Problem 79 - Precalculus with Calculus Previews

Question

\(\mathbf{u}=\langle 1,-1\rangle\) and \(\mathbf{v}=\langle 2,6\rangle\). Find the indicated number. $$ \operatorname{com} p_{\mathbf{u}}(\mathbf{v}-\mathbf{u}) $$

Step-by-Step Solution

Answer

The component is \(-3\sqrt{2}\).

Step 1: Understand Vector Operations

The problem requires finding the component of the vector \( \mathbf{v} - \mathbf{u} \) along \( \mathbf{u} \). First, let's compute \( \mathbf{v} - \mathbf{u} \).

Step 2: Subtracting Vectors

Subtract the vector \( \mathbf{u} = \langle 1, -1 \rangle \) from the vector \( \mathbf{v} = \langle 2, 6 \rangle \):\[ \mathbf{v} - \mathbf{u} = \langle 2, 6 \rangle - \langle 1, -1 \rangle = \langle 2-1, 6+1 \rangle = \langle 1, 7 \rangle. \]

Step 3: Use the Component Formula

The formula to find the component of a vector \( \mathbf{a} \) along another vector \( \mathbf{b} \) is:\[ \operatorname{com} p_{\mathbf{b}}(\mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|}. \]Now, apply this formula using \( \mathbf{a} = \mathbf{v} - \mathbf{u} = \langle 1, 7 \rangle \) and \( \mathbf{b} = \mathbf{u} = \langle 1, -1 \rangle \).

Step 4: Calculate the Dot Product

Calculate \( \mathbf{a} \cdot \mathbf{b} = \langle 1, 7 \rangle \cdot \langle 1, -1 \rangle \).\[ \mathbf{a} \cdot \mathbf{b} = (1)(1) + (7)(-1) = 1 - 7 = -6. \]

Step 5: Compute the Magnitude of \(\mathbf{b}\)

Find the magnitude of \( \mathbf{u} = \langle 1, -1 \rangle \):\[ \|\mathbf{u}\| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}. \]

Step 6: Calculate the Component

Substitute the dot product and the magnitude into the component formula:\[ \operatorname{com} p_{\mathbf{u}}(\mathbf{v} - \mathbf{u}) = \frac{-6}{\sqrt{2}}. \]To simplify:\[ \operatorname{com} p_{\mathbf{u}}(\mathbf{v} - \mathbf{u}) = -3\sqrt{2}. \]

Key Concepts

Vector Operations

Vectors are mathematical quantities that have both magnitude (size) and direction. One of the fundamental operations with vectors is vector subtraction. To subtract vectors, you subtract their corresponding components. For example, consider the vectors \( \mathbf{u} = \langle 1, -1 \rangle \) and \( \mathbf{v} = \langle 2, 6 \rangle \). When we subtract \( \mathbf{u} \) from \( \mathbf{v} \), we calculate each component separately:

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

Thus, the vector \( \mathbf{v} - \mathbf{u} \) becomes \( \langle 1, 7 \rangle \). Vector operations like this are crucial for many areas in mathematics and physics, enabling us to analyze quantities that change in both magnitude and direction.

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors to produce a scalar (a single number). When you compute the dot product of two vectors, you essentially combine them in a way that measures how much one vector goes in the direction of the other. Here's how you do it mathematically:

• Multiply the first components of both vectors together.
• Multiply the second components of both vectors together.
• Sum the two results.

For instance, for vectors \( \mathbf{a} = \langle 1, 7 \rangle \) and \( \mathbf{b} = \langle 1, -1 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = (1)(1) + (7)(-1) = 1 - 7 = -6. \]This result can be used in formulas to find other properties of vectors, like the component in a given direction.

Vector Magnitude

The magnitude of a vector is a measure of its length or size, often denoted using the double bar symbol \( \| \cdot \| \). Computing the magnitude of a vector \( \mathbf{b} = \langle x, y \rangle \) involves using the Pythagorean theorem. Here’s how you find the magnitude:

• Square the first component of the vector.
• Square the second component of the vector.
• Add these squares together.
• Take the square root of the sum.

So, for vector \( \mathbf{u} = \langle 1, -1 \rangle \), the magnitude \( \| \mathbf{u} \| \) is calculated as:\[ \| \mathbf{u} \| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}. \]This magnitude represents the length of the vector in the coordinate space and is crucial when you apply vector formulas, such as for finding directional components.