Problem 11 - Advanced Engineering Mathematics

Question

In Problems \(1-12, \mathbf{a}=\langle 2,-3,4\rangle, \mathbf{b}=\langle-1,2,5\rangle\), and \(\mathbf{c}=\langle 3,6,-1\rangle .\) Find the indicated scalar or vector. \(\left(\frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}}\right) \mathbf{b}\)

Step-by-Step Solution

Answer

The vector is \( \langle -\frac{2}{5}, \frac{4}{5}, 2 \rangle \).

Step 1: Compute the Dot Product \( \mathbf{a} \cdot \mathbf{b} \)

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated by multiplying the corresponding components of the vectors \( \mathbf{a} = \langle 2,-3,4 \rangle \) and \( \mathbf{b} = \langle -1,2,5 \rangle \), and then summing the results. Thus, \( \mathbf{a} \cdot \mathbf{b} = (2)(-1) + (-3)(2) + (4)(5) = -2 - 6 + 20 = 12 \).

Step 2: Compute the Dot Product \( \mathbf{b} \cdot \mathbf{b} \)

To find \( \mathbf{b} \cdot \mathbf{b} \), we take the dot product of vector \( \mathbf{b} \) with itself. Compute as \( \mathbf{b} \cdot \mathbf{b} = (-1)(-1) + (2)(2) + (5)(5) = 1 + 4 + 25 = 30 \).

Step 3: Find the Scalar \( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \)

Now, use the results from Step 1 and Step 2 to find the scalar value. The scalar is calculated as \( \frac{12}{30} = \frac{2}{5} \).

Step 4: Scale Vector \( \mathbf{b} \) by \( \frac{2}{5} \)

Multiply each component of vector \( \mathbf{b} = \langle -1, 2, 5 \rangle \) by the scalar \( \frac{2}{5} \). This gives: \( \left( \frac{2}{5} \right) \mathbf{b} = \left( \frac{2}{5} \right) \langle -1, 2, 5 \rangle = \langle -\frac{2}{5}, \frac{4}{5}, 2 \rangle \).

Key Concepts

Understanding the Dot Product

The dot product is a fundamental operation in vector algebra, used to multiply two vectors and result in a scalar quantity. To compute the dot product of two vectors, you multiply the corresponding components of the vectors and add these products together.
For example, if you have two vectors, \( \mathbf{a} = \langle 2, -3, 4 \rangle \) and \( \mathbf{b} = \langle -1, 2, 5 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is found by:

• Multiply the first components: \( 2 \times -1 = -2 \)
• Multiply the second components: \( -3 \times 2 = -6 \)
• Multiply the third components: \( 4 \times 5 = 20 \)

Sum these results: \( -2 + (-6) + 20 = 12 \).
The dot product indicates the extent to which two vectors "point" in the same direction. A positive dot product indicates that vectors point in similar directions, while a zero or negative result can indicate orthogonality or opposite directions.

Grasping Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a single real number). This operation scales the vector by the scalar, impacting its magnitude but not its direction, unless the scalar is negative, in which case the direction reverses.

• Given a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \)
• A scalar \( k \)

The scalar multiplication results in a scaled vector \( k \mathbf{v} = \langle k v_1, k v_2, k v_3 \rangle \).
For example, multiplying vector \( \mathbf{b} = \langle -1, 2, 5 \rangle \) by a scalar \( \frac{2}{5} \) gives \( \left( \frac{2}{5} \right) \mathbf{b} = \langle -\frac{2}{5}, \frac{4}{5}, 2 \rangle \). This operation adjusts the length of \( \mathbf{b} \) without affecting its direction relative to the origin.

Vector Scaling and Its Effects

Vector scaling, closely related to scalar multiplication, refers to modifying a vector's size by multiplying it with a scalar. This is a common operation when one needs to change how long a vector is while maintaining its direction within the coordinate system.
The formula for scaling a vector can be generalized as \( k \mathbf{v} \), where \( k \) is a scalar value provided. With vector scaling, we aim to adjust only the magnitude of the vector.

• A vector's magnitude or length changes in proportional terms to the scalar.
• If \( k > 1 \), the vector becomes longer.
• If \( 0 < k < 1 \), the vector becomes shorter.

The given exercise illustrates this by using a scalar of \( \frac{2}{5} \), effectively reducing the magnitude of vector \( \mathbf{b} \) but maintaining its direction within the vector space. Consequently, the vector \( \mathbf{b} \) changes to \( \langle -\frac{2}{5}, \frac{4}{5}, 2 \rangle \), smaller in size yet identical in orientation on the vector field.