Problem 22 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-28: Given that \(a=\langle 2,-3\rangle, b=\langle 3,4\rangle\), and \(c=\langle-1,5\rangle\), find the number. (a) \(\mathbf{b} \cdot(\mathbf{a}-\mathbf{c})\) (b) \(\mathbf{b} \cdot \mathbf{a}-\mathbf{b} \cdot \mathbf{c}\)

Step-by-Step Solution

Answer

Both (a) and (b) equal -23.

Step 1: Subtract Vector c from Vector a

Given vectors are \( \mathbf{a} = \langle 2, -3 \rangle \) and \( \mathbf{c} = \langle -1, 5 \rangle \). To find \( \mathbf{a} - \mathbf{c} \), subtract the corresponding components: \( (2 - (-1), -3 - 5) = (2 + 1, -8) = \langle 3, -8 \rangle \).

Step 2: Compute the Dot Product for Part (a)

Now compute the dot product \( \mathbf{b} \cdot (\mathbf{a} - \mathbf{c}) \). Given \( \mathbf{b} = \langle 3, 4 \rangle \) and \( \mathbf{a} - \mathbf{c} = \langle 3, -8 \rangle \), calculate: \( 3 \times 3 + 4 \times (-8) = 9 - 32 = -23 \).

Step 3: Compute the Dot Product \( \mathbf{b} \cdot \mathbf{a} \) for Part (b)

Find \( \mathbf{b} \cdot \mathbf{a} \) using \( \mathbf{b} = \langle 3, 4 \rangle \) and \( \mathbf{a} = \langle 2, -3 \rangle \). Calculate: \( 3 \times 2 + 4 \times (-3) = 6 - 12 = -6 \).

Step 4: Compute the Dot Product \( \mathbf{b} \cdot \mathbf{c} \) for Part (b)

Find \( \mathbf{b} \cdot \mathbf{c} \) using \( \mathbf{b} = \langle 3, 4 \rangle \) and \( \mathbf{c} = \langle -1, 5 \rangle \). Calculate: \( 3 \times (-1) + 4 \times 5 = -3 + 20 = 17 \).

Step 5: Compute \( \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{c} \) for Part (b)

Subtract the two results from Steps 3 and 4: \( -6 - 17 = -23 \). This gives \( \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{c} = -23 \).

Key Concepts

Dot Product

The dot product is an essential operation when dealing with vectors. It is a way of multiplying two vectors to result in a scalar (a single number). The formula for the dot product of two vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \) is:\[\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2\]In simple terms, you multiply each pair of corresponding components and then sum up those products.
In Part (a) of our exercise, we compute \( \mathbf{b} \cdot (\mathbf{a} - \mathbf{c}) \). We find the dot product between vector \( \mathbf{b} = \langle 3, 4 \rangle \) and the result of the subtraction \( \mathbf{a} - \mathbf{c} = \langle 3, -8 \rangle \). Calculating gives:

• \( 3 \times 3 = 9 \)
• \( 4 \times (-8) = -32 \)

Adding these gives \( 9 - 32 = -23 \).
Understanding dot products is crucial in vector mathematics as it can help in understanding angles, lengths, and projections between vectors.

Vector Subtraction

Vector subtraction is another important vector operation. It's all about taking one vector away from another by individually subtracting their corresponding components.
Let's break it down with our vectors \( \mathbf{a} = \langle 2, -3 \rangle \) and \( \mathbf{c} = \langle -1, 5 \rangle \).

• For the first components: subtract -1 from 2, that's \( 2 - (-1) = 3 \).
• For the second components: subtract 5 from -3, that's \( -3 - 5 = -8 \).

Thus, \( \mathbf{a} - \mathbf{c} \) becomes \( \langle 3, -8 \rangle \).
This operation is like following a path; moving from the start of the first vector to the endpoint of the subtracted vector.
Vector subtraction is used extensively in various applications like physics for understanding movement and relative positions.

Component-wise Operations

Component-wise operations are fundamental to working with vectors as they break down complex operations into manageable bits. These operations involve performing arithmetic directly on the same components of vectors parallelly.
In our exercise, we've used component-wise subtraction in Step 1:

• \( 2 - (-1) \): Subtraction is done on the first components.
• \( -3 - 5 \): Subtraction is done on the second components.

Similarly, in Steps 2 through 4, we applied component-wise multiplications when calculating the dot products.
Component-wise operations allow us to manipulate vectors easily and logically. By working on each component independently, we can simplify the problem-solving process substantially.
Grasping component-wise operations helps you become more comfortable with more advanced vector manipulation.