Integral Review Part II

To begin our discussion, we use the word vector to show a displacement. What separates a vector, as a quantity, is that a vector is identified with length (magnitude) and direction. Calculus uses vectors for many other purposes such as velocity or force. In 1-space, a displacement from a point -3 to 5 is shown by a vector which has a magnitude 8 and directed as an arrow with initial point -3 and terminal 5. We use bold fonts to show vectors in this document and a sharp set of brackets to provide the coordinates of a vector such as \boldsymbol{u} =\left \langle u_{1} ,u_{2} ,u_{3}\right \rangle.

Using this description, the magnitude of a vector \boldsymbol{u} must represent the length of the displacement that is the distance between initial and terminal point of the given vector. We show this by \left \Vert \boldsymbol{u}\right \Vert. Any vector with length equal to 1 is said to be a unit vector.

In 2-space, 3-space or more general n-space for any natural number n , using terminal points and initial points for displacement, one can construct such arrows. One needs to review the concept of vectors in general, not only in the sense of displacement. This shall be done in a linear algebra course. In the way that we construct vectors, one can start with a different initial and terminal points and get the same displacement. This is why, vectors with the same magnitude and same direction are considered as one vector (or better, a class of vectors). Vector u and v below are said to be equivalent to each other.

In n-space, addition and multiplication by real numbers of vectors are coordinate-wise. In fact, for any two vectors \boldsymbol{u} and \boldsymbol{v} , we define

(1)   \begin{equation*}\boldsymbol{u} +\boldsymbol{v} \end{equation*}

to be the resultant vector, or sum of the two vectors \boldsymbol{u} and \boldsymbol{v} , obtained by placing the initial point of \boldsymbol{v} at the terminal point of \boldsymbol{u} and connecting the initial of \boldsymbol{u} with terminal point of \boldsymbol{v}.

Suppose k is a real number and given \boldsymbol{u} =\left \langle u_{1} ,u_{2}\right \rangle and \boldsymbol{v} =\langle v_{1} ,v_{2}\rangle,

(2)   \begin{equation*}k\boldsymbol{u} =\left \langle ku_{1} ,ku_{2}\right \rangle \end{equation*}

(3)   \begin{equation*}\boldsymbol{u} +\boldsymbol{v} =\left \langle u_{1} ,u_{2}\right \rangle +\left \langle v_{1} +v_{2}\right \rangle =\left \langle u_{1} +v_{1} ,u_{2} +v_{2}\right \rangle \end{equation*}

Any linear algebra book (or these days any calculus book) provides a list of vector properties. Suppose \boldsymbol{u} ,\boldsymbol{v} and \boldsymbol{w} are vectors in n-space and k and t are real numbers. Here \boldsymbol{0} is the zero vector. We assign the explanation for each of these as an exercise.

1. \boldsymbol{u} +\boldsymbol{v} =\boldsymbol{v} +\boldsymbol{u}

2. \left (\boldsymbol{u} +\boldsymbol{v}\right ) +\boldsymbol{w} =\boldsymbol{u} +\left (\boldsymbol{v} +\boldsymbol{w}\right )

3. \boldsymbol{u} +\boldsymbol{0} =\boldsymbol{u}

4. \left ( -\boldsymbol{u}\right ) +\boldsymbol{u} =\boldsymbol{0}

5. k\left (\boldsymbol{u} +\boldsymbol{v}\right ) =k\boldsymbol{u} +k\boldsymbol{v}

6. \left (k +t\right )\boldsymbol{u} =k\boldsymbol{u} +t\boldsymbol{u}

7. \left (kt\right )\boldsymbol{u} =k\left (t\boldsymbol{u}\right )

8. \left (1\right )\boldsymbol{u} =\boldsymbol{u}

We use \boldsymbol{i} ,\boldsymbol{j} and \boldsymbol{k} to show the basic vectors \left \langle 1 ,0 ,0\right \rangle, \left \langle 0 ,1 ,0\right \rangle and \left \langle 0 ,0 ,1\right \rangle with respect to order.

Example 1.

The notion \left (a ,b\right ) represents a point in 2-space as explained in previous section and \left \langle a ,b\right \rangle is a vector. The vector \left \langle a ,b\right \rangle can be drawn in such a way that it has its initial point at the origin with terminal point \left (a ,b\right ) .

Example 2.

