Problem 14 - Essential Calculus Early Transcendentals

Question

Find \(\mathbf{a}+\mathbf{b}, 2 \mathbf{a}+3 \mathbf{b},|\mathbf{a}|,\) and \(|\mathbf{a}-\mathbf{b}|\) \(\mathbf{a}=4 \mathbf{i}+\mathbf{j}, \quad \mathbf{b}=\mathbf{i}-2 \mathbf{j}\)

Step-by-Step Solution

Answer

\(\mathbf{a} + \mathbf{b} = 5\mathbf{i} - \mathbf{j}\), \(2\mathbf{a} + 3\mathbf{b} = 11\mathbf{i} - 4\mathbf{j}\), \(|\mathbf{a}| = \sqrt{17}\), \(|\mathbf{a} - \mathbf{b}| = 3\sqrt{2}\)."

Step 1: Finding \(\mathbf{a} + \mathbf{b}\)

To add vectors \( \mathbf{a} \) and \( \mathbf{b} \), add their corresponding components. We have \( \mathbf{a} = 4\mathbf{i} + \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} - 2 \mathbf{j} \). So, \( \mathbf{a} + \mathbf{b} = (4\mathbf{i} + \mathbf{j}) + (\mathbf{i} - 2 \mathbf{j}) = (4\mathbf{i} + \mathbf{i}) + (\mathbf{j} - 2 \mathbf{j}) = 5\mathbf{i} - \mathbf{j} \).

Step 2: Finding \(2\mathbf{a} + 3\mathbf{b}\)

First, multiply \( \mathbf{a} = 4\mathbf{i} + \mathbf{j} \) by 2 to get \( 2\mathbf{a} = 2(4\mathbf{i} + \mathbf{j}) = 8\mathbf{i} + 2\mathbf{j} \). Then, multiply \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} \) by 3 to get \( 3\mathbf{b} = 3(\mathbf{i} - 2\mathbf{j}) = 3\mathbf{i} - 6\mathbf{j} \). Now, add these results: \( 2\mathbf{a} + 3\mathbf{b} = (8\mathbf{i} + 2\mathbf{j}) + (3\mathbf{i} - 6\mathbf{j}) = (8\mathbf{i} + 3\mathbf{i}) + (2\mathbf{j} - 6\mathbf{j}) = 11\mathbf{i} - 4\mathbf{j} \).

Step 3: Calculating the Magnitude of \(|\mathbf{a}|\)

The magnitude of a vector \( \mathbf{a} = 4\mathbf{i} + \mathbf{j} \) can be found using \( |\mathbf{a}| = \sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17} \).

Step 4: Finding \(|\mathbf{a} - \mathbf{b}|\)

First, find the difference vector \( \mathbf{a} - \mathbf{b} = (4\mathbf{i} + \mathbf{j}) - (\mathbf{i} - 2\mathbf{j}) = (4\mathbf{i} - \mathbf{i}) + (\mathbf{j} + 2\mathbf{j}) = 3\mathbf{i} + 3\mathbf{j} \). Now, the magnitude is \( |\mathbf{a} - \mathbf{b}| = \sqrt{(3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \).

Key Concepts

Vector Addition

Adding vectors is like combining similar elements. For vectors \( \mathbf{a} = 4\mathbf{i} + \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} \), adding them involves adding their \( \mathbf{i} \) and \( \mathbf{j} \) components separately. This means adding the \( \mathbf{i} \) components \( 4\mathbf{i} \) and \( \mathbf{i} \), which equals \( 5\mathbf{i} \). Then, add the \( \mathbf{j} \) components \( \mathbf{j} \) and \( -2\mathbf{j} \), resulting in \( -\mathbf{j} \). Thus, \( \mathbf{a} + \mathbf{b} = 5\mathbf{i} - \mathbf{j} \). This operation is crucial when you want to determine the resultant vector of two or more vectors acting at the same point.

Scalar Multiplication

When you multiply a vector by a scalar, you're essentially scaling the vector. Consider the vector \( \mathbf{a} = 4\mathbf{i} + \mathbf{j} \). If you multiply this vector by 2, each component gets multiplied separately: \( 2(4\mathbf{i} + \mathbf{j}) = 8\mathbf{i} + 2\mathbf{j} \). Similarly, for vector \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} \), multiplying by 3 gives \( 3(\mathbf{i} - 2\mathbf{j}) = 3\mathbf{i} - 6\mathbf{j} \).
After scaling, if you add these results, you apply vector addition principles: \( 8\mathbf{i} + 3\mathbf{i} = 11\mathbf{i} \) and \( 2\mathbf{j} - 6\mathbf{j} = -4\mathbf{j} \). Hence, \( 2\mathbf{a} + 3\mathbf{b} = 11\mathbf{i} - 4\mathbf{j} \).
This operation allows vectors to be resized without altering their direction, which is useful in adjusting forces or space proportions.

Vector Magnitude

The magnitude of a vector is the length of that vector. For \( \mathbf{a} = 4\mathbf{i} + \mathbf{j} \), its magnitude is calculated using the Pythagorean theorem. Imagine a right-angle triangle where \( 4\mathbf{i} \) and \( \mathbf{j} \) are its legs.
The formula to compute the magnitude is \( |\mathbf{a}| = \sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17} \). This basically sums the squares of each component and then finds their square root, thereby offering the 'distance' of the vector from the origin point.

• The length or magnitude tells us how 'big' or 'strong' a vector is.
• It is always a non-negative quantity.

Vector Subtraction

To subtract vectors, you take away the corresponding components of the second vector from the first. For the vectors \( \mathbf{a} = 4\mathbf{i} + \mathbf{j} \) and \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} \), subtraction involves \( (4\mathbf{i} - \mathbf{i}) = 3\mathbf{i} \) for the \( \mathbf{i} \) components, and \( (\mathbf{j} - (-2\mathbf{j})) = 3\mathbf{j} \) for the \( \mathbf{j} \) components, resulting in \( \mathbf{a} - \mathbf{b} = 3\mathbf{i} + 3\mathbf{j} \).
Calculating the magnitude of this difference vector, \( \mathbf{a} - \mathbf{b} \), involves using the same formula as before: \( |\mathbf{a} - \mathbf{b}| = \sqrt{(3)^2 + (3)^2} = \sqrt{9 + 9} = 3\sqrt{2} \). Vector subtraction helps in finding relative displacements or changes between vectors.