Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all vectors perpendicular to both \(\langle 1,-2,-3\rangle\) and \(\langle-3,2,0\rangle\).

Step-by-Step Solution

Answer

Any vector of the form \( c \cdot \langle 6, 9, 8 \rangle \) is perpendicular.

Step 1: Identify the Problem

We are tasked with finding vectors perpendicular to vectors \( \langle 1, -2, -3 \rangle \) and \( \langle -3, 2, 0 \rangle \). This means finding a vector that is orthogonal to both, typically done by finding their cross product.

Step 2: Calculate the Cross Product

To find a vector perpendicular to both \( \langle 1, -2, -3 \rangle \) and \( \langle -3, 2, 0 \rangle \), we need to take their cross product. Use the determinant formula for the cross product:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -2 & -3 \ -3 & 2 & 0 \end{vmatrix}\]This leads to:\[\mathbf{u} \times \mathbf{v} = \mathbf{i}(0 - (-6)) - \mathbf{j}(0 - 9) + \mathbf{k}(2 - (-6))\]Resulting in the vector: \[\langle 6, 9, 8 \rangle\].

Step 3: Verify Perpendicularity

To ensure \(\langle 6, 9, 8 \rangle\) is perpendicular to both original vectors, check the dot product with each:- \(\langle 1, -2, -3 \rangle \cdot \langle 6, 9, 8 \rangle = 1\cdot6 + (-2)\cdot9 + (-3)\cdot8 \)- Calculating gives us: \(6 - 18 - 24 = -36\). dot product is calculated as,- \(\langle -3, 2, 0 \rangle \cdot \langle 6, 9, 8 \rangle = -3\cdot6 + 2\cdot9 + 0\cdot8 \)- Calculating gives us: \(-18 + 18 + 0 = 0\). Both dot products result in zero, confirming orthogonality.

Step 4: Generalize the Solution

Since any scalar multiple of \( \langle 6, 9, 8 \rangle \) is also perpendicular to the two given vectors, the set of all vectors perpendicular to \( \langle 1, -2, -3 \rangle \) and \( \langle -3, 2, 0 \rangle \) can be written as \( c \cdot \langle 6, 9, 8 \rangle \), where \(c\) is any scalar.

Key Concepts

Cross Product

The cross product is a vector operation used to find a vector that is perpendicular to two given vectors in three-dimensional space.
This is especially useful when you are tasked with a problem requiring a perpendicular or orthogonal vector, such as in this exercise.
The cross product of two vectors \( \mathbf{u} = \langle a_1, b_1, c_1 \rangle \) and \( \mathbf{v} = \langle a_2, b_2, c_2 \rangle \) can be calculated using the following determinant formula:

• \( \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} \)
• This expands to: \( \mathbf{u} \times \mathbf{v} = \mathbf{i}(b_1\cdot c_2 - c_1\cdot b_2) - \mathbf{j}(a_1\cdot c_2 - c_1\cdot a_2) + \mathbf{k}(a_1\cdot b_2 - b_1\cdot a_2) \)

Thus, this process results in a new vector that not only shares no direction with the original vectors but also has a magnitude that is proportional to the area of the parallelogram formed by the original vectors.

Orthogonality

Orthogonality is a core concept in vector calculus, referring to vectors that are perpendicular to each other.
When two vectors are orthogonal, their dot product is zero, indicating that they meet at a right angle.
This property is crucial for various applications in mathematics and physics, such as determining directions that are independent.

• In th exercise, once the cross product \( \langle 6, 9, 8 \rangle \) is found, we need to check if it is orthogonal to both vectors \( \langle 1, -2, -3 \rangle \) and \( \langle -3, 2, 0 \rangle \).
• Taking the dot product of \( \langle 6, 9, 8 \rangle \) with each original vector returns zero, confirming that \( \langle 6, 9, 8 \rangle \) is indeed orthogonal to both.

Orthogonality ensures that these vectors are positioned in a manner that they do not influence each other directionally in 3D space.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors resulting in a scalar.
It is widely used to determine the angle between two vectors and to check for orthogonality.
The formula for the dot product between two vectors \( \mathbf{u} = \langle a_1, b_1, c_1 \rangle \) and \( \mathbf{v} = \langle a_2, b_2, c_2 \rangle \) is:

• \( \mathbf{u} \cdot \mathbf{v} = a_1 \cdot a_2 + b_1 \cdot b_2 + c_1 \cdot c_2 \)

In the exercise, the dot product is used as a verification step.
By taking the dot product of the cross product vector \( \langle 6, 9, 8 \rangle \) and each given vector, if both products equal zero, it confirms the orthogonality.
This is a practical approach to ascertain that vectors are perpendicular.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar, effectively scaling the vector's magnitude without changing its direction.
It's a fundamental operation that enables the creation of a family of vectors pointing in the same direction.
This is particularly useful when using vectors derived from the cross product, which can be multiplied by any scalar to remain perpendicular to the original vectors.

• In the context of the exercise, any vector perpendicular to \( \langle 1, -2, -3 \rangle \) and \( \langle -3, 2, 0 \rangle \) can be represented as \( c \cdot \langle 6, 9, 8 \rangle \), where \( c \) is any real number.
• Thus, scalar multiplication allows us to adjust the size of the vector \( \langle 6, 9, 8 \rangle \) without affecting its perpendicular relationship with the original vectors.

The ability to scale vectors is crucial in both theoretical and applied vector calculus, ensuring that orientations are preserved while magnitudes are manipulated.