Given vectors \boldsymbol{w} =\left \langle 2 ,1\right \rangle and \boldsymbol{v} =\left \langle -4 , -1\right \rangle we draw the vector \boldsymbol{v} -\boldsymbol{w} and find its magnitude to be

(4)   \begin{equation*}\Vert \boldsymbol{v} -\boldsymbol{w}\Vert =\sqrt{\left ( -4 -2\right )^{2} +( -1 -1)^{2}} =\sqrt{40} \end{equation*}

Try to do this similar exercise. Given \boldsymbol{u} =\left \langle 6 , -2\right \rangle and \boldsymbol{v} =\left \langle 1 ,2\right \rangle, find \boldsymbol{u} -2\boldsymbol{v} and \frac{1}{2}\boldsymbol{u} +3\boldsymbol{u} using a geometric argument.

Example 3.

Given a number a and a unit vector \boldsymbol{u} we can construct the vector a\boldsymbol{u} with length a in direction of \boldsymbol{u}. Here, our goal is to find two vectors parallel to the vector \left \langle 3 ,1\right \rangle and length 3. Let us begin with making a unit vector

(5)   \begin{equation*}\frac{1}{\left \Vert \left \langle 3 ,1\right \rangle \right \Vert }\left \langle 3 ,1\right \rangle =\frac{1}{\sqrt{9 +1}}\left \langle 3 ,1\right \rangle =\left \langle \frac{3}{\sqrt{10}} ,\frac{1}{\sqrt{10}}\right \rangle \end{equation*}

Now a new vector 3\left \langle \frac{3}{\sqrt{10}} ,\frac{1}{\sqrt{10}}\right \rangle has length 3 and the desired direction. Observe that the vector with the same magnitude and opposite direction also satisfies the given condition that is -3\left \langle \frac{3}{\sqrt{10}} ,\frac{1}{\sqrt{10}}\right \rangle.

Example 4.

We want to show that \left \langle \frac{1}{2} ,\frac{\sqrt{3}}{2}\right \rangle is a unit vector. For this, all we have to do is to show the magnitude of this vector is equal to 1. Indeed,

(6)   \begin{equation*}\left \Vert \left \langle \frac{1}{2} ,\frac{\sqrt{3}}{2}\right \rangle \right \Vert =\sqrt{\genfrac{(}{)}{}{}{1}{2}^{2} +(\frac{\sqrt{3}}{2})^{2}} =\sqrt{\frac{1}{4} +\frac{3}{4}} =1\end{equation*}

In general, the vector \left \langle \cos \theta ,\sin \theta \right \rangle for any \theta is a unit vector.

Example 5.

Given any real number k , and a vector \boldsymbol{u} in 3-space, show

(7)   \begin{equation*}\left \Vert k\boldsymbol{u}\right \Vert =\vert k|\Vert\boldsymbol{u}\Vert \end{equation*}

We start by writing \boldsymbol{u} =\left \langle u_{1} ,u_{2} ,u_{3}\right \rangle and so

(8)   \begin{align*}\left \Vert k\boldsymbol{u}\right \Vert =\left \Vert \left \langle ku_{1} ,ku_{2} ,ku_{3}\right \rangle \right \Vert \\ =\sqrt{\left (ku_{1}\right )^{2} +(ku_{2})^{2} +(ku_{3})^{2}} \\ =\sqrt{k^{2}\left (u_{1}^{2} +u_{2}^{2} +u_{3}^{2}\right )} \\ =\sqrt{k^{2}}\left \Vert \boldsymbol{u}\right \Vert =\left \vert k\right \vert \left \Vert \boldsymbol{u}\right \Vert \end{align*}

Example 6.

Given any vector \boldsymbol{u} , we want to show \frac{\boldsymbol{u}}{\left \Vert \boldsymbol{u}\right \Vert } is a unit vector. We evaluate its magnitude

(9)   \begin{equation*}\genfrac{\Vert }{\Vert }{}{}{\boldsymbol{u}}{\left \Vert\boldsymbol{u}\right \Vert } =\left \Vert \frac{1}{\left \Vert\boldsymbol{u}\right \Vert }\boldsymbol{u}\right \Vert =\genfrac{\vert }{\vert }{}{}{1}{\left \Vert\boldsymbol{u}\right \Vert }\left \Vert\boldsymbol{u}\right \Vert =\frac{1}{\left \Vert\boldsymbol{u}\right \Vert }\left \Vert\boldsymbol{u}\right \Vert =1 \end{equation*}

Example 7.

