Problem 77 - Precalculus with Calculus Previews

Question

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

Step-by-Step Solution

Answer

The component of \(\mathbf{u}\) along \(\mathbf{v}\) is \(-\frac{\sqrt{10}}{5}\).

Step 1: Understand the Concept of Component

The component of vector \(\mathbf{u}\) along vector \(\mathbf{v}\), denoted as \(\operatorname{comp}_{\mathbf{v}} \mathbf{u}\), is a scalar that represents the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) divided by the magnitude of \(\mathbf{v}\). The formula is \(\operatorname{comp}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}\).

Step 2: Calculate the Dot Product

The dot product of two vectors \(\mathbf{u} = \langle 1, -1 \rangle\) and \(\mathbf{v} = \langle 2, 6 \rangle\) is calculated as follows:\[ \mathbf{u} \cdot \mathbf{v} = 1 \cdot 2 + (-1) \cdot 6 = 2 - 6 = -4 \]

Step 3: Calculate the Magnitude of Vector \(\mathbf{v}\)

The magnitude of vector \(\mathbf{v} = \langle 2, 6 \rangle\) is calculated as follows:\[ \|\mathbf{v}\| = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} \]

Step 4: Find the Component

Now substitute the dot product and magnitude of \(\mathbf{v}\) into the component formula:\[ \operatorname{comp}_{\mathbf{v}} \mathbf{u} = \frac{-4}{2\sqrt{10}} = \frac{-2}{\sqrt{10}} \] \[ \operatorname{comp}_{\mathbf{v}} \mathbf{u} = \frac{-2 \sqrt{10}}{10} = -\frac{\sqrt{10}}{5} \]

Key Concepts

Dot Product

When working with vectors, the dot product is one of the most fundamental operations you can perform. The dot product of two vectors is a scalar, meaning it does not have direction, just magnitude. To calculate the dot product of vectors \(\mathbf{u} = \langle 1, -1 \rangle\) and \(\mathbf{v} = \langle 2, 6 \rangle\), you multiply the corresponding components and then sum these products:

• Multiply the first components: \(1 \times 2 = 2\)
• Multiply the second components: \((-1) \times 6 = -6\)
• Add the results: \(2 + (-6) = -4\)

So, the dot product \(\mathbf{u} \cdot \mathbf{v} \) is \(-4\). This result is useful in numerous calculations involving vectors.The dot product can tell us about the angle between two vectors. A dot product greater than zero indicates an acute angle, less than zero indicates an obtuse angle, and exactly zero implies they are perpendicular. Since our result is \(-4\), the vectors \(\mathbf{u}\) and \(\mathbf{v}\) form an obtuse angle.

Magnitude of a Vector

The magnitude of a vector is essentially its length. To find the magnitude of a vector \(\mathbf{v} = \langle 2, 6 \rangle\), we apply the Pythagorean theorem:Calculate the square root of the sum of the squares of the vector's components:

• First component squared: \(2^2 = 4\)
• Second component squared: \(6^2 = 36\)

Now, add these results and take the square root:\[\|\mathbf{v}\| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\]The magnitude \(2\sqrt{10}\) gives us the length of vector \(\mathbf{v}\). Magnitudes help us understand the scale of a vector in space, which is essential in both two-dimensional and three-dimensional vector math.Knowing the magnitude is crucial when it comes to comparing vector sizes or normalizing vectors (making them unit vectors). It is also an integral part of calculating vector projections.

Projection of a Vector

Projection is a way to show one vector along the direction of another vector. It helps visualize how much of one vector is in the direction of another. To find the projection of vector \(\mathbf{u}\) along another vector \(\mathbf{v}\), we use the component calculation:The formula for the component or projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is:\[\operatorname{comp}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}\]Using our exercise's values:

• Dot product: \(-4\)
• Magnitude of \(\mathbf{v}\): \(2\sqrt{10}\)

This gives:\[\operatorname{comp}_{\mathbf{v}} \mathbf{u} = \frac{-4}{2\sqrt{10}} = \frac{-2}{\sqrt{10}} = -\frac{\sqrt{10}}{5}\]This result is a scalar defining how much of \(\mathbf{u}\) acts in the direction of \(\mathbf{v}\). In practical terms, this projection can show us the influence, force, or movement of one vector along another direction, making it hugely beneficial in physics and engineering tasks.