In Section 9.2, we considered vector addition and scalar multiplication algebraically and interpreted the corresponding geometric meaning. In this section, we will introduce the first of two types of vector multiplication that we will study. This multiplication produces a scalar that tells us a remarkable amount about the geometric relationship between the vectors that are multiplied. We will start investigating these properties in the Preview Activity, but first, we provide the definition of this form of multiplication for two-dimensional vectors.
Consider the two-dimensional vectors \(\vu = \langle u_1,u_2\rangle\) and \(\vv = \langle v_1,v_2\rangle\text{.}\) The dot product of \(\vu\) and \(\vv\) is denoted \(\vu\cdot\vv\) and defined to be
For instance, we have \(\langle 2,-1\rangle\cdot \langle 1,4\rangle = 2\cdot 1 +(-1)\cdot 4 = -2\text{.}\) At first glance, this may feel like we’ve lost information, as we have taken two objects with both magnitude and direction, multiplied them, and obtained a scalar. However, this scalar allows us to draw many useful conclusions, as you will now begin exploring.
Compute each of the dot products listed in the table below. Fill in only the “Value” column at this point. The “Angle Classification” column will be completed in the next part.
When we look at vectors drawn with the same initial point, as you have done with the vectors in Figure 9.3.1, we can consider the angle between two vectors to be the smaller angle formed by looking at them. This results in an angle between \(0\) and \(\pi\) if we measure in radians or between \(0^{\circ}\) and \(180^{\circ}\) if we measure in degrees. Rather than measuring the angles between the vectors you’ve drawn in Figure 9.3.1 precisely, classify the angle between the pairs of vectors in the table in part b by writing acute, right, or obtuse in the “Angle Classification” column.
Look at the values of the dot products that you computed in part b and the classifications of the angles between the vectors that you made in part c. Write a sentence or two to describe what you notice about the relationship between the sign of the dot product and the type of angle between the vectors.
Do you think that knowing only the value of the dot product \(\vv\cdot \vw\) would be enough to determine the exact value of the angle between \(\vv\) and \(\vw\) if you didn’t know the components of \(\vv\) and \(\vw\text{?}\) Write a sentence to explain your reasoning.
Although the angle between a vector and itself is \(0\text{,}\) the dot product of a vector with itself can be computed. Compute \(\vu_1\cdot \vu_1\) and \(\vu_2\cdot \vu_2\) as well as the magnitude of \(\vu_1\) and \(\vu_2\text{.}\) What do you notice about the relationship between these values? Write a sentence explaining why this seems reasonable based on the way the dot product of a vector with itself and the magnitude of that vector are computed.
In Preview Activity 9.3.1, you have started developing an understanding of how the dot product gives us information about the angle between two vectors in the plane. The rest of this section will formalize this idea about the dot product of two vectors and explore properties of this measurement.
The idea of multiplying corresponding components of vectors and summing them can be extended to any dimension, which gives us the definition of the dot product.
Notice that the resulting quantity is a scalar. In the next section, we will explore another product of vectors for which the resulting quantity is a vector. For this reason, you will sometimes see the phrase scalar product used in place of “dot product”.
The dot product gives us a way to define a product of two vectors, and we will soon formalize the ideas from Preview Activity 9.3.1 about what the dot product tells us about the angle between vectors. First, however, we state some useful properties the dot product that show how it behaves in ways that are similar to the product of real numbers. In fact, all of these properties are consequences of properties of addition and multiplication of real numbers.
In Preview Activity 9.3.1, you observed that for two-dimensional vectors, the dot product of a vector with itself is the square of the vector’s magnitude. This is so useful, that we state it below.
If \(\vu=\langle u_1,u_2,\dots,u_n\rangle\) is a vector in \(\R^n\text{,}\) then \(\vu\cdot \vu = \vecmag{\vu}^2\text{.}\) In other words, the dot product of a vector with itself gives the square of the length of the vector.
