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Chapter 11: Problem 50

Orthogonal vectors Let a and b be real numbers. Describe all unit vectors orthogonal to \(\mathbf{v}=\mathbf{i}+\mathbf{j}+\mathbf{k}\).

Short Answer

Expert verified
Question: Find all possible unit vectors orthogonal to the given vector \(\mathbf{v} = \mathbf{i}+\mathbf{j}+\mathbf{k}\). Answer: The unit vectors orthogonal to \(\mathbf{v}\) must satisfy the equation \(a^2 + b^2 + ab = \dfrac{1}{2}\), where \(\mathbf{u} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) and \(c = -a - b\).

Step by step solution

01

Define Orthogonal Condition

For two vectors \(\mathbf{v}\) and \(\mathbf{u}\) to be orthogonal, their dot product must be equal to 0. The dot product of two vectors \(\mathbf{v}\) and \(\mathbf{u}\) is given by: \( \mathbf{v} \cdot \mathbf{u} = \lVert \mathbf{v} \rVert \lVert \mathbf{u} \rVert \cos{\theta}\) where \(\lVert \mathbf{v} \rVert\) and \(\lVert \mathbf{u} \rVert\) are the magnitudes of \(\mathbf{v}\) and \(\mathbf{u}\), respectively, and \(\theta\) represents the angle between \(\mathbf{v}\) and \(\mathbf{u}\). Since \(\mathbf{v}\) and \(\mathbf{u}\) are orthogonal, \(\theta = 90 ^{\circ}\), which means \(\cos{\theta} = 0\). Thus, their dot product must be equal to 0.
02

Set Up the Dot Product

Let \(\mathbf{u} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) be a unit vector orthogonal to \(\mathbf{v} = \mathbf{i}+\mathbf{j}+\mathbf{k}\). Calculate the dot product \(\mathbf{v} \cdot \mathbf{u}\) as follows: \(\mathbf{v} \cdot \mathbf{u} = (1)(a) + (1)(b) + (1)(c) = a + b + c\)
03

Apply the Orthogonal Condition

Since the vectors \(\mathbf{v}\) and \(\mathbf{u}\) are orthogonal, their dot product should be equal to 0. Substitute the dot product found in Step 2 and set it to 0: \(a + b + c = 0\)
04

Apply the Unit Vector Condition

Since \(\mathbf{u}\) is a unit vector, its magnitude is equal to 1. Therefore, we have: \(\lVert \mathbf{u} \rVert = \sqrt{a^2 + b^2 + c^2} = 1\) Squaring both sides, we get: \(a^2 + b^2 + c^2 = 1\)
05

Find All Possible Unit Vectors

Now, we have two equations describing the unit vector \(\mathbf{u}\): 1. \(a + b + c = 0\) 2. \(a^2 + b^2 + c^2 = 1\) To describe all unit vectors orthogonal to \(\mathbf{v}\), combine the two equations: \(a + b + c = 0 \Rightarrow c = -a - b\) Substitute for \(c\) in the second equation: \(a^2 + b^2 + (-a - b)^2 = 1\) Simplify the equation: \(a^2 + b^2 + a^2 + 2ab + b^2 = 1\) Combine like terms: \(2a^2 + 2b^2 + 2ab = 1\) Divide the equation by 2: \(a^2 + b^2 + ab = \dfrac{1}{2}\) This equation describes all unit vectors \(\mathbf{u} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) orthogonal to the given vector \(\mathbf{v} = \mathbf{i}+\mathbf{j}+\mathbf{k}\).

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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 is a way to multiply two vectors, which results in a scalar. This product is given by the formula:
  • For vectors \( \mathbf{v} = x_1\mathbf{i} + y_1\mathbf{j} + z_1\mathbf{k} \) and \( \mathbf{u} = x_2\mathbf{i} + y_2\mathbf{j} + z_2\mathbf{k} \), the dot product is \( \mathbf{v} \cdot \mathbf{u} = x_1x_2 + y_1y_2 + z_1z_2 \).
When calculated, it gives a measure of how much one vector extends in the direction of another. This means if the dot product is zero, the vectors are orthogonal because there is no extension of one vector in the direction of the other. This concept is critical for solving problems involving orthogonal vectors, as shown in the step-by-step solution.
Unit Vector
Unit vectors are vectors with a magnitude of exactly 1. These are often used to indicate direction because they don鈥檛 have any influence of length. The formula to find the length or magnitude of a vector \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is given by:
  • \( \lVert \mathbf{u} \rVert = \sqrt{a^2 + b^2 + c^2} \)
For \( \mathbf{u} \) to be a unit vector, it must satisfy \( \lVert \mathbf{u} \rVert = 1 \). In this problem, this property ensures that the vectors being examined have only unit length, simplifying the problem of finding vectors orthogonal to \( \mathbf{v} \). Unit vectors are essential in maintaining the size-standard while focusing purely on direction.
Orthogonal Condition
The orthogonal condition is a criterion that determines when two vectors are perpendicular to each other. For two vectors, \( \mathbf{v} \) and \( \mathbf{u} \), to be orthogonal, their dot product must be zero:
  • \( \mathbf{v} \cdot \mathbf{u} = 0 \)
Here, using the vectors \( \mathbf{v} = \mathbf{i}+\mathbf{j}+\mathbf{k} \) and \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), we compute the dot product to test for orthogonality by setting \( a + b + c = 0 \). The orthogonal condition effectively tells us how the components of the vectors must relate to each other for one to be perpendicular to the other, and it serves as the starting point for finding unit vectors orthogonal to a given vector.
Magnitude of Vector
The magnitude of a vector measures its length or size and is obtained using the formula:
  • For a vector \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), the magnitude is \( \lVert \mathbf{u} \rVert = \sqrt{a^2 + b^2 + c^2} \).
This is useful when determining if a vector is a unit vector or not. In vector mathematics, normalizing a vector refers to converting it to a unit vector by dividing each component by the vector鈥檚 magnitude. Ensuring the vector \( \mathbf{u} \) in this problem is a unit vector implies that any solution for the orthogonal vector must also satisfy the magnitude condition \( a^2 + b^2 + c^2 = 1 \). This combination of conditions helps us find all possible unit vectors orthogonal to the vector \( \mathbf{v} \).

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Most popular questions from this chapter

Properties of dot products Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle,\) and \(\mathbf{w}=\left\langle w_{1}, w_{2}, w_{3}\right\rangle .\) Prove the following vector properties, where \(c\) is a scalar. $$|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u} \| \mathbf{v}|$$

An object moves along a path given by \(\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t\rangle, \quad\) for \(0 \leq t \leq 2 \pi\) a. What conditions on \(a, b, c,\) and \(d\) guarantee that the path is a circle? b. What conditions on \(a, b, c,\) and \(d\) guarantee that the path is an ellipse?

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The vectors \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\) are parallel for all values of \(t\) in the domain. b. The curve described by the function \(\mathbf{r}(t)=\left\langle t, t^{2}-2 t, \cos \pi t\right\rangle\) is smooth, for \(-\infty

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle e^{2 t}, 1-2 e^{-t}, 1-2 e^{t}\right\rangle ; \mathbf{r}(0)=\langle 1,1,1\rangle$$

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle e^{t}, \sin t, \sec ^{2} t\right\rangle ; \mathbf{r}(0)=\langle 2,2,2\rangle$$
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