \langle -3, -1, 2 \rangle \):\[\mathbf{u} \cdot \mathbf{n} = \langle 3, 2, 1 \rangle \cdot \langle -3, -1, 2 \rangle = -9 - 2 + 2 = -9\]Now compute magnitudes:\[\| \mathbf{u} \| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}, \quad \| \mathbf{n} \| = \sqrt{(-3)^2 + (-1)^2 + 2^2} = \sqrt{14}\]Evaluate \( \cos \phi \):\[\cos \phi = \frac{-9}{\sqrt{14} \cdot \sqrt{14}} = \frac{-9}{14}\]Now, find \( \phi \):\[\phi = \cos^{-1}\left(\frac{-9}{14}\right)\]Finally, the angle \( \theta \) is:\[\theta = 90^\circ - \phi\]

Key Concepts

Understanding the Scalar Triple Product

To find the volume of a parallelepiped, we use the scalar triple product. This product involves three vectors, say \( \mathbf{u}, \mathbf{v}, \text{and} \, \mathbf{w} \). It is expressed as \( \mathbf{u} \cdot ( \mathbf{v} \times \mathbf{w}) \). Here's what it means and how it works:
The scalar triple product combines the dot product and the cross product into one. It's like putting two of your vector operations together and does all the "3D space" thinking for you.

• The cross product \( \mathbf{v} \times \mathbf{w} \) gives us another vector that is perpendicular to both \( \mathbf{v} \) and \( \mathbf{w} \). It's like creating a new direction that stands out from the plane containing \( \mathbf{v} \) and \( \mathbf{w} \).
• Then, the dot product with \( \mathbf{u} \) measures how much of \( \mathbf{u} \) aligns along that new direction. This tells us the volume, essentially asking: "How far does \( \mathbf{u} \) stretch into the height of the parallelepiped?"

So when you compute \( \mathbf{u} \cdot ( \mathbf{v} \times \mathbf{w}) \), you are effectively finding the volume of the parallelepiped outlined by the edges \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \). The calculation gives a scalar (a simple number) that represents this volume.
Don't forget to take the absolute value! Sometimes, the scalar triple product can give a negative number depending on the orientation of the vectors. But volume is a size, which means it's always positive.

Exploring the Cross Product

The cross product is an essential tool for working with vectors, especially in three dimensions. When you take the cross product of two vectors, such as \( \mathbf{u} \) and \( \mathbf{v} \), denoted by \( \mathbf{u} \times \mathbf{v} \), you get a new vector. This new vector is

• Perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \). Think of it as standing straight out from the plane they form together.
• The magnitude represents the area of the parallelogram that \( \mathbf{u} \) and \( \mathbf{v} \) form. That's how we find the area of faces of solid shapes like parallelepipeds.

To compute \( \mathbf{u} \times \mathbf{v} \), use the determinant of a matrix composed of the unit vectors \( \mathbf{i}, \mathbf{j}, \text{and} \mathbf{k} \) as the first row, followed by the components of \( \mathbf{u} \) and \( \mathbf{v} \).
This operation might look intimidating at first, but with practice, it's a handy trick to jump from two given vectors to useful geometric info about them.

Explaining the Angle Between a Vector and a Plane

Finding the angle between a vector and a plane can seem challenging, yet understanding the process simplifies it greatly. The trick is to first find the angle between the vector and a normal (perpendicular) to the plane.
The plane containing vectors \( \mathbf{v} \) and \( \mathbf{w} \) has a normal vector found using the cross product \( \mathbf{v} \times \mathbf{w} \). Let's call this normal \( \mathbf{n} \).
To measure the angle \( \phi \) between \( \mathbf{u} \) and \( \mathbf{n} \), we use the dot product formula:
\[\cos \phi = \frac{\mathbf{u} \cdot \mathbf{n}}{\| \mathbf{u} \| \| \mathbf{n} \|}\]This formula helps us find \( \phi \), the angle between \( \mathbf{u} \) and the normal.
However, the angle \( \theta \) we are ultimately interested in is between \( \mathbf{u} \) and the plane itself, not the normal. So, this angle \( \theta \) is:
\[\theta = 90^\circ - \phi \]By following these steps, finding angles related to planes becomes more straightforward. Just remember: work with the normal first; then adjust for the plane.