Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}=-2 \mathbf{i}+3 \mathbf{j}, \mathbf{b}=2 \mathbf{i}-3 \mathbf{j}\), and \(\mathbf{c}=-5 \mathbf{j}\). Find each of the following: (a) \(2 \mathbf{a}-4 \mathbf{b}\) (b) \(\mathbf{a} \cdot \mathbf{b}\) (c) \(\mathbf{a} \cdot(\mathbf{b}+\mathbf{c})\) (d) \((-2 \mathbf{a}+3 \mathbf{b}) \cdot 5 \mathbf{c}\) (e) \(\|\mathbf{a}\| \mathbf{c} \cdot \mathbf{a}\) (f) \(\mathbf{b} \cdot \mathbf{b}-\|\mathbf{b}\|\)

Step-by-Step Solution

Answer

a) -12i + 18j b) -13 c) -28 d) 375 e) -15√13 f) 13-√13

Step 1: Calculate 2a and 4b

First, we need to find \(2 \mathbf{a}\) and \(4 \mathbf{b}\). The vectors are given as \(\mathbf{a} = -2 \mathbf{i} + 3 \mathbf{j}\) and \(\mathbf{b} = 2 \mathbf{i} - 3 \mathbf{j}\). Thus, we have:\[2 \mathbf{a} = 2(-2 \mathbf{i} + 3 \mathbf{j}) = -4 \mathbf{i} + 6 \mathbf{j}\]\[4 \mathbf{b} = 4(2 \mathbf{i} - 3 \mathbf{j}) = 8 \mathbf{i} - 12 \mathbf{j}\]

Step 2: Find 2a - 4b

Now compute \(2 \mathbf{a} - 4 \mathbf{b}\):\[2 \mathbf{a} - 4 \mathbf{b} = (-4 \mathbf{i} + 6 \mathbf{j}) - (8 \mathbf{i} - 12 \mathbf{j})\]Combine like terms:\[= -4 \mathbf{i} + 6 \mathbf{j} - 8 \mathbf{i} + 12 \mathbf{j} = -12 \mathbf{i} + 18 \mathbf{j}\]

Step 3: Dot Product a and b

Next, find \(\mathbf{a} \cdot \mathbf{b}\):\[\mathbf{a} \cdot \mathbf{b} = (-2 \mathbf{i} + 3 \mathbf{j}) \cdot (2 \mathbf{i} - 3 \mathbf{j})\]The dot product is calculated as follows:\[= (-2)(2) + (3)(-3) = -4 - 9 = -13\]

Step 4: Find b + c

First, calculate \(\mathbf{b} + \mathbf{c}\):\(\mathbf{b} = 2 \mathbf{i} - 3 \mathbf{j}, \quad \mathbf{c} = -5 \mathbf{j}\)\[\mathbf{b} + \mathbf{c} = (2 \mathbf{i} - 3 \mathbf{j}) + (-5 \mathbf{j}) = 2 \mathbf{i} - 8 \mathbf{j}\]

Step 5: Dot Product a with (b + c)

Now find \(\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})\):\[\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = (-2 \mathbf{i} + 3 \mathbf{j}) \cdot (2 \mathbf{i} - 8 \mathbf{j})\]Calculate the dot product:\[= (-2)(2) + (3)(-8) = -4 - 24 = -28\]

Step 6: Scale a and b then Dot with c

First, compute \(-2 \mathbf{a} + 3 \mathbf{b}\):\[-2 \mathbf{a} + 3 \mathbf{b} = -2(-2 \mathbf{i} + 3 \mathbf{j}) + 3(2 \mathbf{i} - 3 \mathbf{j})\]\[= (4 \mathbf{i} - 6 \mathbf{j}) + (6 \mathbf{i} - 9 \mathbf{j}) = 10 \mathbf{i} - 15 \mathbf{j}\]Now dot it with \(5 \mathbf{c} = -25 \mathbf{j}\):\[(10 \mathbf{i} - 15 \mathbf{j}) \cdot (-25 \mathbf{j}) = 0 - 15(-25) = 375\]