Let \(\vu,\vv,\vw\) be vectors in \(\R^n\text{.}\) Suppose that you know that \(\vu\cdot \vw = 10\) and \(\vv\cdot \vw = -3\text{.}\) Compute \((\vu+\vv)\cdot \vw\text{.}\)
Let \(\vu,\vv\) be vectors in \(\R^n\text{.}\) Suppose that you know that \(\vu\cdot \vv = 3\text{,}\)\(\vecmag{\vu} = 4\text{,}\) and \(\vecmag{\vv} = 7\text{.}\) Compute \((\vu+\vv)\cdot (\vu+\vv)\text{.}\)
As we saw in Preview Activity 9.3.1, the sign of the dot product gives us information about the type of angle formed by two vectors. We also saw that the dot product alone does not give us sufficient information to measure the angle. However, with a bit more information about the vectors, we are able to measure the angle, as we will soon see. If we are given two vectors \(\vu\) and \(\vv\text{,}\) there are two angles that these vectors create, as depicted in Figure 9.3.4. We will call \(\theta\text{,}\) the smaller of these angles, the angle between these vectors. Notice that \(\theta\) lies between 0 and \(\pi\) because we would consider the angle from the other side if the angle was greater than \(\pi\text{.}\)
A plot of two vectors with an acute angle between them. The acute angle is labeled \(\theta\) and the other angle formed by the vectors is labeled with \(2\pi-\theta\text{.}\)
A plot of two vectors with an obtuse angle between them. The obtuse angle is labeled \(\theta\) and the other angle formed by the vectors is labeled with \(2\pi-\theta\text{.}\)
Recall that the Law of Cosines says that if we consider any triangle with side lengths \(a,b,c\) and \(\theta\) the measure of the angle formed by the sides of length \(a\) and \(b\text{,}\) then \(c^2=a^2+b^2-2ab\cos(\theta)\text{.}\) To determine the angle \(\theta\) here, we may apply the Law of Cosines to the triangle shown in Figure 9.3.6.
A triangle formed by three vectors. The vectors \(\vu\) and \(\vv\) form an angle marked \(\theta\text{.}\) The third side of the triangle is marked \(\vu-\vv\text{.}\)
Using the fact that the dot product of a vector with itself gives us the square of its length, together with other properties of the dot product, we find:
The powerful conclusion from (9.3.1) is that the dot product is a simple algebraic calculation that provides us with important geometric information about the angle between the vectors that would be difficult to determine otherwise. While you normally need to consider if the angle that comes out of an inverse trigonometric function needs to be adjusted based on the context of a problem, we do not need to worry about that with (9.3.2) because the angle measurement and the range of arccosine are both from \(0\) to \(\pi\text{.}\)
If the angle between the vectors \(\vu\) and \(\vv\) is a right angle, what does the expression \(\vu \cdot \vv = \vecmag{\vu} \vecmag{\vv} \cos(\theta)\) say about their dot product?
If the angle between the vectors \(\vu\) and \(\vv\) is acute—that is, less than \(\pi/2\)—what does the expression \(\vu\cdot\vv=\vecmag{\vu}\vecmag{\vv}\cos(\theta)\) say about their dot product?
If the angle between the vectors \(\vu\) and \(\vv\) is obtuse—that is, greater than \(\pi/2\)—what does the expression \(\vu\cdot\vv=\vecmag{\vu}\vecmag{\vv}\cos(\theta)\) say about their dot product?
The last three parts of Activity 9.3.3 allow us to succinctly state the observations you arrived at in the Preview Activity. The next subsection will state these conclusions explicitly.
Many important relationships between vectors arise when the angle between two vectors is a right angle. In this case, we say the vectors are orthogonal. For instance, orthogonality often plays a role in optimization problems: to determine the shortest path from a point in \(\R^3\) to a given plane, we move along a line orthogonal to the plane. From your prior mathematical experience, you are likely familiar with saying that two objects that form a right angle are perpendicular. When vectors are involved, we prefer to use the term orthogonal because it allows for one (or both) of the vectors to be \(\vzero\text{,}\) which is a case where we would not say the vectors are perpendicular.
As Activity 9.3.3 indicates, the dot product provides a simple means to determine whether two vectors are orthogonal to one another. In this case,\(\vu\cdot\vv=\vecmag{\vu}\vecmag{\vv}\cos(\pi/2) = 0\text{,}\) so we make the following important observation.
