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) \(-12\mathbf{i} + 18\mathbf{j}\), (b) \(-13\), (c) \(-28\), (d) \(375\), (e) \(-15\sqrt{13}\), (f) \(13 - \sqrt{13}\).

Step 1: Calculate 2a - 4b

First, scale vector \( \mathbf{a} \) by 2:\[ 2\mathbf{a} = 2(-2 \mathbf{i} + 3 \mathbf{j}) = -4 \mathbf{i} + 6 \mathbf{j} \]Next, scale vector \( \mathbf{b} \) by 4:\[ 4\mathbf{b} = 4(2 \mathbf{i} - 3 \mathbf{j}) = 8 \mathbf{i} - 12 \mathbf{j} \]Now subtract the two results:\[ 2\mathbf{a} - 4\mathbf{b} = (-4 \mathbf{i} + 6 \mathbf{j}) - (8 \mathbf{i} - 12 \mathbf{j}) = -12 \mathbf{i} + 18 \mathbf{j} \]

Step 2: Find dot product a · b

Compute the dot product of \( \mathbf{a} = -2 \mathbf{i} + 3 \mathbf{j} \) and \( \mathbf{b} = 2 \mathbf{i} - 3 \mathbf{j} \):\[ \mathbf{a} \cdot \mathbf{b} = (-2)(2) + (3)(-3) = -4 - 9 = -13 \]

Step 3: Find a · (b + c)

First, add vectors \( \mathbf{b} \) and \( \mathbf{c} \):\[ \mathbf{b} + \mathbf{c} = (2\mathbf{i} - 3\mathbf{j}) + (-5\mathbf{j}) = 2\mathbf{i} - 8\mathbf{j} \]Next, find the dot product with \( \mathbf{a} \):\[ \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = (-2 \times 2) + (3 \times -8) = -4 - 24 = -28 \]

Step 4: Calculate (-2a + 3b) · 5c

Start by calculating \(-2\mathbf{a} + 3\mathbf{b}\):\[ -2\mathbf{a} = -2(-2\mathbf{i} + 3\mathbf{j}) = 4\mathbf{i} - 6\mathbf{j} \]\[ 3\mathbf{b} = 3(2\mathbf{i} - 3\mathbf{j}) = 6\mathbf{i} - 9\mathbf{j} \]\[ -2\mathbf{a} + 3\mathbf{b} = (4\mathbf{i} - 6\mathbf{j}) + (6\mathbf{i} - 9\mathbf{j}) = 10\mathbf{i} - 15\mathbf{j} \]Now scale vector \( \mathbf{c} \) and compute the dot product:\[ 5\mathbf{c} = 5(-5\mathbf{j}) = -25\mathbf{j} \]\[ (-2\mathbf{a} + 3\mathbf{b}) \cdot 5\mathbf{c} = (10\mathbf{i} - 15\mathbf{j}) \cdot (-25\mathbf{j}) = 0(10 \times 0 + (-15 \times -25)) = 375 \]

Step 5: Compute ||a|| c · a

First, find the magnitude of \( \mathbf{a} \):\[ \|\mathbf{a}\| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]Next, find the dot product \( \mathbf{c} \cdot \mathbf{a} \):\[ \mathbf{c} = -5\mathbf{j} \rightarrow \mathbf{c} \cdot \mathbf{a} = 0(-2) + (-5)(3) = -15 \]Then, compute \( \|\mathbf{a}\| \mathbf{c} \cdot \mathbf{a} \):\[ \sqrt{13}(-15) = -15\sqrt{13} \]

Step 6: Evaluate b · b - ||b||

Start by computing \( \mathbf{b} \cdot \mathbf{b} \) (the dot product of \( \mathbf{b} \) with itself):\[ \mathbf{b} \cdot \mathbf{b} = (2 \times 2) + (-3 \times -3) = 4 + 9 = 13 \]Find the magnitude of \( \mathbf{b} \):\[ \|\mathbf{b}\| = \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]Finally, evaluate \( \mathbf{b} \cdot \mathbf{b} - \|\mathbf{b}\| \):\[ 13 - \sqrt{13} \]

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an important operation in vector calculus. It takes two vectors and returns a single number. This operation involves multiplying corresponding components of the vectors and then summing the results. For example, if we have vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} \), the dot product is calculated as follows:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]
This operation is particularly useful because it has several properties, such as:

• It is commutative, meaning \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \).
• If the dot product is zero, it indicates that the vectors are perpendicular.

In the solution, we used the dot product to calculate interactions between vectors like \( \mathbf{a} \cdot \mathbf{b} = -13 \) which tells us the extent to which these vectors "point in the same direction" or overlap.

Magnitude of a Vector

The magnitude of a vector is a measure of its length. It can be thought of as the distance from the origin to the tip of the vector in a Cartesian coordinate system. The magnitude is computed using the Pythagorean theorem. For a vector \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} \), the magnitude \( \|\mathbf{v}\| \) is:
\[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \]
This scalar value helps us understand how "big" or "long" a vector is. Some key uses for calculating the magnitude include:

• Normalization, where a vector is scaled down to have a magnitude of 1.
• Finding distances in vector spaces.

In our exercise, the magnitude was used to determine \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \). Knowing these lengths provided insight when those vectors were combined with others in operations like the dot product.

Vector Addition

Vector addition is a fundamental operation where two or more vectors are combined to form a resultant vector. This operation is performed by adding corresponding components of the vectors. If you have vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} \), the resultant vector \( \mathbf{r} \) from addition is:
\[ \mathbf{r} = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j} \]
Key points to remember about vector addition include:

• It is commutative, meaning \( \mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a} \).
• It allows for the graphical addition using the tip-to-tail method, visually demonstrating how vectors add to one another.

In this context, by adding vectors like \( \mathbf{b} \) and \( \mathbf{c} \), we easily extend our calculations to more complex expressions like \( \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) \), further expanding the concept into other operations.

Vector Scaling

Vector scaling involves multiplying a vector by a scalar, which stretches or shrinks the vector's length without affecting its direction. When a vector \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} \) is scaled by a scalar \( k \), the resultant vector is:
\[ k\mathbf{v} = (kv_1)\mathbf{i} + (kv_2)\mathbf{j} \]
The important aspects of vector scaling include:

• Scaling by a positive scalar increases or decreases the vector's magnitude, maintaining the direction.
• Scaling by a negative scalar inverts the direction of the vector.

In the solution, vector scaling helped calculate expressions like \( 2\mathbf{a} \) and \( 4\mathbf{b} \), aiding in vector subtraction and addition processes. This elementary operation facilitates the manipulation of vectors in many mathematical and physical applications.