The vector \boldsymbol{u} makes the angle 45^{ \circ } with x-axis and has length 5. We want to write \boldsymbol{u} in terms of \boldsymbol{i} and \boldsymbol{j} . First we can get a direction by defining the unit vector \langle \cos 45^{ \circ } ,\sin 45^{ \circ }\rangle =\langle \frac{1}{\sqrt{2}} ,\frac{1}{\sqrt{2}}\rangle. Next if we multiply this vector by 5 , we get the desired vector \left \langle \frac{5}{\sqrt{2}} ,\frac{5}{\sqrt{2}}\right \rangle. Note that the length of this vector is 5. Now we write this vector as a linear combination of the basic vectors in 2-space

(10)   \begin{align*}\left \langle \frac{5}{\sqrt{2}} ,\frac{5}{\sqrt{2}}\right \rangle =\left \langle \frac{5}{\sqrt{2}} ,0\right \rangle +\left \langle 0 ,\frac{5}{\sqrt{2}}\right \rangle \\ =\frac{5}{\sqrt{2}}\left \langle 1 ,0\right \rangle +\frac{5}{\sqrt{2}}\left \langle 0 ,1\right \rangle \\ =\frac{5}{\sqrt{2}}\boldsymbol{i} +\frac{5}{\sqrt{2}}\boldsymbol{j}\end{align*}

Dot Product.

Given two vectors \boldsymbol{u} and \boldsymbol{v} in n-space, we define the dot product of \boldsymbol{u} and \boldsymbol{v} to be

(11)   \begin{equation*}\boldsymbol{u} \cdot \boldsymbol{v} =u_{1}v_{1} +u_{2}v_{2} +u_{3}v_{3} + . . . +u_{n}v_{n} \end{equation*}

Note that the result of the dot product of two vectors is a number. For example, dot product of \left \langle 1 , -1\right \rangle and \left \langle 3 , -2\right \rangle is

(12)   \begin{equation*}\left \langle 1 , -1\right \rangle \cdot \left \langle 3 , -2\right \rangle =5 \end{equation*}

A set of properties of dot product is below. Suppose \boldsymbol{u} ,\boldsymbol{v} and \boldsymbol{w} are vectors and k is a real number. Here \boldsymbol{0} is the zero vector unless said otherwise.

1. \boldsymbol{u} \cdot \boldsymbol{u} =\left \Vert \boldsymbol{u}\right \Vert ^{2}.

To see this, write \boldsymbol{u} =\left \langle u_{1} ,u_{2} , . . . ,u_{n}\right \rangle . Now using definition of dot product

(13)   \begin{align*}\boldsymbol{u} \cdot\boldsymbol{u} =\left \langle u_{1} ,u_{2} , . . . ,u_{n}\right \rangle \cdot \left \langle u_{1} ,u_{2} , . . . ,u_{n}\right \rangle \\ =u_{1}^{2} +u_{2}^{2} + . . . +u_{n}^{2} \\ =\left \Vert\boldsymbol{u}\right \Vert ^{2}\end{align*}

and we are done.

2. \boldsymbol{u} \cdot \boldsymbol{v} =\boldsymbol{v} \cdot \boldsymbol{u}

3. \boldsymbol{u} \cdot \boldsymbol{v} +\boldsymbol{u} \cdot \boldsymbol{w} =\boldsymbol{u} \cdot \left (\boldsymbol{v} +\boldsymbol{w}\right )

4. \boldsymbol{0} \cdot \boldsymbol{u} =0 Here, the first zero is the zero vector.

5. k\left (\boldsymbol{u} \cdot \boldsymbol{v}\right ) =\left (k\boldsymbol{u}\right ) \cdot \boldsymbol{v}

6. The following statements are equivalent to each other

a) \boldsymbol{u} \cdot \boldsymbol{u} =0 Here, 0 is the number zero.

b) \boldsymbol{u} =\boldsymbol{0} Here, \boldsymbol{0} is the vector zero.

