Problem 37 - Advanced Engineering Mathematics

Question

In Problems 37 and 38, find the component of the given vector in the direction from the origin to the indicated point. \(\mathbf{a}=4 \mathbf{i}+6 \mathbf{j}, \quad P(3,10)\)

Step-by-Step Solution

Answer

The component of vector \(\mathbf{a}\) in the direction to point \(P(3,10)\) is \(\frac{72}{\sqrt{109}}\).

Step 1: Find the Direction Vector

First, we need the directional vector from the origin to the point \(P(3, 10)\). This vector can be represented as \(\mathbf{b} = 3\mathbf{i} + 10\mathbf{j}\).

Step 2: Calculate the Magnitude of the Direction Vector

Compute the magnitude of vector \(\mathbf{b}\) using the formula: \|\mathbf{b}\| = \sqrt{3^2 + 10^2} = \sqrt{9 + 100} = \sqrt{109}\.

Step 3: Find the Unit Vector in the Direction of the Direction Vector

To find the unit vector in the direction of \(\mathbf{b}\), divide \(\mathbf{b}\) by its magnitude: \hat{\mathbf{b}} = \frac{\mathbf{b}}{\|\mathbf{b}\|} = \frac{3\mathbf{i} + 10\mathbf{j}}{\sqrt{109}}\.

Step 4: Compute the Dot Product of Given Vector and Unit Vector

Calculate the dot product of \(\mathbf{a} = 4\mathbf{i} + 6\mathbf{j}\) with \(\hat{\mathbf{b}} = \frac{3\mathbf{i} + 10\mathbf{j}}{\sqrt{109}}\) using the dot product formula: \\mathbf{a} \cdot \hat{\mathbf{b}} = (4\mathbf{i} + 6\mathbf{j}) \cdot \left( \frac{3\mathbf{i} + 10\mathbf{j}}{\sqrt{109}} \right) = \frac{4 \times 3 + 6 \times 10}{\sqrt{109}} = \frac{12 + 60}{\sqrt{109}}\.

Step 5: Simplify the Dot Product

Simplify the dot product calculation: \(\frac{72}{\sqrt{109}}\). This represents the component of vector \(\mathbf{a}\) in the direction from the origin to point \(P(3,10)\).

Key Concepts

Directional Vector

A directional vector gives us a way to describe movement or direction in a space. Think of it as an arrow pointing from one point to another. In the given exercise, we start at the origin, which is the point (0,0), and we are moving towards a point, specifically point \( P(3,10) \). To create the directional vector from the origin to \( P \), we find out how much we move horizontally and vertically. For this, we use the coordinates of the point \( P \).

• Horizontal distance = 3, represented by \( 3\mathbf{i} \)
• Vertical distance = 10, represented by \( 10\mathbf{j} \)

Thus, the directional vector \( \mathbf{b} \) is \( 3\mathbf{i} + 10\mathbf{j} \). This vector tells us exactly how far and in which direction we need to go to reach point \( P \) from the origin.

Magnitude of a Vector

The magnitude of a vector gives us the length of the vector, much like using a ruler to measure the length of a line. For the vector \( \mathbf{b} = 3\mathbf{i} + 10\mathbf{j} \), its magnitude can be found using the Pythagorean theorem. This theorem lets us calculate the distance using the horizontal and vertical components of the vector.
The formula for finding the magnitude \( \|\mathbf{b}\| \) is:\[ \|\mathbf{b}\| = \sqrt{(3)^2 + (10)^2} \]
Breaking it down:

• Square the horizontal component: \( 3^2 = 9 \)
• Square the vertical component: \( 10^2 = 100 \)
• Add these squares: \( 9 + 100 = 109 \)
• Take the square root: \( \sqrt{109} \)

This results in the magnitude \( \sqrt{109} \), which is the length of \( \mathbf{b} \). It represents how far the point \( P \) is from the origin, not just horizontally or vertically, but directly as the crow flies.

Unit Vector

A unit vector is a vector with a magnitude of 1, and it is used to indicate direction without regard to distance. To convert a vector into a unit vector, we divide the vector by its own magnitude.

For directional vector \( \mathbf{b} = 3\mathbf{i} + 10\mathbf{j} \), we have already calculated its magnitude as \( \sqrt{109} \). To find the unit vector \( \hat{\mathbf{b}} \), perform the following calculations:\[ \hat{\mathbf{b}} = \frac{\mathbf{b}}{\|\mathbf{b}\|} = \frac{3\mathbf{i} + 10\mathbf{j}}{\sqrt{109}} \]

• This formula shows that each component of the vector is divided by \( \sqrt{109} \).
• The result is a direction-preserving vector with a unit length.

The unit vector \( \hat{\mathbf{b}} \) helps us focus purely on direction, without the influence of how long the original vector was.

Dot Product

The dot product of two vectors is a way to measure how much one vector goes in the direction of another. It can also help determine the angle between the vectors. The formula to calculate the dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) in component form is:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]This method multiplies corresponding components and produces a scalar (a single number).

In the problem, vector \( \mathbf{a} = 4\mathbf{i} + 6\mathbf{j} \) is dotted with the unit vector \( \hat{\mathbf{b}} = \frac{3\mathbf{i} + 10\mathbf{j}}{\sqrt{109}} \).\[ \mathbf{a} \cdot \hat{\mathbf{b}} = \frac{(4 \times 3) + (6 \times 10)}{\sqrt{109}} = \frac{72}{\sqrt{109}} \]

• Multiply \( 4 \) with \( 3 \), and \( 6 \) with \( 10 \)
• Add: \( 12 + 60 = 72 \)
• Divide by \( \sqrt{109} \) to get the result \( \frac{72}{\sqrt{109}} \).

This result tells us the component of \( \mathbf{a} \) in the direction of \( \mathbf{b} \), essentially measuring how much of \( \mathbf{a} \) points in the same direction as \( \mathbf{b} \). It's a practical tool often used in physics and engineering fields.