More generally, the sign of the dot product gives us useful information about the relative orientation of the vectors. Since the magnitudes in equation (9.3.1) are positive, recalling that
\begin{align*}
\cos(\theta) \amp \gt 0 \mathstrut \amp \amp \text{ if } \theta \text{ is an acute angle,}\\
\cos(\theta) \amp = 0 \mathstrut \amp\amp \text{ if } \theta \text{ is a right angle, and}\\
\cos(\theta) \amp \lt 0 \mathstrut \amp\amp \text{ if } \theta \text{ is an obtuse angle,}\\
\end{align*}
then we see that for nonzero vectors \(\vu\) and \(\vv\text{,}\)
\begin{align*}
\vu\cdot\vv \amp \gt 0 \mathstrut \amp\amp \text{ if } \theta \text{ is an acute angle,}\\
\vu\cdot\vv \amp = 0 \mathstrut \amp\amp \text{ if }
\theta \text{ is a right angle, and}\\
\vu\cdot\vv \amp \lt 0 \mathstrut \amp\amp \text{ if }
\theta \text{ is an obtuse angle.}
\end{align*}
These different orientations of vectors will lead us to seeking to understand how to construct a vector that represents the portion of one vector that is in the direction of (or parallel to) another vector. We examine this question in depth in the next subsection.
Return to Preview Activity 9.3.1 and use the results of this subsection to determine which pairs of vectors are orthogonal. Also classify the angles between nonorthogonal pairs of vectors as acute or obtuse.
Suppose we want to split the vector \(\vu\) into parts that are parallel to \(\vv\) and orthogonal to \(\vv\text{,}\) which we will call \(\vw_1\) and \(\vw_2\) respectively. In other words, we want to write \(\vu=\vw_1+\vw_2\) where \(\vw_1=k \vv\) and \(\vw_2 \cdot \vv =0\text{.}\) Geometrically, this will look like one of the possibilities shown in Figure 9.3.8.
Vector \(\vv\) points horizontally with \(\vu\) above it making an acute angle \(\theta\text{.}\) The vector \(\vw_2\) is orthogonal to \(\vv\text{,}\) pointing upward so that the tip of \(\vw_2\) coincides with the tip of \(\vu\text{.}\) Vector \(\vw_1\) has the same tail as \(\vv\) and its tip coincides with the tail of \(\vw_2\text{.}\)
Vector \(\vv\) points horizontally with \(\vu\) above it making an obtuse angle \(\theta\text{.}\) The vector \(\vw_2\) is orthogonal to the horizontal axis, which passes through \(\vv\text{,}\) and \(\vw_2\) points upward so that the tip of \(\vw_2\) coincides with the tip of \(\vu\text{.}\) Vector \(\vw_1\) has the same tail as \(\vv\) and its tip coincides with the tail of \(\vw_2\text{.}\)
Notice that if the angle between \(\vu\) and \(\vv\) is obtuse, then the amount of \(\vu\) that is parallel to \(\vv\) will be in the direction opposite \(\vv\text{.}\) Before we look at how to calculate the vectors \(\vw_1\) and \(\vw_2\text{,}\) we will do an activity about drawing the graphical representation of this splitting of a vector into parts that are parallel to and orthogonal to another vector.
In Figure 9.3.8, you can see illustrations of splitting \(\vu\) into two parts: \(\vw_1\text{,}\) which is parallel to \(\vv\text{,}\) and \(\vw_2\text{,}\) which is orthogonal to \(\vv\text{.}\) Use this figure for reference as you do the following.
We know from the previous subsection that there is a third configuration of vectors, which occurs when \(\vu\) and \(\vv\) are orthogonal. Suppose that \(\vu\) and \(\vv\) are nonzero orthogonal vectors. What would \(\vw_1\) and \(\vw_2\) be in this case?
We want to switch the roles of \(\vu\) and \(\vv\) for the examples in the previous parts. Specifically, for these configuration of vectors, we want to split \(\vv\) into parts that are parallel to \(\vu\text{,}\) which we will call \(\vz_1\text{,}\) and orthogonal to \(\vu\text{,}\) which we will call \(\vz_2\text{.}\) On Figure 9.3.9, draw \(\vz_1\) and \(\vz_2\) for each configuration.
Now that we have visually explored the relationship between \(\vu\text{,}\)\(\vv\text{,}\)\(\vw_1\text{,}\) and \(\vw_2\text{,}\) we will see how we can find formulas that allow us to compute \(\vw_1\) and \(\vw_2\) in terms of \(\vu\) and \(\vv\text{.}\)
Suppose that \(\vu\) and \(\vv\) are vectors in \(\R^n\text{.}\) We wish to find \(\vw_1\) and \(\vw_2\) so that \(\vw_1\) is parallel to \(\vv\) and \(\vw_2\) is orthogonal to \(\vv\text{,}\) as shown in Figure 9.3.8.
