Problem 13 - Essential Calculus Early Transcendentals

Question

Find \(\mathbf{a}+\mathbf{b}, 2 \mathbf{a}+3 \mathbf{b},|\mathbf{a}|,\) and \(|\mathbf{a}-\mathbf{b}|\) \(\mathbf{a}=\langle 5,-12\rangle, \quad \mathbf{b}=\langle- 3,-6\rangle\)

Step-by-Step Solution

Answer

\( \mathbf{a} + \mathbf{b} = \langle 2, -18 \rangle \), \( 2\mathbf{a} + 3\mathbf{b} = \langle 1, -42 \rangle \), \(|\mathbf{a}| = 13 \), \(|\mathbf{a} - \mathbf{b}| = 10 \).

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

To find \( \mathbf{a} + \mathbf{b} \), add the corresponding components of \( \mathbf{a} \) and \( \mathbf{b} \): \( \langle 5, -12 \rangle + \langle -3, -6 \rangle = \langle 5 + (-3), -12 + (-6) \rangle = \langle 2, -18 \rangle \).

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

First, find \( 2\mathbf{a} = 2 \times \langle 5, -12 \rangle = \langle 10, -24 \rangle \) and \( 3\mathbf{b} = 3 \times \langle -3, -6 \rangle = \langle -9, -18 \rangle \). Then add them: \( \langle 10, -24 \rangle + \langle -9, -18 \rangle = \langle 1, -42 \rangle \).

Step 3: Calculate \(|\mathbf{a}|\)

The magnitude \(|\mathbf{a}|\) is calculated by \( \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \).

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

First find \( \mathbf{a} - \mathbf{b} = \langle 5, -12 \rangle - \langle -3, -6 \rangle = \langle 5 + 3, -12 + 6 \rangle = \langle 8, -6 \rangle \). Then, calculate the magnitude: \( \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \).

Key Concepts

Vector Addition

Vector addition is a fundamental operation used in mathematics and physics that involves combining two or more vectors. When you add vectors, you're essentially summing up each of their corresponding components.
To perform vector addition, take vector \( \mathbf{a} = \langle 5, -12 \rangle \) and vector \( \mathbf{b} = \langle -3, -6 \rangle \). Add the components step by step:

• First, add the x-components: \( 5 + (-3) = 2 \)
• Then, add the y-components: \( -12 + (-6) = -18 \)

This gives the resultant vector \( \langle 2, -18 \rangle \). The visual outcome is like moving one vector's endpoint to the start of another, creating a straight path from the start of the first vector to the end of the last. This can be very helpful in solving various problems in physics where multiple forces, represented by vectors, need to be combined.

Vector Magnitude

The magnitude of a vector is essentially its "length" or "size" in space. It often represents quantities that have both direction and size, such as displacement or force. Calculating vector magnitude involves using the Pythagorean theorem in the context of vector components.
For vector \( \mathbf{a} = \langle 5, -12 \rangle \), the magnitude \(|\mathbf{a}|\) is determined by:

• The formula: \( \sqrt{x^2 + y^2} \)
• Applying the components: \( \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)

So, the magnitude of vector \( \mathbf{a} \) is 13. This tells us how "long" the vector \( \mathbf{a} \) is, without considering its direction. Understanding magnitude is crucial, especially when comparing the sizes of different vectors.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a real number (a scalar), effectively scaling its length without altering its direction, unless the scalar is negative, which also reverses the direction.
Let's examine this using vectors \( \mathbf{a} \) and \( \mathbf{b} \):

• Multiply \( \mathbf{a} = \langle 5, -12 \rangle \) by 2: \( 2\mathbf{a} = \langle 10, -24 \rangle \)
• Multiply \( \mathbf{b} = \langle -3, -6 \rangle \) by 3: \( 3\mathbf{b} = \langle -9, -18 \rangle \)

After scaling them, the resultant vector can be added as per vector addition rules. These aspects show that scalar multiplication can modify a vector's magnitude, but the direction only changes with negative scalars. This operation is widely used across physics and engineering, for example in adjusting a force vector.

Subtraction of Vectors

Subtracting vectors is quite similar to vector addition, but instead of adding, you subtract each corresponding component of the vectors. It helps in finding the difference between two vectors.
For example, subtract vector \( \mathbf{b} = \langle -3, -6 \rangle \) from vector \( \mathbf{a} = \langle 5, -12 \rangle \) as follows:

• Subtract the x-components: \( 5 - (-3) = 5 + 3 = 8 \)
• Subtract the y-components: \( -12 - (-6) = -12 + 6 = -6 \)

This results in vector \( \langle 8, -6 \rangle \). Calculating the magnitude of this difference vector gives you its size:
\(|\mathbf{a} - \mathbf{b}| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\).
Subtracting vectors is an essential method in physics to find relative velocity or displacement.