We show this part:

a) implies b) Suppose we know \boldsymbol{u} \cdot\boldsymbol{u} =0. Then \left \Vert\boldsymbol{u}\right \Vert ^{2} =0 so \sqrt{u_{1}^{2} +u_{2}^{2} + . . . +u_{n}^{2}} =0. This means

(14)   \begin{equation*}u_{1}^{2} +u_{2}^{2} + . . . +u_{n}^{2} =0 \end{equation*}

The only way that sum of n non-negative numbers is equal to zero is if all of them are zero. This means all coordinates of \boldsymbol{u} are equal to zero.

b) implies a) Suppose \boldsymbol{u} =\boldsymbol{0}. Then all coordinates of \boldsymbol{u} are zero that is u_{1} =u_{2} =u_{3} = . . . =u_{n} =0. Use definition of dot product to get

(15)   \begin{equation*}0^{2} +0^{2} + . . . +0^{2} =0 \end{equation*}

And we are done.

Example 8.

Consider the two vectors shown below.

A difference vector is shown in red. The law of cosines provides us with the fact that

(16)   \begin{equation*}\left \Vert \boldsymbol{u} -\boldsymbol{v}\right \Vert ^{2} =\left \Vert \boldsymbol{u}\right \Vert ^{2} +\left \Vert \boldsymbol{v}\right \Vert ^{2} -2\Vert \boldsymbol{u}\Vert \Vert\boldsymbol{v}\Vert \cos \theta \end{equation*}

On the left of this equality we have

(17)   \begin{align*}\left \Vert \boldsymbol{u} -\boldsymbol{v}\right \Vert ^{2} =\left (\boldsymbol{u} -\boldsymbol{v}\right ) \cdot \left (\boldsymbol{u} -\boldsymbol{v}\right ) \\ =\boldsymbol{u} \cdot \boldsymbol{u} -\boldsymbol{u} \cdot \boldsymbol{v} -\boldsymbol{v} \cdot \boldsymbol{u} +\boldsymbol{v} \cdot \boldsymbol{v} \\ =\left \Vert \boldsymbol{u}\right \Vert ^{2} -2\boldsymbol{u} \cdot \boldsymbol{v} +\left \Vert \boldsymbol{v}\right \Vert ^{2}\end{align*}

Setting this equal to the right side of equality (6) one gets

(18)   \begin{equation*}\left \Vert \boldsymbol{u}\right \Vert ^{2} -2\boldsymbol{u} \cdot \boldsymbol{v} +\left \Vert \boldsymbol{v}\right \Vert ^{2} =\left \Vert \boldsymbol{u}\right \Vert ^{2} +\left \Vert \boldsymbol{v}\right \Vert ^{2} -2\Vert \boldsymbol{u}\Vert \Vert\boldsymbol{v}\Vert \cos \theta \end{equation*}

(19)   \begin{equation*} -2\boldsymbol{u} \cdot \boldsymbol{v} = -2\Vert \boldsymbol{u}\Vert \Vert\boldsymbol{v}\Vert \cos \theta \end{equation*}

that is

(20)   \begin{equation*}\boldsymbol{u} \cdot \boldsymbol{v} =\Vert \boldsymbol{u}\Vert \Vert\boldsymbol{v}\Vert \cos \theta \end{equation*}

One can use this relation to notice, saying that two vectors are perpendicular to each other is exactly saying that their dot product is equal to zero. Therefore, we have a tool to identify if two vectors are perpendicular to each other.

Example 9.

Consider the following two vectors \boldsymbol{u} and \boldsymbol{v}; more, consider the vector shadow of \boldsymbol{u} on \boldsymbol{v} if light is going downward perpendicular to \boldsymbol{v}.


Observe that to find the coordinate of this shadow vector, we need its magnitude. This is because, we can multiply this magnitude to \frac{\boldsymbol{v}}{\left \Vert \boldsymbol{v}\right \Vert } and get the shadow vector. We call this shadow vector, vector projection of \boldsymbol{u} onto \boldsymbol{v} and show it by \text{proj}_{\boldsymbol{v}}\boldsymbol{u}. Now, if \theta is the angle between \boldsymbol{u} and \boldsymbol{v}