Compute \(\vu\cdot \vv\) as \((\vw_1+\vw_2)\cdot \vv\text{.}\) Simplify your answer as much as possible, using the fact that \(\vw_2\) is orthogonal to \(\vv\text{.}\)
Since \(\vw_1\) is parallel to \(\vv\text{,}\) there is a scalar \(k\) so that \(\vw_1 = k\vv\text{.}\) Substitute \(k\vv\) for \(\vw_1\) in your answer to the previous part and then solve for \(k\text{.}\)
The vector \(\vw_1\) we have been considering in this subsection is so useful that it has specific notation. We use the previous activity to state the following definition.
Using the property that the dot prodcut of a vector with itself is the square of the vector’s magnitude, we can write \(\proj_{\vv} \vu = \left(\frac{\vu \cdot \vv}{\vecmag{\vv}}\right) \frac{\vv}{\vecmag{\vv}}\text{.}\) Since \(\vv/\vecmag{\vv}\) is a unit vector, we can see from this form that the magnitude of \(\proj_{\vv}\vu\) is \((\vu\cdot \vv)/\vecmag{\vv}\text{.}\) We call this scalar the component of \(\vu\) in the direction of \(\vv\) and denote it by
Recalling from earlier in the section that \(\vu\cdot \vv = \vecmag{\vu}\vecmag{\vv}\cos(\theta)\text{,}\) where \(\theta\) is the angle between \(\vu\) and \(\vv\text{,}\) we can also write
Vector subtraction also gives us a way to solve for the piece \(\vw_2\) of \(\vu\) that is orthogonal to \(\vv\text{.}\) We find \(\vw_2=\vu-\vw_1= \vu-\frac{\vu \cdot \vv}{\vv \cdot \vv} \vv\text{.}\) The orthogonal projection of \(\vu\) onto \(\vv\) is
These relationships (for one configuration of \(\vu\) and \(\vv\)) are shown in Figure 9.3.11. Remember that the length of \(\proj_{\vv}\vu\) is \(\comp_{\vv}\vu\text{.}\) Also notice how the arrangement of the vectors provides an illustration of the fact that \(\vu = \proj_{\vv}\vu + \proj_{\perp \vv}{\vu}\text{.}\)
Vector \(\vv\) points horizontally with \(\vu\) above it making an acute angle \(\theta\text{.}\) The vector \(\proj_{\perp \vv}\vu\) is orthogonal to \(\vv\text{,}\) pointing upward so that the tip of \(\vw_2\) coincides with the tip of \(\vu\text{.}\) Vector \(\proj_{\vv}\) has the same tail as \(\vv\) and its tip coincides with the tail of \(\vw_2\text{.}\)
Figure9.3.11.The relationship between \(\vu\text{,}\)\(\vv\text{,}\) the projection of \(\vu\) onto \(\vv\) and the orthogonal projection of \(\vu\) onto \(\vv\text{.}\)
The component of \(\vu\) in the direction of \(\vv\) will be a scalar measure of how much of \(\vu\) is parallel to \(\vv\text{.}\) In fact, this shows up in the second formula for \(\proj_{\vv} \vu\text{.}\) Specifically, we can write \(\proj_{\vv} \vu = (\comp_{\vv} \vu) \frac{\vv}{\vecmag{\vv}}\text{,}\) which separates the projection of \(\vu\) onto \(\vv\) into how much of \(\vu\) is parallel to \(\vv\) (a scalar) times the unit vector in the direction of \(\vv\text{.}\) This useful idea of separating the magnitude of a vector and the direction of a vector (as described by a unit vector) will be used in the analysis of many vector calculations. The first expression for \(\proj_{\vv} \vu = \frac{\vu \cdot \vv}{\vv\cdot\vv} \vv\) can be interpreted conceptually as saying the projection of \(\vu\) onto \(\vv\) must be parallel to \(\vv\) and \(\frac{\vu \cdot \vv}{\vv\cdot\vv}\) is closely related to the length of \(\proj_{\vv} \vu\text{.}\)
In this example we will look at calculating and making sense of the different parts of the projection of \(\langle 4,3\rangle \) onto \(\langle 0,2 \rangle\text{.}\) For notational purposes, we will let \(\vu = \langle 4,3\rangle \) and \(\vv = \langle 0,2 \rangle\text{.}\) Applying Definition 9.3.10, we find
Note that the scalar \((3)\) in the last equation is \(\comp_\vv \vu\) and the vector \(\langle 0,1\rangle\) is the unit vector in the direction of \(\vv\text{.}\)
The projection of \(\vu\) onto \(\vv\) can be split into the scalar \(3\text{,}\) the component of \(\vu\) in the direction of \(\vv\text{,}\) and the unit vector in the direction of \(\vv\text{,}\)\(\langle 0,1 \rangle \text{.}\)
The orthogonal projection of \(\vu\) onto \(\vv\) will be given by how much of \(\vu\) is horizontal because we subtract out all of the vertical component of \(\vu\text{.}\)
Going forward, we will use the projection more often than the dot product by itself. This is because the projection precisely measures how much of one vector is in a particular direction rather than the much vauger measurement of the type of angle between the vectors.
