1.2. Dot Product#
In this section we will consider other (geometric) properties of vectors, like the length of a vector and the angle between two vectors. When the angle between two vectors is equal to \(\frac12\pi\), two vectors are perpendicular, which is also known as orthogonal. These properties can all be expressed using a new operator: the inner product or dot product.
We will start by considering vectors in \(\mathbb{R}^2\) and \(\mathbb{R}^3\). The translation of the concepts to the general space \(\mathbb{R}^n\) will then become more or less immediate.
1.2.1. Length and perpendicularity in \(\mathbb{R}^2\) and \(\mathbb{R}^3\)#
The length of a vector
in the plane, which we denote by \(\norm{\mathbf{v}}\), can be computed using the Pythagorean theorem:
Using this theorem twice we find a similar formula for the length of a vector
in \(\mathbb{R}^3\). Look at Figure 1.2.2. There are two right triangles: \(\Delta OPQ\) where \(\angle OPQ\) is right, and \(\Delta OQA\) where \(\angle OQA\) is right.
From
where for two points \(A\) and \(B\), by \(AB\) we denote the length of the vector \(\overrightarrow{AB}\), and
we find that
Let us now turn our attention to another important geometric concept, namely that of
perpendicularity. It is clear from Figure 1.2.3 that the vectors \(\begin{bmatrix}2\\3\end{bmatrix}\) and \(\begin{bmatrix}-3\\2\end{bmatrix}\) are perpendicular, whereas the vectors \(\begin{bmatrix}2\\3\end{bmatrix}\) and \(\begin{bmatrix}-1\\3\end{bmatrix}\) are not.
There is another way to look at this, which will be useful for the definition of perpendicularity in higher dimensions. To that end, consider Figure 1.2.4. Here you see two vectors \(\vect{v}\) and \(\vect{w}\) and the parallelogram they span. You also see the diagonals of this parallelogram, which are given by \(\vect{v}+\vect{w}\) and \(\vect{v}-\vect{w}\). Two vectors are perpendicular if and only if the parallelogram they span is a rectangle, and this is exacty the situation where the diagonals have the same length, i.e.,
In the picture on the right the vectors are not perpendicular and
So far we have been talking about two (nonzero) vectors in the plane, i.e., in \(\mathbb{R}^2\). However, two vectors in \(\mathbb{R}^3\) form a parallelogram as well, which also becomes a rectangle if and only if the vectors are perpendicular. We introduce a notation for this: if \( \mathbf{v}\) and \(\mathbf{w}\) are perpendicular, we write this as
Taking squares in Equation (1.2.3), we see that the following holds both in \(\mathbb{R}^2\) and in \(\mathbb{R}^3\):
If we write this out for two arbitrary vectors \(\mathbf{v}=\begin{bmatrix} a_{1}\\a_{2}\end{bmatrix},\mathbf{w}=\begin{bmatrix} b_{1}\\b_{2}\end{bmatrix}\) in \(\mathbb{R}^2\) we get the following:
Likewise, for vectors \(\mathbf{v}=\begin{bmatrix} a_{1}\\a_{2}\\a_{3}\end{bmatrix},\,\mathbf{w}=\begin{bmatrix} b_{1}\\b_{2}\\b_{3}\end{bmatrix}\) in \(\mathbb{R}^3\):
The derivation is completely analogous to the one above, only now we have one extra term. So to check ‘algebraically’ whether two vectors are perpendicular we just have to compute \(a_1b_1 +a_2b_2\, (\,+\,a_3b_3\,)\) and see whether this is equal to 0.
This expression is called the dot product (or inner product) of the vectors \(\mathbf{v}\) and \(\mathbf{w}\). We denote it by \(\mathbf{v}\ip\mathbf{w}\). Note that the dot product of a general vector \(\mathbf{v}=\begin{bmatrix} a_{1}\\a_{2}\\a_{3}\end{bmatrix}\) in \(\mathbb{R}^3\) with itself gives
so the length of a vector can be expressed as follows using the dot product
Using the dot product the concepts length and perpendicular easily carry over to any \(\mathbb{R}^n\), \(n \geq 4\). Let’s do it one by one, starting by generalizing the dot product in the next subsection.