(21)   \begin{equation*}\cos \theta =\frac{\left \Vert\text{proj}_{\boldsymbol{v}}\boldsymbol{u}\right \Vert }{\left \Vert \boldsymbol{u}\right \Vert } \end{equation*}

and so

(22)   \begin{align*}\text{proj}_{\boldsymbol{v}}\boldsymbol{u} =\left \Vert\text{proj}_{\boldsymbol{v}}\boldsymbol{u}\right \Vert \frac{\boldsymbol{v}}{\left \Vert \boldsymbol{v}\right \Vert } \\ =\left \Vert \boldsymbol{u}\right \Vert \cos \theta \frac{\boldsymbol{v}}{\left \Vert \boldsymbol{v}\right \Vert } \\ =\left \Vert \boldsymbol{u}\right \Vert \frac{\boldsymbol{u} \cdot \boldsymbol{v}}{\Vert \boldsymbol{u}\Vert \Vert\boldsymbol{v}\Vert }\frac{\boldsymbol{v}}{\left \Vert \boldsymbol{v}\right \Vert } \\ =\frac{\boldsymbol{u} \cdot \boldsymbol{v}}{\boldsymbol{v} \cdot \boldsymbol{v}}\boldsymbol{v}\end{align*}

Example 10.

Suppose \boldsymbol{u} and \boldsymbol{v} are any two vectors. Then it is a fact that for the vector \boldsymbol{u}\left \Vert\boldsymbol{v}\right \Vert +\boldsymbol{v}\left \Vert\boldsymbol{u}\right \Vert we have

(23)   \begin{equation*}\left \Vert\boldsymbol{u}\left \Vert\boldsymbol{v}\right \Vert +\boldsymbol{v}\left \Vert\boldsymbol{u}\right \Vert \right \Vert ^{2} \geq 0 \end{equation*}

this means

(24)   \begin{equation*}\left (\boldsymbol{u}\left \Vert\boldsymbol{v}\right \Vert +\boldsymbol{v}\left \Vert\boldsymbol{u}\right \Vert \right ) \cdot \left (\boldsymbol{u}\left \Vert\boldsymbol{v}\right \Vert +\boldsymbol{v}\left \Vert\boldsymbol{u}\right \Vert \right ) =2\left \Vert\boldsymbol{u}\right \Vert ^{2}\left \Vert\boldsymbol{v}\right \Vert ^{2} +2\boldsymbol{u} \cdot\boldsymbol{v}\left \Vert\boldsymbol{u}\right \Vert \left \Vert\boldsymbol{v}\right \Vert \geq 0 \end{equation*}

(25)   \begin{equation*}2\left \Vert\boldsymbol{u}\right \Vert ^{2}\left \Vert\boldsymbol{v}\right \Vert ^{2} \geq -2\boldsymbol{u} \cdot\boldsymbol{v}\left \Vert\boldsymbol{u}\right \Vert \left \Vert\boldsymbol{v}\right \Vert \end{equation*}

Simplify to get

(26)   \begin{equation*} -\boldsymbol{u} \cdot\boldsymbol{v} \leq \left \Vert\boldsymbol{u}\right \Vert \left \Vert\boldsymbol{v}\right \Vert \end{equation*}

In fact using the same process with the vector \boldsymbol{u}\left \Vert\boldsymbol{v}\right \Vert -\boldsymbol{v}\left \Vert\boldsymbol{u}\right \Vert instead, provides

(27)   \begin{equation*}\boldsymbol{u} \cdot\boldsymbol{v} \leq \left \Vert\boldsymbol{u}\right \Vert \left \Vert\boldsymbol{v}\right \Vert \end{equation*}

From these two inequality, we get the well-known Cauchy-Schwarz Inequality

(28)   \begin{equation*}\left \vert\boldsymbol{u} \cdot\boldsymbol{v}\right \vert \leq \left \Vert\boldsymbol{u}\right \Vert \left \Vert\boldsymbol{v}\right \Vert \end{equation*}

Example 11.

Suppose \boldsymbol{u} and \boldsymbol{v} are two given vectors. Then

(29)   \begin{align*}(\left \Vert\boldsymbol{u}\right \Vert +\left \Vert\boldsymbol{v}\right \Vert )^{2} =\left \Vert\boldsymbol{u}\right \Vert ^{2} +2\left \Vert\boldsymbol{u}\right \Vert \left \Vert\boldsymbol{v}\right \Vert +\left \Vert\boldsymbol{v}\right \Vert ^{2} \\ \geq \left \Vert\boldsymbol{u}\right \Vert ^{2} +2\boldsymbol{u} \cdot\boldsymbol{v} +\left \Vert\boldsymbol{v}\right \Vert ^{2} \\ =(\boldsymbol{u} +\boldsymbol{v}) \cdot \left (\boldsymbol{u} +\boldsymbol{v}\right ) \\ =\left \Vert\boldsymbol{u} +\boldsymbol{v}\right \Vert ^{2}\end{align*}

This provides the well-known Triangle Inequality which implies on your way back home from school, you should not go anywhere else.

