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Chapter 3: Problem 27

Find the vector component of \(u\) along a and the vector component of \(u\) orthogonal to a. $$\mathbf{u}=(2,1,1,2), \mathbf{a}=(4,-4,2,-2)$$

Short Answer

Expert verified
The vector component of \( \mathbf{u} \) along \( \mathbf{a} \) is \((0.2, -0.2, 0.1, -0.1)\). The vector component of \( \mathbf{u} \) orthogonal to \( \mathbf{a} \) is \((1.8, 1.2, 0.9, 2.1)\).

Step by step solution

01

Find the Dot Product of u and a

To find the component of \( \mathbf{u} \) along \( \mathbf{a} \), we first calculate the dot product \( \mathbf{u} \cdot \mathbf{a} \). This is done by multiplying corresponding components of \( \mathbf{u} \) and \( \mathbf{a} \) and then summing them up:\[ \mathbf{u} \cdot \mathbf{a} = 2 \cdot 4 + 1 \cdot (-4) + 1 \cdot 2 + 2 \cdot (-2) = 8 - 4 + 2 - 4 = 2. \]
02

Find the Magnitude Squared of a

Next, we calculate the magnitude squared of \( \mathbf{a} \), denoted as \( \| \mathbf{a} \|^2 \). This is the sum of the squares of the components of \( \mathbf{a} \):\[ \| \mathbf{a} \|^2 = 4^2 + (-4)^2 + 2^2 + (-2)^2 = 16 + 16 + 4 + 4 = 40. \]
03

Calculate the Vector Component of u along a

The vector component of \( \mathbf{u} \) along \( \mathbf{a} \) is given by:\[ \text{proj}_{\mathbf{a}} \mathbf{u} = \left( \frac{\mathbf{u} \cdot \mathbf{a}}{\| \mathbf{a} \|^2} \right) \mathbf{a}. \]Substituting the values from Step 1 and Step 2:\[ \text{proj}_{\mathbf{a}} \mathbf{u} = \left( \frac{2}{40} \right) \mathbf{a} = \frac{1}{20} \cdot (4, -4, 2, -2) = \left( \frac{4}{20}, \frac{-4}{20}, \frac{2}{20}, \frac{-2}{20} \right) = \left( 0.2, -0.2, 0.1, -0.1 \right). \]
04

Calculate the Vector Component of u Orthogonal to a

The vector component of \( \mathbf{u} \) orthogonal to \( \mathbf{a} \) is found by subtracting the vector component along \( \mathbf{a} \) from \( \mathbf{u} \):\[ \mathbf{u}_{\perp \mathbf{a}} = \mathbf{u} - \text{proj}_{\mathbf{a}} \mathbf{u}. \]Substituting in the vectors:\[ \mathbf{u}_{\perp \mathbf{a}} = (2, 1, 1, 2) - (0.2, -0.2, 0.1, -0.1) = (2 - 0.2, 1 + 0.2, 1 - 0.1, 2 + 0.1) = (1.8, 1.2, 0.9, 2.1). \]

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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 is a way to multiply two vectors to yield a scalar. To compute the dot product of vectors \( \mathbf{u} \) and \( \mathbf{a} \), you multiply corresponding components and add the results.
For instance:
  • Multiply the first components: \(2 \times 4 = 8\)
  • Multiply the second, including signs: \(1 \times (-4) = -4\)
  • Third components: \(1 \times 2 = 2\)
  • Fourth components: \(2 \times (-2) = -4\)
Then, sum those products: \(8 - 4 + 2 - 4\). The result is the dot product \( \mathbf{u} \cdot \mathbf{a} = 2 \).
This scalar provides a measure of how aligned the two vectors are. If the dot product is positive, vectors point in relatively similar directions. If it鈥檚 zero, they are orthogonal, meaning they interact at right angles. Negative indicates opposing orientations.
Magnitude
Magnitude refers to the length or size of a vector. To find it, compute the square root of the sum of the squares of its components. For example, vector \( \mathbf{a} = (4, -4, 2, -2) \) would have a magnitude represented as \( ||\mathbf{a}|| \).
The steps are:
  • Square each component: \(4^2 = 16\), \((-4)^2 = 16\), \(2^2 = 4\), \((-2)^2 = 4\)
  • Sum those squares: \(16 + 16 + 4 + 4 = 40\)
The formula to derive the actual magnitude is \( \| \mathbf{a} \| = \sqrt{40} \). However, in some cases like projections, we use the squared magnitude \( \|\mathbf{a}\|^2 \) directly. Remember, magnitude establishes the "length" of a vector, indicating its scale without direction.
Orthogonal Vectors
Orthogonal vectors mean they are at right angles (90 degrees) to each other. When two vectors are orthogonal, their dot product is zero, which is key. This property is used to find the component of a vector that doesn't align with another.
In our exercise, to find the vector component of \( \mathbf{u} \) orthogonal to \( \mathbf{a} \), we subtract the projection of \( \mathbf{u} \) onto \( \mathbf{a} \) from \( u \).
Here's how:
  • Complete the projection calculation to find \( \text{proj}_{\mathbf{a}} \mathbf{u} = (0.2, -0.2, 0.1, -0.1) \)
  • Subtract this from \( \mathbf{u} = (2, 1, 1, 2) \):
Subtract component-wise:
  • \(2 - 0.2 = 1.8\)
  • \(1 + 0.2 = 1.2\)
  • \(1 - 0.1 = 0.9\)
  • \(2 + 0.1 = 2.1\)
Thus, the orthogonal component of \( \mathbf{u} \) with respect to \( \mathbf{a} \) is \( (1.8, 1.2, 0.9, 2.1) \). This vector sits at a right angle to \( \mathbf{a} \).

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Most popular questions from this chapter

Find the vector component of \(u\) along a and the vector component of \(u\) orthogonal to a. $$\mathbf{u}=(6,2), \mathbf{a}=(3,-9)$$

Let \(u\) and \(v\) be nonzero vectors in 2 - or 3 -space, and let \(k=\|\mathbf{u}\|\) and \(l=\|\mathbf{v}\| .\) Show that the vector \(\mathbf{w}=l \mathbf{u}+k \mathbf{v}\) bisects the angle between u and v.

Find the volume of the parallelepiped with sides \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\). $$\mathbf{u}=(2,-6,2), \mathbf{v}=(0,4,-2), \mathbf{w}=(2,2,-4)$$

Let \(\mathbf{u}=(5,-1,0,3,-3), \mathbf{v}=(-1,-1,7,2,0),\) and \(\mathbf{w}=(-4,2,-3,-5,2) .\) Find the components of (a) \(w-u\) (b) \(2 v+3 u\) (c) \(-w+3(v-u)\) (d) \(5(-v+4 u-w)\) (e) \(-2(3 \mathbf{w}+\mathbf{v})+(2 \mathbf{u}+\mathbf{w})\) (f) \(\frac{1}{2}(w-5 v+2 u)+v\)

Find a unit vector that has the same direction as the given vector. (a) \((-4,-3)\) (b) \((1,7)\) (c) \((-3,2, \sqrt{3})\) (d) \((1,2,3,4,5)\)
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