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Let a and b be real numbers. Find two vectors that are orthogonal to \langle 0,1,1\rangle and to each other.

Short Answer

Expert verified
Question: Find two vectors that are orthogonal to each other and are orthogonal to the given vector ⟨0,1,1⟩. Answer: The two orthogonal vectors are ⟨0, 1, 0⟩ and ⟨1, 0, 0⟩.

Step by step solution

01

Perform cross product with a suitable basis vector

Let's consider the basis vectors ⟨1,0,0⟩, ⟨0,1,0⟩, and ⟨0,0,1⟩. We want to avoid using ⟨0,1,1⟩ as one of the basis vectors in our cross product since it would result in a zero vector. Instead, we can use any of the other two basis vectors, e.g., ⟨1,0,0⟩. The cross product of ⟨1,0,0⟩ and ⟨0,1,1⟩ will give us our first orthogonal vector: ⟨1,0,0⟩ × ⟨0,1,1⟩ = ⟨0, 0-(-1), 0⟩ = ⟨0, 1, 0⟩ So our first orthogonal vector is ⟨0, 1, 0⟩.
02

Perform cross product with the given vector and first orthogonal vector

Now that we have found the first orthogonal vector, we can perform another cross product with the given vector ⟨0,1,1⟩ and the first orthogonal vector ⟨0, 1, 0⟩ to find the second orthogonal vector: ⟨0,1,1⟩ × ⟨0, 1, 0⟩ = ⟨1, 0, 0⟩ So, our second orthogonal vector is ⟨1, 0, 0⟩.
03

State the two orthogonal vectors

We have found two vectors that are orthogonal to each other and are orthogonal to the given vector ⟨0,1,1⟩: First orthogonal vector: ⟨0, 1, 0⟩ Second orthogonal vector: ⟨1, 0, 0⟩

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

cross product
The cross product is a fundamental operation in vector algebra. It is used to find a vector that is perpendicular to two given vectors in three-dimensional space. Let’s look at how this is done step-by-step.

A cross product between two vectors \(\mathbf{a} = \langle a_{1}, a_{2}, a_{3} \rangle\) and \(\mathbf{b} = \langle b_{1}, b_{2}, b_{3} \rangle\) results in a new vector \(\mathbf{c} = \langle c_{1}, c_{2}, c_{3} \rangle\). The components of \(\mathbf{c}\) are found using:
\[c_{1} = a_{2}b_{3} - a_{3}b_{2},\]
\[c_{2} = a_{3}b_{1} - a_{1}b_{3},\]
\[c_{3} = a_{1}b_{2} - a_{2}b_{1}.\]

The cross product is only defined in three dimensions. This makes it especially useful in many areas of physics and engineering where 3D space is considered.

In the solution above, we found two orthogonal vectors by strategically choosing basis vectors and performing cross products. By using the cross product, we made sure to find vectors that are not only perpendicular to \(\langle 0,1,1\rangle\) but also to each other.
basis vectors
Basis vectors are the building blocks of vector spaces. In three-dimensional space, the standard basis vectors are \(\langle 1,0,0 \rangle\), \(\langle 0,1,0 \rangle\), and \(\langle 0,0,1 \rangle\).

These vectors are mutually orthogonal and they form the framework upon which any vector in 3D space can be constructed. You can think of them as axes in a coordinate system, providing directions along the x, y, and z axes.

When solving problems involving orthogonality, basis vectors can be invaluable. They offer a simple way to perform operations like cross products due to their straightforward components. This simplicity helps avoid errors when calculating more complex vector relations.

In the solution provided, we used the basis vector \(\langle 1,0,0 \rangle\) initially. This choice allowed us to find an orthogonal vector quickly without getting a zero vector, which would provide no use in finding orthogonality.
vector algebra
Vector algebra is a field of mathematics that deals with vectors and operations on them. This subject provides the tools needed to work with vectors in various dimensions.

Basic operations include addition, subtraction, and scalar multiplication. Here vectors can be added or subtracted component-wise, producing another vector.

A key operation within vector algebra is the dot product. It results in a scalar and can tell us if two vectors are orthogonal. If their dot product is zero, the vectors are orthogonal.

In this exercise, however, the emphasis was on the cross product because it directly returns an orthogonal vector. Vector algebra underpins many areas of science and engineering, providing a rich framework to handle physical quantities like forces and velocities, which are naturally vector quantities.

The step-by-step solution above showcases an application of vector algebra to find orthogonal vectors efficiently, highlighting how these mathematical operations can be used to solve real-world problems.

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Most popular questions from this chapter

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(1,1,-1) ; \ell\) has the direction of $$\langle-6,8,3\rangle$$.

Prove the following identities. Assume that \(\mathbf{u}, \mathbf{v}, \mathbf{w}\) and \(\mathbf{x}\) are nonzero vectors in \(\mathbb{R}^{3}\). $$(\mathbf{u} \times \mathbf{v}) \cdot(\mathbf{w} \times \mathbf{x})=(\mathbf{u} \cdot \mathbf{w})(\mathbf{v} \cdot \mathbf{x})-(\mathbf{u} \cdot \mathbf{x})(\mathbf{v} \cdot \mathbf{w})$$

Imagine three unit spheres (radius equal to 1 ) with centers at \(O(0,0,0), P(\sqrt{3},-1,0)\) and \(Q(\sqrt{3}, 1,0) .\) Now place another unit sphere symmetrically on top of these spheres with its center at \(R\) (see figure). a. Find the coordinates of \(R\). (Hint: The distance between the centers of any two spheres is 2.) b. Let \(\mathbf{r}_{i j}\) be the vector from the center of sphere \(i\) to the center of sphere \(j .\) Find \(\mathbf{r}_{O P}, \mathbf{r}_{O Q}, \mathbf{r}_{P Q}, \mathbf{r}_{O R},\) and \(\mathbf{r}_{P R}\).

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and fare real numbers. It can be shown that this curve lies in a plane. Find a general expression for a nonzero vector orthogonal to the plane containing the curve. $$\begin{aligned} \mathbf{r}(t)=&(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j} \\\ &+(e \cos t+f \sin t) \mathbf{k} \end{aligned}$$ where \(\langle a, c, e\rangle \times\langle b, d, f\rangle \neq \mathbf{0}.\)

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