Let \(\vv = \langle 4, -8 \rangle\text{.}\) Find \(\proj_{\vv} \vu\text{,}\)\(\comp_{\vv} \vu\text{,}\) and \(\proj_{\perp \vv} \vu\text{.}\) Draw a picture to illustrate the vectors involved. Finally, express \(\vu\) as the sum of two vectors where one is parallel to \(\vv\) and the other is perpendicular to \(\vv\text{.}\)
Now let \(\vw = \langle -2,4 \rangle \text{.}\) Add \(\vw\) to the picture you drew in the previous part. Without doing any calculations, find \(\proj_{\vw} \vu\text{.}\) Explain your reasoning.
Find a vector \(\vw\) not parallel to \(\vz = \langle 3,4 \rangle \) such that \(\proj_{\vz} \vw\) has length \(10\text{.}\) Note that there are infinitely many different answers!
It is important to note that in general \(\proj_{\vv} \vw \neq \proj_{\vw} \vv\text{.}\) You can explore in Exercise 9.3.6.12 how switching the vectors in the projection formula will give a different results.
In physics, work is a measure of the energy required to apply a force to an object through a displacement. In Figure 9.3.14, we can see a diagram showing the force applied to a refrigerator (in orange) that is used to move the refrigerator from point \(A\) to point \(B\text{.}\) The change in location, \(\overrightarrow{AB}\text{,}\) is called the displacement. Notice that not all of the force is in the direction of the displacement. The work done on the refrigerator only depends on the amount of force that is in the direction of \(\overrightarrow{AB}\text{.}\)
A diagram showing the force applied to the refrigerator as the refrigerator is moved from \(A\) to \(B\text{.}\) The momement is rightward and parallel to the ground, but the force is rightward and slightly upward.
This means that the work is determined only by the magnitude of the force applied parallel to the displacement and the length of the displacement. In fact, work is closely related to the projection of the force vector onto the displacement vector. (More precisely, it is the product of the component of the force vector in the direction of the displacement vector times the magnitude of the displacement vector.)
The units commonly used for work are Joules or Newton-meters in the metric system and foot-pounds in customary units. (In electrical settings, units such as coulomb-volts or kilowatt-hours may be used.) Note that these units come from work being the dot product of a force vector, using units of Newtons or pounds, and a displacement vector, using units of meters or feet.
The dot product of two vectors in \(\R^n\text{,}\)\(\vu = \langle u_1,
u_2, \ldots, u_n \rangle\) and \(\vv = \langle v_1, v_2, \ldots, v_n
\rangle\text{,}\) is the scalar
Two vectors are orthogonal if the angle between them is \(\pi/2\text{.}\) In terms of the dot product, the vectors \(\vu\) and \(\vv\) are orthogonal if and only if \(\vu \cdot \vv = 0\text{.}\)
Perform the following operations on the vectors \(\vec{u} = \left\lt 3,-1,-3\right>\text{,}\)\(\vec{v} = \left\lt -3,1,-2\right>\text{,}\) and \(\vec{w} = \left\lt 1,1,0\right>\text{.}\)
Find \({ \mathbf a \cdot \mathbf b }\) if \(\left| {\mathbf a} \right|\) = 9, \(\left| {\mathbf b} \right|\) = 1, and the angle between \({\mathbf a}\) and \({\mathbf b}\) is \(\frac{\pi}{2}\) radians.