(30)   \begin{equation*}\left \Vert\boldsymbol{u} +\boldsymbol{v}\right \Vert \leq \left \Vert\boldsymbol{u}\right \Vert +\left \Vert\boldsymbol{v}\right \Vert \end{equation*}

Example 12.

Find a nonzero vector that is perpendicular to \left \langle \frac{1}{3} ,\frac{1}{33} ,\frac{1}{333}\right \rangle. Our job is to find a vector \left \langle - , - , -\right \rangle such that

(31)   \begin{equation*}\left \langle \frac{1}{3} ,\frac{1}{33} ,\frac{1}{333}\right \rangle \cdot \left \langle - , - , -\right \rangle =0 \end{equation*}

We cannot choose all entries zero by the condition of this problem. We cannot choose 2 entries zero, since we will be forced to choose the last one zero as well. But we can choose one of them zero, say the last one. Now, choose the first entry any number other than zero, say 1. We need to choose the second entry such that

(32)   \begin{equation*}(\frac{1}{3})(1) +\genfrac{(}{)}{}{}{1}{33}( -) =0 \end{equation*}

Solve and get -11. So

(33)   \begin{equation*}\left \langle \frac{1}{3} ,\frac{1}{33} ,\frac{1}{333}\right \rangle \cdot \left \langle 1 , -11 ,0\right \rangle =\frac{1}{3}\left (1\right ) +\frac{1}{33}\left ( -11\right ) +\frac{1}{333}(0) =0 \end{equation*}

Example 13.

Given \boldsymbol{u} =\left \langle a ,a^{2} ,1\right \rangle and \boldsymbol{v} =\left \langle 1 , -1 , -1\right \rangle find \boldsymbol{w} =\text{proj}_{\boldsymbol{v}}\boldsymbol{u} and \boldsymbol{z} =\text{proj}_{\boldsymbol{u}}\boldsymbol{v} . We can use the formula that we built here

(34)   \begin{align*}\boldsymbol{z} =\frac{\boldsymbol{v} \cdot\boldsymbol{u}}{\boldsymbol{u} \cdot\boldsymbol{u}}\boldsymbol{u} =\frac{\left \langle a ,a^{2} ,1\right \rangle \cdot \left \langle 1 , -1 , -1\right \rangle }{\left \langle a ,a^{2} ,1\right \rangle \cdot \left \langle a ,a^{2} ,1\right \rangle }\left \langle a ,a^{2} ,1\right \rangle \\ =\frac{a -a^{2} -1}{a^{2} +a^{4} +1}\left \langle a ,a^{2} ,1\right \rangle \end{align*}

(35)   \begin{align*}\boldsymbol{w} =\frac{\boldsymbol{u} \cdot\boldsymbol{v}}{\boldsymbol{v} \cdot\boldsymbol{v}}\boldsymbol{v} =\frac{\left \langle 1 , -1 , -1\right \rangle \cdot \left \langle a ,a^{2} ,1\right \rangle }{\left \langle 1 , -1 , -1\right \rangle \cdot \left \langle 1 , -1 , -1\right \rangle }\left \langle 1 , -1 , -1\right \rangle \\ =\frac{a -a^{2} -1}{3}\left \langle 1 , -1 , -1\right \rangle \end{align*}

Example 14.

Find a such that a triangle with vertices A =\left ( -1 ,2 ,3\right ), B\left (1 ,1 ,1\right ) and C\left (a , -2 , -2\right ) is a right triangle. We choose B to be the vertex where the angle of the triangle is 90^{ \circ } . Then the vectors AB and BC must be perpendicular to each other. Vector AB is \left \langle 1 +1 ,1 -2 ,1 -3\right \rangle =\left \langle 2 , -1 , -2\right \rangle and vector BC is \left \langle a -1 , -2 -1 , -2 -1\right \rangle =\left \langle a -1 , -3 , -3\right \rangle. Therefore

(36)   \begin{equation*}\left \langle 2 , -1 , -2\right \rangle \cdot \left \langle a -1 , -3 , -3\right \rangle =0 \end{equation*}

(37)   \begin{align*}2\left (a -1\right ) +3 +6 =0 \\ 2a -2 +3 +6 =2a +7 =0 \\ a = -\frac{7}{2}\end{align*}

See the second part of vector fundamentals.

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