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Chapter 9: Problem 14

Find the component of \(\mathbf{u}\) along \(\mathbf{v}\) if a) \(\mathbf{u}=(0,7), \mathbf{v}=(6,8)\) b) \(\mathbf{u}=\left(\begin{array}{c}-\frac{1}{2} \\ 2\end{array}\right), \mathbf{v}=\left(\begin{array}{c}-2 \\ \frac{1}{2}\end{array}\right)\)

Short Answer

Expert verified
The component of \( \mathbf{u} \) along \( \mathbf{v} \) is 5.6 for part (a) and \( \frac{4}{\sqrt{17}} \) for part (b).

Step by step solution

01

Calculate the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) for part (a)

The dot product of two vectors \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \) is given by \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). For part (a), \( \mathbf{u} = (0, 7) \) and \( \mathbf{v} = (6, 8) \). Thus, the dot product is \( 0 \cdot 6 + 7 \cdot 8 = 56 \).
02

Calculate the magnitude of \( \mathbf{v} \) for part (a)

The magnitude of a vector \( \mathbf{v} = (v_1, v_2) \) is given by \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2} \). For part (a), \( \mathbf{v} = (6, 8) \). The magnitude is \( |\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).
03

Compute the component of \( \mathbf{u} \) along \( \mathbf{v} \) for part (a)

The component of \( \mathbf{u} \) along \( \mathbf{v} \) is calculated as \( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} \). From step 1, \( \mathbf{u} \cdot \mathbf{v} = 56 \). From step 2, \( |\mathbf{v}| = 10 \). Thus, the component is \( \frac{56}{10} = 5.6 \).
04

Calculate the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) for part (b)

For part (b), \( \mathbf{u} = \left( -\frac{1}{2}, 2 \right) \) and \( \mathbf{v} = \left( -2, \frac{1}{2} \right) \). The dot product is \( -\frac{1}{2} \cdot (-2) + 2 \cdot \frac{1}{2} = 1 + 1 = 2 \).
05

Calculate the magnitude of \( \mathbf{v} \) for part (b)

Now, calculate the magnitude of \( \mathbf{v} = \left( -2, \frac{1}{2} \right) \). The magnitude is \( |\mathbf{v}| = \sqrt{(-2)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{4 + \frac{1}{4}} = \sqrt{\frac{16}{4} + \frac{1}{4}} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \).
06

Compute the component of \( \mathbf{u} \) along \( \mathbf{v} \) for part (b)

The component of \( \mathbf{u} \) along \( \mathbf{v} \) is \( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} \). From step 4, \( \mathbf{u} \cdot \mathbf{v} = 2 \). From step 5, \( |\mathbf{v}| = \frac{\sqrt{17}}{2} \). Thus, the component is \( \frac{2}{\frac{\sqrt{17}}{2}} = \frac{4}{\sqrt{17}} \).

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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, also known as the scalar product, is a way to multiply two vectors to obtain a scalar (a single number). This is particularly useful when working with vectors because it gives us a measure of how much one vector goes in the direction of another.
  • To find the dot product of two vectors \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \), you use the formula: \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 \).
  • This operation takes the corresponding components of the vectors, multiplies them together, and then sums the results.
For example, if \( \mathbf{u} = (0, 7) \) and \( \mathbf{v} = (6, 8) \), their dot product would be \( 0 \cdot 6 + 7 \cdot 8 = 56 \). The result of the dot product can reveal if the vectors are perpendicular (if the dot product is zero) or indicate how "aligned" or "opposing" they are based on the sign and magnitude of the result.
Magnitude of a Vector
The magnitude of a vector is essentially its length. You can think of it as the distance from the origin to the point represented by the vector in the coordinate plane.
  • The magnitude of a vector \( \mathbf{v} = (v_1, v_2) \) is calculated using the formula: \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2} \).
  • This is derived from the Pythagorean theorem, where the vector components \( v_1 \) and \( v_2 \) form a right triangle with the vector as the hypotenuse.
For instance, if \( \mathbf{v} = (6, 8) \), the magnitude is \( |\mathbf{v}| = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \). Calculating the magnitude is fundamental before proceeding to find other properties like the component of one vector along another. It is analogous to finding the length of a line segment defined by the vector.
Component of One Vector Along Another Vector
Finding the component of one vector along another gives us a measure of how much one vector "projects" onto another. This is useful in physics and engineering for decomposing forces and velocities.
  • The component of vector \( \mathbf{u} \) along vector \( \mathbf{v} \) is found using the formula: \( \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|} \).
  • This formula uses the dot product, which gives a measure of alignment, and scales it by the magnitude of the vector onto which it is projecting.
For example, with \( \mathbf{u} = (0, 7) \) and \( \mathbf{v} = (6, 8) \), first compute the dot product \( \mathbf{u} \cdot \mathbf{v} = 56 \). Then, divide by the magnitude of \( \mathbf{v} \), which is 10, giving us a component of \( 5.6 \). This tells us that vector \( \mathbf{u} \) has a strength or influence of 5.6 along the direction of \( \mathbf{v} \), demonstrating how much of \( \mathbf{u} \) aligns with \( \mathbf{v} \).

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Most popular questions from this chapter

Find the direction angle for each vector. a) \(\mathbf{u}=(2,0)\) b) \(\mathbf{v}=(0,3)\) c) \(\mathbf{w}=(-3,0)\) d) \(\mathbf{u}+\mathbf{v}\) e) \(\mathbf{v}+\mathbf{w}\)

Let \(\mathbf{u}=(1,5)\) and \(\mathbf{v}=(3,-4),\) Find the vector \(\mathbf{x}\) such that \(2 \mathbf{u}-3 \mathbf{x}+\mathbf{v}=5 \mathbf{x}-2 \mathbf{v}\).

Find the work done by the force \(\mathbf{F}\) in moving an object between points \(M\) and \(N\). a) \(\mathbf{F}=400 \mathbf{i}-50 \mathbf{j}, M(2,3), N(12,43)\) b) \(\mathbf{F}=30 \mathbf{i}+150 \mathbf{j}, M(0,30), N(15,70)\) c) \(\mathbf{F}=\left(\begin{array}{c}5 \\ 25\end{array}\right), M(0,0), N(1,6)\)

Find a vector of magnitude 3 that is parallel to \(\mathbf{u}=2 \mathbf{i}+3 \mathbf{j}\)

Note: \(\left(\begin{array}{l}8 \\ 46\end{array}\right)\) is said to be written as a linear combination of \(\left(\begin{array}{l}1 \\ 9\end{array}\right)\) and \(\left(\begin{array}{c}1 \\ -4\end{array}\right)\) Write (5,-5) as a linear combination of (1,-1) and (-1,1).
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