1.2.2. Dot product in \(\mathbb{R}^n\)#
The dot product (or inner product) of two vectors \(\mathbf{v}=\begin{bmatrix}a_{1}\\a_{2}\\ \vdots\\a_{n}\end{bmatrix}\) and \(\mathbf{w}=\begin{bmatrix}b_{1}\\b_{2}\\ \vdots\\b_{n}\end{bmatrix}\) in \(\mathbb{R}^n\) is defined as
The dot product of the two vectors
is given by
And the dot product of the two vectors
is not defined. In fact, the dot product of a vector \(\mathbf{v}\) in \(\mathbb{R}^m\) and a vector \(\mathbf{w}\) in \(\mathbb{R}^n\) is only defined if \(m = n\).
We state the characteristic rules of the dot product in \(\mathbb{R}^n\), which in the sequel we will use time and again, in the following proposition.
The following properties hold for any vectors \(\mathbf{v},\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\) in \(\mathbb{R}^n\) and scalars \(c \in \mathbb{R}\):
i. \(\mathbf{v}_1\ip\mathbf{v}_2 = \mathbf{v}_2\ip\mathbf{v}_1\).
ii. \((c\mathbf{v}_1)\ip\mathbf{v}_2 = c(\mathbf{v}_1\ip\mathbf{v}_2) = \mathbf{v}_1\ip(c \mathbf{v}_2)\).
iii. \((\mathbf{v}_1+\mathbf{v}_2)\ip\mathbf{v}_3 = \mathbf{v}_1\ip\mathbf{v}_3+\mathbf{v}_2\ip\mathbf{v}_3\).
iv. \(\mathbf{v}\ip\mathbf{v} \geq 0\), and \(\mathbf{v}\ip\mathbf{v} = 0 \iff \mathbf{v} = \mathbf{0}\).
Proof of Proposition 1.2.1
The first three properties follow from the corresponding properties of real numbers. For instance, for the first rule we simply use that \(ab = ba\) holds for the product of real numbers \(a\) and \(b\).
i. Let
be two arbitrary vectors in \(\mathbb{R}^n\). Then
ii. For two vectors \(\vect{v}_1 = \begin{bmatrix}a_{1} \\ a_{2}\\ \vdots\\ a_{n}\end{bmatrix}\), \(\vect{v}_2 = \begin{bmatrix}b_{1} \\ b_{2}\\ \vdots\\ b_{n}\end{bmatrix}\)
and constants \(c\) we see that
iii. Is proved in the same way as (ii).
iv. This consists of two statements. For the first, we note that
is the sum of squares of real numbers, so it is nonnegative. That is,
To prove the second statement, we see that
if and only if all the squares are 0, which only happens if each entry \(a_i\) is equal to zero, that is, if \(\mathbf{v} = \mathbf{0}\).
Prove property iii.
Solution to Exercise 1.2.1
Let
be three arbitrary vectors in \(\mathbb{R}^n\). Then
Prove the identity
Solution to Exercise 1.2.2
First of all, because of rule i. and rule iii. of Proposition 1.2.1 it holds that
and it also follows from ii. and iii. that
Then the statement is proved by the following chain of identities
Prove the identity
and explain why it is called the parallelogram rule.
Solution to Exercise 1.2.3
Again it’s a chain of identities using basic properties of the dot product.
1.2.3. Orthogonality#
In \(\mathbb{R}^2\) and \(\mathbb{R}^3\) the dot product gives an easy way to check whether two vectors are perpendicular:
We use this identity to define the concept of perpendicularity in \(\mathbb{R}^n\). It seems a bit ‘academic’, but in this more general setting the term orthogonal is used.
Two vectors \(\mathbf{v}\) and \(\mathbf{w}\) in \(\mathbb{R}^n\) are called orthogonal if \(\mathbf{v}\ip\mathbf{w} = 0\). As before, we denote this by \(\mathbf{v}\perp\mathbf{w}\).
Let \(\mathbf{u} = \begin{bmatrix} 1\\2\\-1\\-1\end{bmatrix}\), \(\mathbf{v} = \begin{bmatrix} 3\\-1\\2\\-1\end{bmatrix}\) and \(\mathbf{w} = \begin{bmatrix} 2\\2\\-1\\2\end{bmatrix}\).
We compute
and conclude that \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal, \(\mathbf{u}\) and \(\mathbf{w}\) are not orthogonal,
\(\mathbf{v}\) and \(\mathbf{w}\) are orthogonal.
To compute some dot products in \(\R^2, \R^3, \R^4\).
