Question

Find the volume of the parallelepiped defined by the given vectors. \(\vec{u}=\langle-1,2,1\rangle, \quad \vec{v}=\langle 2,2,1\rangle, \quad \vec{w}=\langle 3,1,3\rangle\)

Short answer

The volume of the parallelepiped is 23.

Step-by-step solution

Vector Cross Product

To find the volume of the parallelepiped, we first need to compute the cross product of two vectors. We'll use the vectors \(\vec{v} = \langle 2, 2, 1 \rangle\) and \(\vec{w} = \langle 3, 1, 3 \rangle\). The cross product \(\vec{v} \times \vec{w}\) is calculated by finding the determinant of the matrix: \[\vec{v} \times \vec{w} =\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \2 & 2 & 1 \3 & 1 & 3\end{vmatrix}\]The resulting vector is \(\vec{v} \times \vec{w} = \langle (2\cdot3 - 1\cdot1), -(2\cdot3 - 1\cdot1), (2\cdot1 - 2\cdot3) \rangle = \langle 5, -7, -4 \rangle\).

Dot Product Calculation

The volume of the parallelepiped is given by the absolute value of the scalar triple product \(\vec{u} \cdot (\vec{v} \times \vec{w})\). Now, we calculate the dot product of \(\vec{u} = \langle -1, 2, 1 \rangle\) and the result from Step 1, \(\vec{v} \times \vec{w} = \langle 5, -7, -4 \rangle\):\[\vec{u} \cdot (\vec{v} \times \vec{w}) = (-1)\cdot5 + 2\cdot(-7) + 1\cdot(-4)= -5 - 14 - 4 = -23\]

Compute Volume

The volume of the parallelepiped is the absolute value of the result of the dot product calculated in Step 2. Here, the volume is:\[|\vec{u} \cdot (\vec{v} \times \vec{w})| = |-23| = 23\]

Key concepts

Cross Product

The cross product is a fundamental operation in vector calculus, especially useful when working in three-dimensional space. It takes two vectors and returns another vector that is perpendicular to the plane containing the initial vectors. The resulting perpendicular vector is particularly important in physics and engineering for determining rotational directions and surface normals.

For two vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), the cross product \( \vec{a} \times \vec{b} \) is found using the determinant of a matrix involving the unit vectors \( \hat{i}, \hat{j}, \hat{k} \):

• The \(i\)-component: \( (a_2b_3 - a_3b_2) \)
• The \(j\)-component: \( -(a_1b_3 - a_3b_1) \)
• The \(k\)-component: \( (a_1b_2 - a_2b_1) \)

Using this method ensures consistency across mathematical and physical applications where knowing the direction orthogonal to two vectors is crucial. As seen in the solution, for vectors \( \vec{v} \) and \( \vec{w} \), their cross product was computed as \( \langle 5, -7, -4 \rangle \), demonstrating this process.

Dot Product

The dot product, sometimes called the scalar product, is another essential vector operation that outputs a scalar. Unlike the cross product, which results in a vector, the simplicity of the dot product is its ability to quantify the amount of one vector in the direction of another.

When you compute the dot product of two vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \), you're essentially multiplying their corresponding components and summing them up as follows:

• \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

This product gives the projection of one onto the other and can tell us if two vectors are orthogonal (perpendicular) since their dot product would be zero. In the original exercise, the dot product is part of computing the scalar triple product necessary for determining volume, where \( \vec{u} \cdot (\vec{v} \times \vec{w}) = -23 \) involves taking trilateral elements from the vectors.

Scalar Triple Product

The scalar triple product is a combination of the cross product and the dot product. It gives the volume of the parallelepiped defined by three vectors, an application common in geometry and physics. The free parameters in such formulas sustain the necessity of this triple product.

The scalar triple product itself is calculated as \( \vec{u} \cdot (\vec{v} \times \vec{w}) \), where the result from the cross product of \( \vec{v} \) and \( \vec{w} \) is then dotted with a third vector \( \vec{u} \). In simpler terms, it can be understood as a determinant comprised of these vectors:

• A positive result indicates a counter-clockwise orientation looking down the axis of \( \vec{u} \).
• A negative result similarly indicates a clockwise orientation.

The beauty of the scalar triple product is that it numerically represents three-dimensional volumes stored in vector form. In our case, the calculated volume was \(|-23| = 23\), representing the absolute value of this orientation-dependent result, giving scalar magnitude of space legally occupied by vectors.