Step 7: Magnitude of a and Dot with c

Find \(\|\mathbf{a}\|\) first:\[\|\mathbf{a}\| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}\]Now calculate \(\mathbf{c} \cdot \mathbf{a}\):\[\mathbf{c} \cdot \mathbf{a} = (-5 \mathbf{j}) \cdot (-2 \mathbf{i} + 3 \mathbf{j})\]The calculation shows:\[= 0 + (-15) = -15\]Finally, compute \(\|\mathbf{a}\| \mathbf{c} \cdot \mathbf{a}\):\[\sqrt{13} \times (-15) = -15 \sqrt{13}\]

Step 8: Magnitude and Dot Product of b

Calculate \(\mathbf{b} \cdot \mathbf{b}\) first:\[\mathbf{b} \cdot \mathbf{b} = (2 \mathbf{i} - 3 \mathbf{j}) \cdot (2 \mathbf{i} - 3 \mathbf{j}) = 4 + 9 = 13\]\(\|\mathbf{b}\|\) is also \(\sqrt{13}\), so the difference is:\[13 - \sqrt{13}\]

Key Concepts

Dot Product

In vector calculus, the dot product is a crucial tool used to determine the degree of alignment between two vectors. It's a scalar quantity, which means it results in a single number rather than a vector. To calculate the dot product, you multiply corresponding components of the two vectors and then sum these products together. This is best illustrated through the example of vectors \( \mathbf{a} = -2 \mathbf{i} + 3 \mathbf{j} \) and \( \mathbf{b} = 2 \mathbf{i} - 3 \mathbf{j} \).

• First, multiply the \( x \)-components: \( -2 \times 2 = -4 \).
• Next, multiply the \( y \)-components: \( 3 \times -3 = -9 \).
• Add these products together: \( -4 + (-9) = -13 \).

The result, \( -13 \), indicates the vectors are not aligned in the same direction, as it's a negative value. The dot product is also key in determining the angle between vectors, with further applications in physics like finding work done by a force.

Vector Magnitude

Vector magnitude, often referred to as the length or norm of a vector, measures how long a vector is regardless of its direction. Magnitude is always a non-negative scalar, akin to the length of a segment in geometry. You calculate it using the Pythagorean theorem applied in vector form. Let's look at vector \( \mathbf{a} = -2 \mathbf{i} + 3 \mathbf{j} \).

• Square each component: \( (-2)^2 = 4 \) and \( 3^2 = 9 \).
• Add these squares together: \( 4 + 9 = 13 \).
• Take the square root of this sum: \( \sqrt{13} \).

This \( \sqrt{13} \) is the magnitude of vector \( \mathbf{a} \), symbolically written as \( \| \mathbf{a} \| \). Measuring magnitude is essential for understanding a vector's influence in directions, helping in calculations like finding distances and normalizing vectors (scaling them to unit length).

Vector Addition

Vector addition combines two or more vectors to form a resultant vector. The key rule is to add corresponding components : \( x \)-components with \( x \)-components and \( y \)-components with \( y \)-components. This can be seen in the problem when adding vectors \( \mathbf{b} = 2 \mathbf{i} - 3 \mathbf{j} \) and \( \mathbf{c} = -5 \mathbf{j} \).

• Add the \( x \)-components: \( 2 + 0 = 2 \) (since \( \mathbf{c} \) lacks an \( x \)-component).
• Add the \( y \)-components: \( -3 + (-5) = -8 \).

Thus, the resultant vector is \( 2 \mathbf{i} - 8 \mathbf{j} \). This concept is widely used in physics to find net forces or velocities when multiple vectors act on a body.

Vector Subtraction

Vector subtraction is similar to vector addition but involves finding the difference between vectors. It operates by subtracting corresponding components of vectors, similar to reversing the direction and adding. In our exercise, this process is seen in the computation of \( 2 \mathbf{a} - 4 \mathbf{b} \). From vectors \( 2 \mathbf{a} = -4 \mathbf{i} + 6 \mathbf{j} \) and \( 4 \mathbf{b} = 8 \mathbf{i} - 12 \mathbf{j} \), the steps are:

• Subtract the \( x \)-components: \( -4 - 8 = -12 \).
• Subtract the \( y \)-components: \( 6 - (-12) = 18 \).

The result, \( -12 \mathbf{i} + 18 \mathbf{j} \), illustrates how vector subtraction is applied to solve for directions or differences in quantities, such as displacement or changes in force.