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In \(\mathbb{R}^2\), two nonzero vectors that are orthogonal to the same nonzero vector \(\mathbf{v}\) are automatically multiples of each other (i.e. have either the same or the opposite direction). In \(\mathbb{R}^n\) with \(n \geq 3\) this no longer holds. In the previous example both vectors \(\mathbf{u}\) and \(\mathbf{w}\) are orthogonal to the vector \(\mathbf{v}\), but \(\mathbf{u} \neq c\mathbf{w}\).
By definition the zero vector is orthogonal to any vector, since \(\mathbf{0}\ip\mathbf{v} = 0\). Moreover, the zero vector is the only vector that is orthogonal to itself, which is the content of the next proposition.
Suppose \(\mathbf{v} \in \mathbb{R}^n\). Then \(\mathbf{v}\perp\mathbf{v} \iff \mathbf{v} = \mathbf{0}\).
Proof of Proposition 1.2.2
By definition
In Proposition 1.2.1 iv. it was stated that the last equality only holds for \(\mathbf{v} = \mathbf{0}\).
The fact that the zero vector is orthogonal to any vector is an immediate consequence of the definition, but it may seem counterintuitive to you. The following example illustrates a situation where this orthogonality leads to a much nicer outcome.
Let \(\mathbf{n}\) be any nonzero vector in the plane. The set of vectors that are orthogonal to \(\mathbf{n}\) all lie on a line \(\mathcal{L}\) through the origin. (See Figure 1.2.5.) If we agree that \(\mathbf{0}\perp\mathbf{n}\), it will be the whole line. The vector \(\mathbf{n}\) is often said to be a normal vector to the line.
We conclude this subsection with another concept that we will come across later in a much more general context. Informally, it is the (orthogonal) projection of a vector onto another vector. More precisely, it is the orthogonal projection of a vector \(\mathbf{w}\) onto the line \(\mathcal{L}\) generated by the nonzero vector \(\mathbf{v}\), by which we mean \(\mathcal{L}= \{ c\mathbf{v}: c \in \mathbb{R}\}\).
See Figure 1.2.6.
The orthogonal projection of a vector \(\mathbf{w}\) onto the nonzero vector \(\mathbf{v}\) is the vector \(\mathbf{\hat{w}} = c\mathbf{v} \) for which
Another notation for this vector is
In the definition above the vector \(\mathbf{\hat{w}}\) with these properties is unique and it is given by
Proof of Proposition 1.2.3
With the rules of the dot product the vector \(\mathbf{w}\) is easily constructed.
Starting from
and
it follows that we must have
So \(c\) is uniquely given by
and indeed \(\mathbf{\hat{w}}\) must be as stated.
We compute the orthogonal projection of the vector
onto the vector
We proceed as follows
We verify the orthogonality:
so indeed
as required.
Computing the projection of a vector \(\vect{w}\) onto a vector \(\vect{v}\).
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Suppose \(\text{proj}_{\mathbf{v}}(\mathbf{w}_1) = \text{proj}_{\mathbf{v}}(\mathbf{w}_2) \), for three nonzero vectors \(\mathbf{v}, \,\mathbf{w}_1,\,\mathbf{w}_2\) in \(\mathbb{R}^n\). What does this say about the relative positions of the three vectors?
Verify your statement for the following three vectors
Solution to Exercise 1.2.4
Suppose \(\text{proj}_{\mathbf{v}}(\mathbf{w}_1) = \text{proj}_{\mathbf{v}}(\mathbf{w}_2) \). Thus \(\dfrac{\mathbf{w}_1\ip\mathbf{v}}{\mathbf{v}\ip\mathbf{v}} \mathbf{v} = \dfrac{\mathbf{w}_2\ip\mathbf{v}}{\mathbf{v}\ip\mathbf{v}} \mathbf{v}\).
Since \(\mathbf{v}\) is not the zero vector this implies that \(\mathbf{w}_1\ip\mathbf{v} = \mathbf{w}_2\ip\mathbf{v}\). In other words,
which expresses that \((\mathbf{w}_1 - \mathbf{w}_2)\perp \vect{v}\).
For the given vectors \(\mathbf{v}, \mathbf{w}_1, \mathbf{w}_2\) we find
and
We see \((\mathbf{w}_1 - \mathbf{w}_2)\ip \mathbf{v} = 1 - 2 + 10 + 9 = 0\), so indeed \((\mathbf{w}_1 - \mathbf{w}_2)\) and \(\vect{v}\) are orthogonal.
Figure 1.2.7 shows what is going on.
