Problem 40 - Precalculus Mathematics for Calculus

Question

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$\mathbf{u}=\langle- 6,6\rangle, \quad \mathbf{v}=\langle- 2,-1\rangle$$

Step-by-Step Solution

Answer

\(|\mathbf{u}| = 6\sqrt{2}, |\mathbf{v}| = \sqrt{5}, |2\mathbf{u}| = 12\sqrt{2}, \left|\frac{1}{2}\mathbf{v}\right| = \frac{\sqrt{5}}{2}, |\mathbf{u}+\mathbf{v}| = \sqrt{89}, |\mathbf{u}-\mathbf{v}| = \sqrt{65}, |\mathbf{u}| - |\mathbf{v}| = 6\sqrt{2} - \sqrt{5}\).

Step 1: Magnitude of Vector \( \mathbf{u} \)

The magnitude of a vector \( \mathbf{u} = \langle a, b \rangle \) is given by the formula \( |\mathbf{u}| = \sqrt{a^2 + b^2} \). For \( \mathbf{u} = \langle -6, 6 \rangle \), calculate:\[|\mathbf{u}| = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\]

Step 2: Magnitude of Vector \( \mathbf{v} \)

For \( \mathbf{v} = \langle -2, -1 \rangle \), use the same formula:\[|\mathbf{v}| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}\]

Step 3: Magnitude of Scalar Multiplication of \( \mathbf{u} \)

The magnitude of \( 2\mathbf{u} \) is calculated by multiplying the magnitude of \( \mathbf{u} \) by 2:\[|2\mathbf{u}| = 2|\mathbf{u}| = 2 \times 6\sqrt{2} = 12\sqrt{2}\]

Step 4: Magnitude of Scalar Multiplication of \( \mathbf{v} \)

For \( \frac{1}{2} \mathbf{v} \), divide the magnitude of \( \mathbf{v} \) by 2:\[\left|\frac{1}{2} \mathbf{v}\right| = \frac{1}{2} |\mathbf{v}| = \frac{1}{2} \times \sqrt{5} = \frac{\sqrt{5}}{2}\]

Step 5: Magnitude of Sum of Vectors \( \mathbf{u} + \mathbf{v} \)

To find \( \mathbf{u} + \mathbf{v} \), add the components:\[\mathbf{u} + \mathbf{v} = \langle -6 + (-2), 6 + (-1) \rangle = \langle -8, 5 \rangle\]Then calculate the magnitude:\[|\mathbf{u} + \mathbf{v}| = \sqrt{(-8)^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89}\]

Step 6: Magnitude of Difference of Vectors \( \mathbf{u} - \mathbf{v} \)

Subtract the components to obtain \( \mathbf{u} - \mathbf{v} \):\[\mathbf{u} - \mathbf{v} = \langle -6 - (-2), 6 - (-1) \rangle = \langle -4, 7 \rangle\]Then calculate the magnitude:\[|\mathbf{u} - \mathbf{v}| = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}\]

Step 7: Difference of Magnitudes \( |\mathbf{u}| - |\mathbf{v}| \)

Simply subtract the magnitudes:\[|\mathbf{u}| - |\mathbf{v}| = 6\sqrt{2} - \sqrt{5}\]

Key Concepts

Vector Addition

Vector addition involves connecting two vectors end-to-end to form a new vector, known as the resultant. When adding two vectors, each component of the vectors is added separately. For example, with vectors \( \mathbf{u} = \langle -6, 6 \rangle \) and \( \mathbf{v} = \langle -2, -1 \rangle \), their sum is \( \mathbf{u} + \mathbf{v} = \langle -6 + (-2), 6 + (-1) \rangle = \langle -8, 5 \rangle \).
This means:

• Add the first components: \(-6 + (-2) = -8\)
• Add the second components: \(6 + (-1) = 5\)

Vector addition is straightforward but forms the basis for numerous vector operations and applications.

Scalar Multiplication

Scalar multiplication involves multiplying each component of a vector by a scalar (a constant number) to form a new vector. If you have vector \( \mathbf{u} = \langle a, b \rangle \) and want to scale it by 2, every component is multiplied by 2, resulting in \( 2 \mathbf{u} = \langle 2a, 2b \rangle \).
For the given vector \( \mathbf{u} = \langle -6, 6 \rangle \), scalar multiplication by 2 leads to \( 2 \mathbf{u} = \langle -12, 12 \rangle \).

• Multiplying by 2 elongates the vector double its original size.
• This keeps the direction same but changes the magnitude.

The operation can also shrink a vector, as in the case of \( \frac{1}{2} \mathbf{v} = \langle -1, -0.5 \rangle \), effectively halving its size.

Difference of Vectors

The difference of vectors involves subtracting the components of one vector from another. This operation results in a new vector. For example, subtracting vector \( \mathbf{v} = \langle -2, -1 \rangle \) from vector \( \mathbf{u} = \langle -6, 6 \rangle \) gives us \( \mathbf{u} - \mathbf{v} = \langle -6 - (-2), 6 - (-1) \rangle = \langle -4, 7 \rangle \).
Here's how this works:

• Subtract the first components: \(-6 - (-2) = -4\)
• Subtract the second components: \(6 - (-1) = 7\)

This vector effectively "points" from the head of vector \( \mathbf{v} \) to the head of vector \( \mathbf{u} \). It helps to analyze how a change can occur between two points.

Pythagorean Theorem

The Pythagorean theorem is used to calculate the magnitude (or length) of a vector from its components in a 2-dimensional space. The formula is \( |\mathbf{v}| = \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the components of a vector \( \mathbf{v} = \langle a, b \rangle \).
For example:

• Vector \( \mathbf{u} = \langle -6, 6 \rangle \) has a magnitude \( |\mathbf{u}| = \sqrt{(-6)^2 + 6^2} = 6\sqrt{2} \).
• Vector \( \mathbf{v} = \langle -2, -1 \rangle \) has a magnitude \( |\mathbf{v}| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{5} \).

The theorem is crucial for determining the size of vectors and plays a foundational role in trigonometry and geometry.