Problem 11 - Calculus: Early Transcendentals

Question

Define the triple product of three vectors, \(\boldsymbol{x}, \boldsymbol{y},\) and \(z,\) to be the scalar \(\boldsymbol{x} \cdot(\boldsymbol{y} \times \boldsymbol{z}) .\) Show that three vectors lie in the same plane if and only if their triple product is zero. Verify that \(\langle 1,5,-2\rangle,\) \langle 4,3,0\rangle and \langle 6,13,-4\rangle all lie in the same plane.

Step-by-Step Solution

Answer

The vectors are coplanar since their triple product is zero.

Step 1: Understanding the Triple Product

The expression \( \boldsymbol{x} \cdot(\boldsymbol{y} \times \boldsymbol{z}) \) is called the scalar triple product. It results in a scalar and gives the volume of the parallelepiped formed by the vectors \( \boldsymbol{x} \), \( \boldsymbol{y} \), and \( \boldsymbol{z} \).

Step 2: Condition for Coplanar Vectors

The three vectors lie in the same plane if the volume of the parallelepiped they form is zero. Therefore, the triple product being zero, \( \boldsymbol{x} \cdot(\boldsymbol{y} \times \boldsymbol{z}) = 0\), is a condition for the vectors to be coplanar.

Step 3: Calculating the Cross Product

Calculate the cross product \( \boldsymbol{y} \times \boldsymbol{z} \). Let \( \boldsymbol{y} = \langle 4, 3, 0 \rangle \) and \( \boldsymbol{z} = \langle 6, 13, -4 \rangle \). Using the determinant form: \[ \boldsymbol{y} \times \boldsymbol{z} = \begin{vmatrix} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \ 4 & 3 & 0 \ 6 & 13 & -4 \end{vmatrix} \].

Step 4: Evaluating the Determinant

The determinant is calculated as: \[ \boldsymbol{y} \times \boldsymbol{z} = \boldsymbol{i}(3 \cdot (-4) - 0 \cdot 13) - \boldsymbol{j}(4 \cdot (-4) - 0 \cdot 6) + \boldsymbol{k}(4 \cdot 13 - 3 \cdot 6) \]. This simplifies to \( \langle -12, 16, 34 \rangle \).

Step 5: Calculating the Dot Product

Now, compute \( \boldsymbol{x} \cdot (\boldsymbol{y} \times \boldsymbol{z}) \) where \( \boldsymbol{x} = \langle 1, 5, -2 \rangle \). Dot product: \( 1(-12) + 5(16) + (-2)(34) = -12 + 80 - 68 = 0 \).

Step 6: Conclusion of Coplanarity

Since the scalar triple product \( \boldsymbol{x} \cdot (\boldsymbol{y} \times \boldsymbol{z}) \) equals zero, the vectors \( \langle 1, 5, -2 \rangle \), \( \langle 4, 3, 0 \rangle \), and \( \langle 6, 13, -4 \rangle \) are coplanar.

Key Concepts

Vector Cross Product

The vector cross product, also known as the cross product, is a vital operation in vector mathematics. It involves two vectors to generate a third vector perpendicular to the plane formed by the first two. Given two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), their cross product is defined as:

• \( \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \)

This new vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), and its magnitude represents the area of the parallelogram formed by these vectors.

When calculating the cross product using the determinant method, as shown in the provided exercise, we arrange the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of the vectors in a 3x3 matrix format. This assists in achieving the correct coefficients for the resulting vector.

Understanding cross product is crucial when exploring concepts like torque in physics or finding normals to surfaces in computer graphics.

Dot Product

The dot product, sometimes referred to as the scalar product, is another foundational concept in vector operations. It takes two vectors and results in a single scalar value. 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:

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

The dot product measures the extent to which two vectors point in the same direction. It can determine orthogonality; if the dot product is zero, the vectors are perpendicular.

Additionally, the dot product relates to the cosine of the angle between the two vectors, given by:

• \( \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta \)

Here, \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \), and \( \| \mathbf{a} \| \) and \( \| \mathbf{b} \| \) are the magnitudes of the vectors. This relationship demonstrates how the dot product can indicate the angular separation between two vectors.

Mastering the dot product is beneficial in fields like physics for work and energy calculations, and in computer graphics for detecting surface visibility.

Coplanarity of Vectors

Vectors are said to be coplanar if they lie within the same plane. The concept of coplanarity is crucial when dealing with spatial geometry since it helps determine relative positions and alignments. According to vector algebra, three vectors \( \boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z} \) are coplanar if their scalar triple product is zero.

• \( \boldsymbol{x} \cdot (\boldsymbol{y} \times \boldsymbol{z}) = 0 \)

This scalar triple product represents the volume of the parallelepiped spanned by the vectors. If the volume is zero, it implies that the shape collapses onto a plane, indicating that the vectors are coplanar.

The calculation involves first taking the cross product of two vectors to form a perpendicular vector, then taking the dot product with the third vector to find the volume. As shown in the exercise, verifying coplanarity is crucial in solving geometric problems and aids in understanding spatial arrangements in fields like computer-aided design and robotics.

Gaining a grasp on coplanarity aids in many applications, from determining structural integrity to optimizing data visualization.