1.2.4. Norm in \(\mathbb{R}^n\)#
The length of a vector in the plane can be computed using the dot product: for \(\mathbf{v}=\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}\) in \(\mathbb{R}^2\) we have seen that
The identity \(\norm{\mathbf{v}} = \sqrt{\mathbf{v}\ip\mathbf{v}}\) also holds in \(\mathbb{R}^3\).
It seems natural to extend the concept to \(\mathbb{R}^n\). Again, for this more general space a new word is introduced.
The norm of a vector \(\mathbf{v}\) in \(\mathbb{R}^n\), denoted by \(\norm{\mathbf{v}}\), is defined by
Expressed in the entries of \(\mathbf{v}\) this yields
so for vectors in \(\mathbb{R}^2\) and \(\mathbb{R}^3\) the norm of a vector is just the length of the vector.
As we might expect the norm has many properties in common with length.
For any \(\mathbf{v}, \,\mathbf{w} \in \mathbb{R}^{n}\) and all \(c \in \mathbb{R}\) the following holds:
i. \(\norm{\mathbf{v}}\geq 0\), and \(\norm{\mathbf{v}} = 0\) only for \(\mathbf{v}=\mathbf{0}\);
ii. Scaling property:
iii. Triangle Inequality:
The first two of these properties are very easy to prove. The proof of the triangle inequality we postpone until the end of the section. Figure 1.2.8 explains the name.
We compute the norms of the vectors
We find
and
The last norm can also be found via
The distance between two vectors in \(\R^n\) is defined by
For the vectors \(\vect{u} = \begin{bmatrix}1 \\ 3 \\ 2 \\ 4 \end{bmatrix}\) and \(\vect{v} = \begin{bmatrix}5 \\ 1 \\ 3 \\ 4 \end{bmatrix}\) in \(\R^4\)
the distance is given by
Computing the distance between two vectors in \(\R^3\).
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From the rules of the norm the following rules of the distance function can be deduced.
For any three vectors \(\mathbf{u}, \mathbf{v}\) and \(\mathbf{w} \in \mathbb{R}^{n}\) the following statements hold.
i. \(\text{dist}(\vect{u},\vect{v}) = \text{dist}(\vect{v},\vect{u})\);
ii. \(\text{dist}(\vect{u},\vect{v}) = 0 \iff \vect{u}=\vect{v}\);
iii. \(\text{dist}(\vect{u},\vect{w}) \leq \text{dist}(\vect{u},\vect{v}) + \text{dist}(\vect{v},\vect{w})\).
Rule iii. is again called the Triangle Inequality.
Check the three properties of the distance function as stated in Proposition 1.2.5. For Rule iii., only show how it follows from the corresponding Rule iii. in that same proposition.
With the tools so far we can define a notion that comes in handy later.
A unit vector is a vector of norm 1.
Moreover, for any nonzero vector \(\mathbf{v}\), the vector
is called the unit vector in the direction of \(\mathbf{v}\).
For a nonzero vector \(\mathbf{v}\)
is the unique vector \(\mathbf{u}\) of norm 1 such that
Proof of Proposition 1.2.6
Assume that \(\mathbf{v} \neq \mathbf{0}\). For \(\mathbf{u} = k\mathbf{v}\), with \(\norm{\mathbf{u}} = 1\) and \(k > 0\) to hold, we must have
We see that
and consequently
We compute the unit vector \(\mathbf{u}\) in the direction of the vector \(\mathbf{v} = \begin{bmatrix}1 \\ 2 \\ 4 \\ -2 \end{bmatrix}\) in \(\mathbb{R}^4\).
As follows:
so
Interestingly, Pythagoras’ theorem also holds in \(\mathbb{R}^n\).
For any two vectors \(\mathbf{v}\) and \(\mathbf{w}\) in \(\mathbb{R}^n\) we have
Proof of Theorem 1.2.1
This follows quite straightforwardly from the properties of the dot product.
Let us start from the identity on the left and work our way to the conclusion on the right, making sure that each step is reversible. Note that from the definition of the norm it follows immediately that \(\norm{\mathbf{v}}^2 = \mathbf{v}\ip\mathbf{v}\).
Next we subtract \(\mathbf{v}\ip\mathbf{v} + \mathbf{w}\ip\mathbf{w}\) from both sides. Thus the last identity is equivalent to
And then we are almost there:
We verify the equality for the vectors \(\mathbf{v} = \begin{bmatrix} 2 \\ -3\\ 3 \\ 1 \end{bmatrix}\) and \(\mathbf{w} = \begin{bmatrix} 2 \\ 4 \\ 1 \\ 5 \end{bmatrix}\) in \(\mathbb{R}^4\).