Find \({ \mathbf a \cdot \mathbf b }\) if \(\left| {\mathbf a} \right|\) = 2, \(\left| {\mathbf b} \right|\) = 4, and the angle between \({\mathbf a}\) and \({\mathbf b}\) is \(- \frac{\pi}{1}\) radians.
A constant force \({\mathbf F} = -6 {\mathbf i} - 9 {\mathbf j} + 7 {\mathbf k}\) moves an object along a straight line from point \((-6, -3, 8)\) to point \((-8, -13, 18)\text{.}\)
A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 7 feet long and inclined at an angle of 30 degrees above the horizontal.
Let \(\vz = \langle 2, 1, 0 \rangle\text{.}\) Find a unit vector \(\vr\) in \(\R^3\) such that \(\vr\) is perpendicular to both \(\vy\) and \(\vz\text{.}\) How many such vectors are there? Explain your process.
In this exercise, we will show that the roles of \(\vu\) and \(\vv\) are not symmetric in projection formula (Definition 9.3.10). We will use the same values as Example 9.3.12, namely \(\vu = \langle 4,3\rangle \) and \(\vv = \langle 0,2 \rangle\text{,}\) but we will be looking at the projection of \(\vv\) onto \(\vu\text{.}\)
Will \(proj_\vu \vv\) be parallel to \(\vu\) or \(\vv\text{?}\) Use this answer to explain how you know that the result of \(proj_\vu \vv\) must be different than \(proj_\vv \vu\) (which we calculated in Example 9.3.12).
Let \(\vu\) and \(\vv\) be vectors in \(\R^5\) with \(\vu \cdot \vv = -1\text{,}\)\(\Vert \vu \Vert = 2\text{,}\)\(\Vert \vv \Vert = 3\text{.}\) Use the properties of the dot product to find each of the following.
One of the properties of the dot product is that \((\vu+\vv) \cdot \vw = (\vu \cdot \vw) + (\vv \cdot \vw)\text{.}\) That is, the dot product distributes over vector addition on the right. Here we investigate whether the dot product distributes over vector addition on the left.
Let \(\vu = \langle 1,2,-1 \rangle\text{,}\)\(\vv = \langle 4,-3,6 \rangle\text{,}\) and \(\vw = \langle 4,7,2 \rangle\text{.}\) Calculate
When running a sprint, the racers may be aided or slowed by the wind. The wind assistance is a measure of the wind speed that is helping push the runners down the track. It is much easier to run a very fast race if the wind is blowing hard in the direction of the race. So that world records aren’t dependent on the weather conditions, times are only recorded as record times if the wind aiding the runners is less than or equal to 2 meters per second. Wind speed for a race is recorded by a wind gauge that is set up close to the track. It is important to note, however, that weather is not always as cooperative as we might like. The wind does not always blow exactly in the direction of the track, so the gauge must account for the angle the wind makes with the track. Suppose a 4 mile per hour wind is blowing to aid runners by making a \(38^{\circ}\) angle with the race track. Determine if any times set during such a race would qualify as records.
Molecular geometry is the geometry determined by arrangements of atoms in molecules. Molecular geometry includes measurements like bond angle, bond length, and torsional angles. These attributes influence several properties of molecules, such as reactivity, color, and polarity.
As an example of the molecular geometry of a molecule, consider the methane \(\text{CH}_4\) molecule, as illustrated in Figure 9.3.15. According to the Valence Shell Electron Repulsion (VSEPR) model, atoms that surround single different atoms do so in a way that positions them as far apart as possible. This means that the hydrogen atoms in the methane molecule arrange themselves at the vertices of a regular tetrahedron. The bond angle for methane is the angle determined by two consecutive hydrogen atoms and the central carbon atom. To determine the bond angle for methane, we can place the center carbon atom at the point \(\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right)\) and the hydrogen atoms at the points \((0,0,0)\text{,}\)\((1,1,0)\text{,}\)\((1,0,1)\text{,}\) and \((0,1,1)\text{.}\) Find the bond angle for methane to the nearest tenth of a degree.
