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Chapter 3: Problem 24

Find the vector component of \(u\) along a and the vector component of \(u\) orthogonal to a. $$\mathbf{u}=(1,0,0), \mathbf{a}=(4,3,8)$$

Short Answer

Expert verified
\(\text{Proj}_{\mathbf{a}} \mathbf{u} = \left( \frac{16}{89}, \frac{12}{89}, \frac{32}{89} \right)\); \(\text{Orth}_{\mathbf{a}} \mathbf{u} = \left( \frac{73}{89}, -\frac{12}{89}, -\frac{32}{89} \right)\).

Step by step solution

01

Find Dot Product of u and a

To find the vector component of \(\mathbf{u}\) along \(\mathbf{a}\), we first need the dot product \(\mathbf{u} \cdot \mathbf{a}\). The dot product is computed as follows: \( \mathbf{u} \cdot \mathbf{a} = (1)(4) + (0)(3) + (0)(8) = 4 \).
02

Calculate Magnitude of a

Before proceeding to find the vector component, calculate the magnitude of \(\mathbf{a}\) which is needed for normalization. This is given by \( \| \mathbf{a} \| = \sqrt{4^2 + 3^2 + 8^2} = \sqrt{16 + 9 + 64} = \sqrt{89} \).
03

Calculate Vector Component of u Along a

The vector component of \(\mathbf{u}\) along \(\mathbf{a}\) is calculated using the formula:\[\text{Proj}_{\mathbf{a}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{a}}{\| \mathbf{a} \|^2} \mathbf{a} = \frac{4}{89} (4, 3, 8)\]Evaluating gives:\[\text{Proj}_{\mathbf{a}} \mathbf{u} = \left( \frac{16}{89}, \frac{12}{89}, \frac{32}{89} \right)\]
04

Calculate Vector Component of u Orthogonal to a

The vector component of \(\mathbf{u}\) orthogonal to \(\mathbf{a}\) is computed by subtracting the projection from \(\mathbf{u}\):\[\text{Orth}_{\mathbf{a}} \mathbf{u} = \mathbf{u} - \text{Proj}_{\mathbf{a}} \mathbf{u} = (1, 0, 0) - \left( \frac{16}{89}, \frac{12}{89}, \frac{32}{89} \right) = \left( \frac{73}{89}, -\frac{12}{89}, -\frac{32}{89} \right)\]

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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 fundamental operation in vector algebra. It helps us understand the relationship between two vectors. In simple terms, it gives a measure of how much two vectors point in the same direction.
To compute the dot product of vectors \(\mathbf{u} = (1,0,0)\) and \(\mathbf{a} = (4,3,8)\), you multiply corresponding components of the vectors and add them up:\[\mathbf{u} \cdot \mathbf{a} = 1 \times 4 + 0 \times 3 + 0 \times 8 = 4\] This result tells us that vector \(\mathbf{u}\) projects with a magnitude of 4 in the direction of \(\mathbf{a}\).
  • It's a scalar value, not a vector.
  • If the dot product is zero, vectors are perpendicular.
Magnitude of a Vector
The magnitude of a vector, often referred to as the 'length' of the vector, is a measure of how long the vector is irrespective of its direction. This is a crucial component when normalizing vectors or measuring distances.
For our vector \(\mathbf{a} = (4,3,8)\), the magnitude is calculated using the formula:\[\| \mathbf{a} \| = \sqrt{4^2 + 3^2 + 8^2} = \sqrt{16 + 9 + 64} = \sqrt{89}\]
This value is necessary for finding the projection of one vector onto another.
  • A vector's magnitude is always positive.
  • It is equivalent to the Euclidean distance of the vector from the origin.
Projection of a Vector
Finding the projection of a vector \(\mathbf{u}\) along another vector \(\mathbf{a}\) gives us a vector pointing in the direction of \(\mathbf{a}\) that describes the extent of \(\mathbf{u}\) along \(\mathbf{a}\). This process involves a formula that includes the dot product and magnitude:
\[\text{Proj}_{\mathbf{a}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{a}}{\| \mathbf{a} \|^2} \mathbf{a} = \frac{4}{89} (4, 3, 8)\]
Evaluating gives:
\[\text{Proj}_{\mathbf{a}} \mathbf{u} = \left( \frac{16}{89}, \frac{12}{89}, \frac{32}{89} \right)\]
This resultant vector aligns with \(\mathbf{a}\) but has its magnitude scaled by the fraction of \(\mathbf{u}'s\) dot product with \(\mathbf{a}\).
  • Projections shrink or expand the vector depending on direction and magnitude.
  • The resulting vector is parallel to \(\mathbf{a}\).
Orthogonal Vector Component
The orthogonal vector component is the part of the vector \(\mathbf{u}\) that is perpendicular to vector \(\mathbf{a}\). To find this, we subtract the projection of \(\mathbf{u}\) along \(\mathbf{a}\) from \(\mathbf{u}\). This tells us what remains of \(\mathbf{u}\) after removing its alignment with \(\mathbf{a}\):
\[\text{Orth}_{\mathbf{a}} \mathbf{u} = \mathbf{u} - \text{Proj}_{\mathbf{a}} \mathbf{u} = (1, 0, 0) - \left( \frac{16}{89}, \frac{12}{89}, \frac{32}{89} \right) = \left( \frac{73}{89}, -\frac{12}{89}, -\frac{32}{89} \right)\]
This vector lies entirely in the direction perpendicular to \(\mathbf{a}\), meaning it exerts no influence along \(\mathbf{a}\).
  • Orthogonal components are crucial in resolving motion into independent perpendicular directions.
  • This component and the projected component add up to give the original vector \(\mathbf{u}\).

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Most popular questions from this chapter

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) \(\frac{1}{2}(w-5 v+2 u)+v\)

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$$\text { Simplify }(\mathbf{u}+\mathbf{v}) \times(\mathbf{u}-\mathbf{v})$$.

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.
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