Problem 29 - Algebra and Trigonometry

Question

Three vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) are given. (a) Find their scalar triple product \(\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) .\) (b) Are the vectors coplanar? If not, find the volume of the parallelepiped that they determine. $$ \mathbf{a}=\langle 1,2,3\rangle, \quad \mathbf{b}=\langle- 3,2,1\rangle, \quad \mathbf{c}=\langle 0,8,10\rangle $$

Step-by-Step Solution

Answer

(a) Scalar triple product is 0. (b) Vectors are coplanar.

Step 1: Find the Cross Product

Calculate the cross product of \( \mathbf{b} \times \mathbf{c} \). The formula for the cross product of vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) is given by:\[\mathbf{u} \times \mathbf{v} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{array} \right|\]Substitute the components of \( \mathbf{b} \) and \( \mathbf{c} \) into the determinant:\[\mathbf{b} \times \mathbf{c} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 2 & 1 \ 0 & 8 & 10 \end{array} \right| = (2 \cdot 10 - 1 \cdot 8)\mathbf{i} - (-3 \cdot 10 - 1 \cdot 0)\mathbf{j} + (-3 \cdot 8 - 2 \cdot 0)\mathbf{k}\]This simplifies to:\[= (20 - 8)\mathbf{i} + 30\mathbf{j} + (-24)\mathbf{k} = 12\mathbf{i} + 30\mathbf{j} - 24\mathbf{k}\]Thus, \( \mathbf{b} \times \mathbf{c} = \langle 12, 30, -24 \rangle \).

Step 2: Calculate the Scalar Triple Product

Compute the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \). Use the dot product formula:\[\mathbf{a} \cdot \mathbf{v} = a_1v_1 + a_2v_2 + a_3v_3\]With \( \mathbf{a} = \langle 1, 2, 3 \rangle \) and \( \mathbf{b} \times \mathbf{c} = \langle 12, 30, -24 \rangle \):\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1 \cdot 12 + 2 \cdot 30 + 3 \cdot (-24)\]Simplifying the expression gives:\[= 12 + 60 - 72 = 0\]

Step 3: Interpret the Scalar Triple Product

Check the result of the scalar triple product to determine coplanarity. If \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \), the vectors are coplanar. In this case, the product is indeed 0, indicating that the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) lie in the same plane.

Key Concepts

Cross Product

When you're dealing with vectors, the cross product is a fundamental operation that finds a vector perpendicular to two given vectors. If you have vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), their cross product is determined using the following formula:

• \( \mathbf{u} \times \mathbf{v} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{array} \right| \)

The result is a new vector that is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \), and its length represents the area of the parallelogram formed by these vectors.
To apply this to our vectors \( \mathbf{b} = \langle -3, 2, 1 \rangle \) and \( \mathbf{c} = \langle 0, 8, 10 \rangle \), we compute:

• \( \mathbf{b} \times \mathbf{c} = (2 \cdot 10 - 1 \cdot 8)\mathbf{i} - (-3 \cdot 10 - 1 \cdot 0)\mathbf{j} + (-3 \cdot 8 - 2 \cdot 0)\mathbf{k} \)

This simplifies to \( \langle 12, 30, -24 \rangle \). Understanding the cross product is vital for vector operations in physics and engineering.

Dot Product

The dot product is another key operation in vector algebra. It returns a scalar, not a vector, and is calculated using the components of two vectors. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), it is given by:

• \( \mathbf{a} \cdot \mathbf{v} = a_1v_1 + a_2v_2 + a_3v_3 \)

The dot product is essentially a measure of how much one vector extends in the direction of another. If the dot product is zero, this indicates that the two vectors are orthogonal (perpendicular).
In our problem, we calculate the scalar triple product using the dot product: \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \). With \( \mathbf{a} = \langle 1, 2, 3 \rangle \) and \( \mathbf{b} \times \mathbf{c} = \langle 12, 30, -24 \rangle \), we find:

• \( 1 \cdot 12 + 2 \cdot 30 + 3 \cdot (-24) = 12 + 60 - 72 = 0 \)

The result is zero, showing that the vector \( \mathbf{a} \) is orthogonal to the cross product of \( \mathbf{b} \) and \( \mathbf{c} \). This leads us to our next concept.

Vector Coplanarity

Vector coplanarity is determined by checking if three vectors lie on the same plane. This can be tested using the scalar triple product of three vectors: if \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) equals zero, the vectors are coplanar.
In our exercise, we found that the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \). This implies that the vectors \( \mathbf{a}, \mathbf{b}, \) and \( \mathbf{c} \) do indeed lie in the same plane, and thus, they are coplanar.
Vector coplanarity is essential in many mathematical and physics applications, such as analyzing forces in a system or describing the orientation of objects in 3D space.

Parallelepiped Volume

The volume of a parallelepiped, a three-dimensional figure formed by six parallelograms, can be determined using the scalar triple product. Specifically, the absolute value of \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) gives this volume.
A parallelepiped’s volume helps model and solve real-world problems involving 3D objects, forces, and movements. However, if the scalar triple product is zero, as we found in our problem, it means the volume is zero and the figure is flat, confirming coplanarity.
Understanding this concept can be quite useful in design and engineering contexts where spatial reasoning and volumetric calculations are essential.