First of all
so \(\mathbf{v}\perp \mathbf{w}\), and second
Furthermore
and we see that indeed
One of the most basic properties, also one with a wide range of applications, is the so-called Cauchy-Schwarz Inequality.
(Cauchy-Schwarz Inequality)
For any two vectors in \(\mathbb{R}^n\)
Proof of Theorem 1.2.2 (Cauchy-Schwarz Inequality)
There are many ways to prove the Cauchy-Schwarz inequality. There is even a whole book devoted to it: “Cauchy Schwarz master class” by J.M. Steele.
The following proof is based on orthogonal projection and Pythagoras’ Theorem.
If \(\mathbf{v} = \mathbf{0}\), the zero vector, then the inequality obviously holds; in fact it becomes an equality:
and also
So now suppose \(\mathbf{v} \neq \mathbf{0}\).
Let
be the projection of \(\mathbf{w}\) onto \(\mathbf{v}\). Then we can apply Pythagoras’ Theorem!
It follows that
Substitution of the expression for \(\mathbf{\hat{w}}\) leads to
Using
we deduce that
Taking square roots we may conclude that indeed
We verify that the inequality holds for the vectors \(\mathbf{v} = \begin{bmatrix} 1 \\ -2\\ 3 \\ -4 \end{bmatrix}\) and \(\mathbf{w} = \begin{bmatrix} -5 \\ 4 \\-3 \\ 0 \end{bmatrix}\) in \(\mathbb{R}^4\).
As follows
and we see that indeed
With this inequality established, the Triangle Inequality (1.2.9) is easily proved. Let’s repeat it, and prove it.
For any two vectors in \(\mathbb{R}^n\),
Proof of Theorem 1.2.3
Since all terms involved are non-negative we may as well show that the inequality holds for the squares:
and this, apart from the factor 2, is the Cauchy-Schwarz Inequality.
We verify the inequality for the vectors \(\mathbf{v} = \begin{bmatrix} -1 \\ 2\\ 3 \end{bmatrix}\) and \(\mathbf{w} = \begin{bmatrix} 4 \\ -4\\ 3 \end{bmatrix}\):
and indeed
1.2.5. Angles in \(\mathbb{R}^n\)#
The first motivation to consider the dot product came from the question of perpendicularity of two vectors in the plane or in \(\R^3\). Perpendicularity of two vectors means that the angle between them is equal to \(\frac12\pi\). Below we will show that it is possible to express the angle between any two (nonzero) vectors into dot products. And use this to define the concept of angle in a general space \(\R^n\).
First we will show a geometrical characterization of the dot product that holds in \(\mathbb{R}^2\) as well as in \(\mathbb{R}^3\).
For two nonzero vectors \(\mathbf{v}\) and \(\mathbf{w}\) in either \(\mathbb{R}^2\) or \(\mathbb{R}^3\) the following identity holds:
where \(\varphi\) is the angle between \(\mathbf{v}\) and \(\mathbf{w}\).
Note that this is in line with the special case of two perpendicular vectors:
The angle between two nonzero vectors \(\mathbf{v}\) and \(\mathbf{w}\) is thus determined by dot products in the following way
The value of \(\varphi\) between \(0\) and \(\pi\) is then uniquely determined by
Proof of Proposition 1.2.7
We will derive formula (1.2.10). Assume that \(\mathbf{v}\) and \(\mathbf{w}\) are nonzero vectors. Recall the formula of the orthogonal projection of \(\mathbf{w}\) onto \(\mathbf{v}\),
Let \(\varphi \in[0,\pi]\) denote the angle between two nonzero vectors \(\mathbf{v}\) and \(\mathbf{w}\).
From Figure 1.2.10 it is clear that the factor
is positive if the angle is acute, zero if the angle is right, and negative if the angle is obtuse.
In the case of an acute angle, by considering the right triangle \(\Delta OAB\), where \(A\) is the end point of \(\mathbf{\hat{w}}\) and \(B\) is the end point of \(\mathbf{w}\), we see that on the one hand
and on the other hand
So we may conclude that
In the case of an obtuse angle, we use that the projection of \(\mathbf{w}\) onto \(\mathbf{v}\) is equal to the projection of \(\mathbf{w}\) onto \(-\mathbf{v}\), as it is in fact the projection onto the line consisting of all multiples of \(\mathbf{v}\). Now look at the picture on the right of Figure 1.2.10 . There you see that \(\mathbf{w}\) and \(-\mathbf{v}\) make an acute angle \(\psi = \pi - \varphi\), so we can apply Equation (1.2.11) to \(\mathbf{w}\) and \(-\mathbf{v}\):
Note that the absolute value of the expression
is the length of the orthogonal projection of \(\vect{w}\) onto \(\vect{v}\).
