Problem 61 - Precalculus with Calculus Previews

Question

In Problems \(49-62, \mathbf{u}=\langle 2,-3\rangle, \mathbf{v}=\langle-1,5\rangle,\) and \(\mathbf{w}=\langle 3,-2\rangle .\) Find the indicated scalar or vector. $$ \left(\frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}\right) \mathbf{v} $$

Step-by-Step Solution

Answer

The resulting vector is \(\langle \frac{17}{26}, -\frac{85}{26} \rangle\).

Step 1: Find the Dot Product \(\mathbf{u} \cdot \mathbf{v}\)

To find the dot product \(\mathbf{u} \cdot \mathbf{v}\), use the formula \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2\). For vectors \(\mathbf{u} = \langle 2, -3 \rangle\) and \(\mathbf{v} = \langle -1, 5 \rangle\), compute:\[\mathbf{u} \cdot \mathbf{v} = 2 \times (-1) + (-3) \times 5 = -2 - 15 = -17\]

Step 2: Find the Dot Product \(\mathbf{v} \cdot \mathbf{v}\)

Find the dot product \(\mathbf{v} \cdot \mathbf{v}\) using the formula \(\mathbf{v} \cdot \mathbf{v} = v_1^2 + v_2^2\). For \(\mathbf{v} = \langle -1, 5 \rangle\), compute:\[\mathbf{v} \cdot \mathbf{v} = (-1)^2 + 5^2 = 1 + 25 = 26\]

Step 3: Divide the Dot Products

Divide the result of \(\mathbf{u} \cdot \mathbf{v}\) by \(\mathbf{v} \cdot \mathbf{v}\) to find the scalar:\[\frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} = \frac{-17}{26}\]

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

Multiply the scalar \(-\frac{17}{26}\) by vector \(\mathbf{v} = \langle -1, 5 \rangle\) to find the resulting vector:\[\left(\frac{-17}{26}\right) \mathbf{v} = \left(\frac{-17}{26}\right) \langle -1, 5 \rangle = \langle \frac{17}{26}, -\frac{85}{26} \rangle\]

Step 5: Simplify the Resulting Vector

Simplify each component of the vector if possible. In this case, the components are \(\frac{17}{26}\) and \(-\frac{85}{26}\), which cannot be simplified further:The final vector is \[\langle \frac{17}{26}, -\frac{85}{26} \rangle\].

Key Concepts

Scalar Multiplication

When dealing with vectors in mathematics, scalar multiplication is a fundamental operation. It's the process of multiplying a vector by a scalar (a real number). This action scales the vector by the given value. In the exercise, a scalar is formed through division of dot products, specifically \(-\frac{17}{26}\). Here's how it works:

• Multiply each component of the vector by the scalar. If the vector is \( \mathbf{v} = \langle a, b \rangle\) and the scalar is \(-\frac{17}{26}\), the result is \(-\frac{17}{26} \times \langle a, b \rangle = \langle -\frac{17}{26}a, -\frac{17}{26}b \rangle\).

After performing the multiplication, you change both the direction and magnitude of the original vector according to the scalar factor. In our example, \( \langle 17/26, -85/26 \rangle \) represents the new vector after scaling \(-\mathbf{v} \) by \(-17/26\). This method is helpful for vector adjustments in various applications like mechanics and robotics.

Vector Computation

The concept of vector computation involves operations that transform vectors, producing new ones based on specific mathematical procedures. A central part of vector computation is finding the dot product, as seen in the exercise.

• To calculate the dot product \( \mathbf{u} \cdot \mathbf{v} \), multiply corresponding components and add them up, such as \( 2 \times (-1) + (-3) \times 5 \).

This operation is crucial, producing a scalar that reflects the vectors' alignment. When vectors are perpendicular, the dot product equals zero, signifying no alignment, and when parallel, the product is maximized or minimized depending on their directions. In the exercise, computing both \( \mathbf{u} \cdot \mathbf{v} \) and \( \mathbf{v} \cdot \mathbf{v} \) lays the groundwork for deriving an effective scalar for further vector manipulation. Vector computation thus serves as the backbone for many spatial calculations relevant in engineering, physics, and computer graphics.

Simplifying Vectors

Once you obtain a vector through operations like scalar multiplication, the next step is simplifying or refining the expression. Simplification involves breaking down results, such as fractions, to their most basic form. In this exercise, the scalar multiplication resulted in the vector components \( \left \langle \frac{17}{26}, -\frac{85}{26} \right \rangle \). Let's examine what simplification entails:

• Look for common factors in fractional components to reduce them to simpler terms, if possible.
• Evaluate both components individually to find any possible reductions.

In this example, no further simplification was achievable since the components were already in their simplest forms. Simplifying ensures that the vector is in a readily interpretable form for further mathematical operations or practical applications. Understanding vector simplification helps refine results, making them more usable in advanced mathematical modeling and analysis.