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Chapter 12: Problem 1

Find a. \(\mathbf{v} \cdot \mathbf{u},|\mathbf{v}|,|\mathbf{u}|\) b. the cosine of the angle between \(\mathbf{v}\) and \(\mathbf{u}\) c. the scalar component of \(u\) in the direction of \(v\) d. the vector proj, u. $$ \mathbf{v}=2 \mathbf{i}-4 \mathbf{j}+\sqrt{5} \mathbf{k}, \quad \mathbf{u}=-2 \mathbf{i}+4 \mathbf{j}-\sqrt{5} \mathbf{k} $$

Short Answer

Expert verified
a. \( \mathbf{v} \cdot \mathbf{u} = -25 \), \(|\mathbf{v}| = 5 \), \(|\mathbf{u}| = 5 \) b. \( \cos \theta = -1 \) c. Scalar component is \(-5 \) d. \( \text{proj}_{\mathbf{v}} \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \)

Step by step solution

01

Compute Dot Product

The dot product \( \mathbf{v} \cdot \mathbf{u} \) is calculated using the formula: \( \mathbf{v} \cdot \mathbf{u} = v_1 u_1 + v_2 u_2 + v_3 u_3 \). For the given vectors, \( \mathbf{v} = 2\mathbf{i} - 4\mathbf{j} + \sqrt{5}\mathbf{k} \) and \( \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \), this gives:
02

Continued: Calculate Dot Product

Now, compute each term individually:- \(2 \times (-2) = -4\)- \((-4) \times 4 = -16\)- \(\sqrt{5} \times (-\sqrt{5}) = -5\)Sum these results: \(-4 - 16 - 5 = -25\). So, \( \mathbf{v} \cdot \mathbf{u} = -25 \).
03

Compute Magnitude of \( \mathbf{v} \)

The magnitude of vector \( \mathbf{v} \) is calculated as \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \). Here, compute as follows:- \(v_1^2 = 2^2 = 4\)- \(v_2^2 = (-4)^2 = 16\)- \(v_3^2 = (\sqrt{5})^2 = 5\)Now add them: \(4 + 16 + 5 = 25\). Thus, \(|\mathbf{v}| = \sqrt{25} = 5\).
04

Compute Magnitude of \( \mathbf{u} \)

The magnitude of vector \( \mathbf{u} \) is calculated as \( |\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2} \). Here, compute as follows:- \(u_1^2 = (-2)^2 = 4\)- \(u_2^2 = 4^2 = 16\)- \(u_3^2 = (-\sqrt{5})^2 = 5\)Now add them: \(4 + 16 + 5 = 25\). Thus, \(|\mathbf{u}| = \sqrt{25} = 5\).
05

Calculate Cosine of the Angle

Cosine of the angle \( \theta \) between \( \mathbf{v} \) and \( \mathbf{u} \) is given by:\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \]Substitute the computed values: \(-25/(5 \cdot 5) = -25/25 = -1\). So, \( \cos \theta = -1 \).
06

Scalar Component of \( \mathbf{u} \) in the Direction of \( \mathbf{v} \)

The scalar component of \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is given by:\[ \text{scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \]Substitute the values: \(-25/5 = -5\). So, the scalar component is \(-5\).
07

Vector Projection of \( \mathbf{u} \) onto \( \mathbf{v} \)

The vector projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is given by:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \]Calculate \( \mathbf{v} \cdot \mathbf{v} = 25 \), then:\[ \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{-25}{25} \right) \mathbf{v} = -\mathbf{v} = - (2 \mathbf{i} - 4 \mathbf{j} + \sqrt{5} \mathbf{k}) \]Hence, \( \text{proj}_{\mathbf{v}} \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product, often called the scalar product, is an important operation between two vectors. It combines two vectors to produce a scalar quantity. This means you get a single number rather than another vector. The formula for the dot product of two vectors \( \mathbf{v} \cdot \mathbf{u} \) is:
  • \( \mathbf{v} \cdot \mathbf{u} = v_1 u_1 + v_2 u_2 + v_3 u_3 \)
For example, find the dot product of vectors \( \mathbf{v} = 2\mathbf{i} - 4\mathbf{j} + \sqrt{5}\mathbf{k} \) and \( \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \).
  • Compute the products of corresponding components: \(2 \times (-2), (-4) \times 4, \text{and} \sqrt{5} \times (-\sqrt{5})\).
  • Sum these products: \(-4 - 16 - 5 = -25\).
Thus, the dot product \( \mathbf{v} \cdot \mathbf{u} = -25 \).

This operation is central in determining other vector-related quantities.
Vector Magnitude
Vector magnitude is a measure of a vector's "length." Think of it as the distance from the origin to the point described by the vector in a coordinate system. The magnitude of a vector \( \mathbf{v} = (v_1, v_2, v_3) \) is given by the formula:
  • \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)
Let's calculate the magnitude of vector \( \mathbf{v} = 2\mathbf{i} - 4\mathbf{j} + \sqrt{5}\mathbf{k} \):

  • \(v_1^2 = 2^2 = 4\)
  • \(v_2^2 = (-4)^2 = 16\)
  • \(v_3^2 = (\sqrt{5})^2 = 5\)
  • Add them: \(4 + 16 + 5 = 25\)
  • Take the square root: \(\sqrt{25} = 5\)
Therefore, \(|\mathbf{v}| = 5\). The magnitude tells us how "big" the vector is in space.
Angle between Vectors
Understanding the angle between two vectors helps describe their orientation in relation to one another. The cosine of the angle \( \theta \) between two vectors \( \mathbf{v} \) and \( \mathbf{u} \) is computed using the dot product and magnitudes:
  • \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \)
In our case, \( \mathbf{v} \cdot \mathbf{u} = -25 \), and both magnitudes \(|\mathbf{v}|\) and \(|\mathbf{u}|\) are 5.
By substituting these into the formula, we obtain:

  • \( \cos \theta = \frac{-25}{5 \times 5} = -1 \)
This implies that the vectors are collinear, pointing in exactly opposite directions since \( \cos \theta = -1 \) corresponds to an angle of 180 degrees.

This concept is key in physics and engineering for understanding directionality.
Scalar Component
The scalar component describes how much of one vector goes in the direction of another. For vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \), the scalar component is computed as:
  • \( \text{scalar component} = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}|} \)
With \( \mathbf{v} \cdot \mathbf{u} = -25 \) and \(|\mathbf{v}| = 5\), the calculation proceeds:

  • \( \text{scalar component} = \frac{-25}{5} = -5 \)
This scalar component signifies how much of \( \mathbf{u} \) stretches along \( \mathbf{v} \) and is particularly useful in physics to determine work done in the direction of force.
Vector Projection
Vector projection translates this scalar component into a vector. It tells us the vector \( \mathbf{u} \)'s "shadow" along \( \mathbf{v} \). The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is represented by:
  • \( \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \)
Given \( \mathbf{v} \cdot \mathbf{u} = -25 \) and \( \mathbf{v} \cdot \mathbf{v} = 25 \), we find:
  • \( \text{proj}_{\mathbf{v}} \mathbf{u} = \left( \frac{-25}{25} \right) \mathbf{v} = -\mathbf{v} \)
  • This means: \( \text{proj}_{\mathbf{v}} \mathbf{u} = -2 \mathbf{i} + 4 \mathbf{j} - \sqrt{5} \mathbf{k} \)
The projection helps us understand how much of \( \mathbf{u} \) aligns in the direction of \( \mathbf{v} \), and is applicable in various scientific disciplines where decomposition of forces and directions are needed.

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Most popular questions from this chapter

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