In a methane molecule \(\ce{CH_4}\) the four \(\ce{H}\)-atoms are positioned in a perfectly symmetrical way around the \(\ce{C}\)-atom. We can model this as follows: put the \(\ce{C}\)-atom at the origin of \(\mathbb{R}^3\), and the \(\ce{H}\)-atoms at the positions/vectors
Then all four points have the same distance \(\sqrt{3}\) to the origin, and all points have the same distance to each other, namely
The angle between, for instance, \(\mathbf{v}_1\) and \(\mathbf{v}_3\) is determined by
So
Since we have defined the dot product and the norm in \(\mathbb{R}^n\), we can use the last formula to also define the angle between two vectors in \(\mathbb{R}^n\).
For two nonzero vectors \(\mathbf{v}\) and \(\mathbf{w}\) in \(\mathbb{R}^n\), the angle between the vectors is defined as
This definition makes sense, since the Cauchy-Schwarz inequality (Theorem 1.2.2) implies
Note that just as before in the plane and in three-dimensional space, for nonzero vectors \(\mathbf{v}\) and \(\mathbf{w}\) we have
Let \(\mathbf{e_1}\) be the vector in \(\mathbb{R}^n\) with first entry equal to 1 and all other entries equal to 0, and \(\mathbf{v}\) be the vector with all entries equal to 1. We find the angle between \(\mathbf{e}_1\) and \(\mathbf{v}\) in all cases \(n = 2, 3, 4,\ldots\)
For each \(n\geq2\) we write \(\varphi_n = \angle(\mathbf{e}_1,\mathbf{v})\). Then
So:
For \(n=1\) we find \(\cos(\varphi_1) = 1\), so \(\varphi_1 = 0\), which makes sense, and for \(n=2\), \(\cos(\varphi_2) = \frac{1}{\sqrt{2}}\), so \(\varphi_2 = \frac14\pi\), which you can check by a sketch in the plane.
For \(n\geq3\) we don’t get easy answers, but as \(\frac{1}{\sqrt{n}} \downarrow 0\) when \(n\) gets large, we may conclude that for large \(n\) in \(\mathbb{R}^n\) the two vectors are ‘almost’ orthogonal.
1.2.6. Grasple Exercises#
To compute dot products in \(\R^2\), \(\R^3\) and \(\R^4\).
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To find a vector orthogonal to a given vector in \(\R^2\).
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To find a vector orthogonal to two given vectors in \(\R^2\).
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To find a vector orthogonal to a given vector in \(\R^3\).
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To find a vector orthogonal to a given vector in \(\R^4\).
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To compute the norms of vectors in \(\R^2\), \(\R^3\), \(\R^4\).
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To find the norm of the ‘all one’ vector in \(\mathbb{R}^n\).
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To find the distance between two vectors in \(\mathbb{R}^4\).
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Find \(h\) such that the distance between two points has a given value \(d\).
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To find a unit vector on a given line through \((0,0)\).
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Find \(\text{proj}_{\mathbf{v}}(\mathbf{w})\) in \(\R^2\).
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Find \(\text{proj}_{\mathbf{v}}(\mathbf{w})\) in \(\R^4\).
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Regarding norm and orthogonality of \(\vect{u}\), \(\vect{v}\), \(\vect{u}-\vect{v}\) and \(\vect{u}+\vect{v}\).
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To show that \((\vect{v}+\vect{w})\ip(\vect{v}-\vect{w}) = \norm{\vect{v}}^2 - \norm{\vect{w}}^2\).
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To prove that \((\vect{v}+\vect{w})\ip\vect{x} = \vect{v}\ip\vect{x}+\vect{w}\ip\vect{x}\).
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What to conclude from \(\norm{\vect{v}+\vect{w}} = \norm{\vect{v}}+\norm{\vect{w}}\)?
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To prove that \(-1 \leq \dfrac{\vect{u}\ip\vect{v}}{\norm{\vect{u}} \norm{\vect{v}}} \leq 1\).
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What to conclude from \(\text{proj}_{\mathbf{v}}(\mathbf{w}_1 ) = \text{proj}_{\mathbf{v}}(\mathbf{w}\_2)\)?