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Vector Product

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What is the multiplication of \(4\) by \(5\), it is \(20\), easy. Multiplication of numbers is pretty easy and everyone can do it but what if you were asked to multiply two vectors, \(\vec{a}=2\hat{i} + 3\hat{j}\) and \(\vec{b}=\hat{i} - 3\hat{k}\), well, it’s not that simple. Vectors are a very distinct entity, the classical rules of numeric multiplications are of not much use here. To get the product of two or more vectors, we need to understand the concept of a** Vector Product**.

The conventional multiplication of two numbers is well-defined at this point. But to multiply two vectors is not as straightforward as that. The product of two vectors can be carried out in two different ways; either by Scalar product or by Vector product. The scalar product of two vectors results in a scalar quantity, as the name suggests. Whereas the vector product of two vectors yields a vector, which is the reason it is named Vector product.

The vector product of two vectors gives a vector that is in the direction perpendicular to the plane formed by the two vectors. The magnitude of the resulting vector is the product of the magnitudes of the two original vectors and the sine of the angle between their directions.

Suppose, there are two vectors that form the xy pane between them. The vector product of these vectors will be perpendicular to the xy plane, i.e. the z-axis.

Let there be two vectors, \(\vec{A}\) and \(\vec{B}\), and \(\theta\) be the angle between them.

The vector product between them is denoted by \(\vec{A}\times\vec{B}\) (‘\(\times\)’ should not be mistaken as a multiplication sign, here it denotes the vector product), due to the cross symbol, vector product is also equivalently known as **Cross Product**. Now that the vectors and the angle between them are established, let us look at the definition once again, but in a mathematical manner:

The vector product of two non-zero vectors \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{A}\times\vec{B}=AB\space sin\theta\space\hat{n}\) where \(A\) and \(B\) are the magnitudes of the vectors and \(\theta\) is the acute angle between the two vectors. And \(\hat{n}\) represents a unit vector in the direction perpendicular to the plane formed by the two vectors.

It is important to specify that none of the vectors are null vectors, i.e. zero vectors. In that case, the vector product has no geometric significance. The vector product results in a vector that has magnitude \(AB\space sin\theta\) and pointing in the direction which is perpendicular to the plane formed by the two vectors, i.e. perpendicular to both the vectors simultaneously.

To demonstrate how this looks, consider the diagram below:

Suppose the vectors are oriented in such a way that they lie on the xy plane, i.e. the z component of the vectors is 0. If we take the vector product, it can be imagined as turning the screw in the direction of \(\vec{A}\) to \(\vec{B}\), which is the anti-clockwise direction. The direction in which the screw would proceed is vertically upwards, i.e. positive z-axis. The vertical arrow is precisely the vector formed by their vector product. In the diagram above, \(\vec{A}\) and \(\vec{B}\) have the same starting point, but this is not necessary. In fact, they can be any two vectors, it is only shown here to visually represent the concept.

Let us denote the resulting vector as \(\vec{C}\), then according to the definition:

$$\vec{C}=AB\space sin\theta\space\hat{n}$$

and the magnitude is given by simply taking the absolute value on both sides,

$$|\vec{C}|=AB\space sin\theta$$

The magnitude of the unit vector becomes \(1\) and all we are left with is \(AB\space sin\theta\). Remember that \(|\vec{C}|=C\), it is just a matter of convenience on which notation one prefers (the same goes for \(\vec{A}\) and \(\vec{B}\), or any other vector).

To visualize the direction in which the resulting vector will point, the right-hand screw rule is of great help. Simply turn the fingers in the rotating vector direction and observe the direction of the thumb. Rotate the fingers from \(\vec{A}\) towards \(\vec{B}\) and the direction of the thumb gives the direction of the resulting vector.

Until now, you have learned what a vector product means geometrically. Now, we shall see its algebraic form in more detail. To generalize the vector product:

Let A be a vector such that \(\vec{A}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}\) where \(a_{1},a_{2} \space and \space a_{3}\) are real-valued constants such that they are not simultaneously \(0\) (otherwise it would be a null vector) and \(\vec{B}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}\) , where \(b_{1},b_{2} \space and \space b_{3}\) are real-valued constants and are again not simultaneously \(0\). Here, \(\hat{i}, \hat{j} \space and \space \hat{k}\) represent the unit vectors in x-axis, y-axis and z-axis respectively. Both the vectors and the respective unit vectors are shown in the following diagram:

As we have studied before,

$$\vec{A}\times\vec{B}=AB \space sin\theta \space \hat{n}$$

Before we evaluate the vector component using the \(x, y \space and \space z\) components, we need to understand how the constituting unit vectors resolve. For instance, consider \(\hat{i}\times\hat{i}\), when evaluated,

$$\hat{i}\times\hat{i}=(1)(1) \space sin0^{o}=\vec{0}$$

Because \(sin0^{o}=0\) and the magnitude of all the unit vectors is \(1\). Similarly, \(\hat{j}\times\hat{j}=\vec{0}\) and \(\hat{k}\times\hat{k}=\vec{0}\) where \(\vec{0}\) represents the null vector. Evaluating the other relations, for instance,

$$\hat{i}\times\hat{j}=sin\left(\frac{\pi}{2}\right)\hat{k}=\hat{k}$$

Because the angle between the unit vector along the x-axis and the unit vector along the y-axis is \(\frac{\pi}{2}\) radians, which gives the sine value of \(1\). Similarly, we can calculate the other entities, which are as follows:

$$\hat{j}\times\hat{i}=-\hat{k}, \space \hat{i}\times\hat{k}=-\hat{j}, \space \hat{k}\times\hat{i}=\hat{j}, \space \hat{j}\times\hat{k}=\hat{i} \space and \space \hat{k}\times\hat{j}=-\hat{i}$$

Now, using the above entities for unit vectors, we can calculate \(\vec{A} \times \vec{B}\) as follows:

$$\vec{A}\times\vec{B}=(a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k})\times(b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k})$$

$$\vec{A}\times\vec{B}=(a_{2}b_{3}-b_{2}a_{3})\hat{i}-(a_{1}b_{3}-a_{3}b_{1})\hat{j}+(a_{1}b_{2}-a_{2}b_{1})\hat{k}$$

The cross products which resulted in a null vector are eliminated, and all we are left with is the above equation. The above equation can be a bit tedious to remember at times and not very convenient to write. As a result, the above equation is often represented by a \(3\) by \(3\) determinant, which is:

$$\vec{A}\times\vec{B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\b_{1} & b_{2} & b_{3} \end{vmatrix}$$

which eventually equates to the equation derived earlier. But the determinant is relatively easy to remember. To verbally express the elements of the determinant, the first row is made of the unit vectors, the second row and third rows are formed by the scalar components of the vectors. Let's look at an example to get a grip over calculating such determinants, eventually, the vector product.

Vector product is not restricted to two vectors, but can also be extended to three distinct vectors. Let there be three non-zero vectors \(\vec{A}, \space \vec{B} \space and \space \vec{C}\) and we want to evaluate the vector triple product between them: \(\vec{A} \times (\vec{B} \times \vec{C})\) where the order of the vector product is very important as \(\vec{A} \times (\vec{B} \times \vec{C}) \neq (\vec{A} \times \vec{B}) \times \vec{C}\) i.e. Vector triple product is not associative. There are essentially two ways by which the vector triple product can be calculated: Firstly by calculating the vector product of \(\vec{A}\) and \(\vec{B}\) and then doing its vector product with \(\vec{C}\). This involves carrying out two vector products, one after the another.

Another way of calculating the vector product is using the dot product:

$$ (\vec{A} \times \vec{B}) \times \vec{C}=(\vec{A} \cdot \vec{C}) \vec{B}-(\vec{B} \cdot \vec{C}) \vec{A} $$

As it can be seen, this formula involves the dot product of \(\vec{A}\) with \(\vec{C}\) and then \(\vec{B}\) with \(\vec{C}\). In practice, both the above methods are equally convenient.

For the vectors: \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=2 \hat{i}-\hat{k}\) and \(\vec{c}=\hat{j}-\hat{k}\), find the Vector Triple Product: \((\vec{a} \times \vec{b}) \times \vec{c}\)

**Solution: **

We shall use the following formula

$$ (\vec{a} \times \vec{b}) \times \vec{c}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a} $$

**Step 1:** Evaluating \(\vec{a} \cdot \vec{c}\) :

$$ \begin{aligned} \vec{a} \cdot \vec{c} &=(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{j}-\hat{k}) \\ &=0+1-1 \\ \therefore \vec{a} \cdot \vec{c} &=0 \end{aligned} $$

**Step 2:** Evaluating \(\vec{b} \cdot \vec{c}\) :

$$ \begin{aligned} \vec{b} \cdot \vec{c} &=(2 \hat{i}-\hat{k}) \cdot(\hat{j}-\hat{k}) \\ &=0+0+1 \\ &=1 \end{aligned} $$

**Step 3: **Substituting for \(\vec{a} \cdot \vec{c}\) and \(\vec{b} \cdot \vec{c}\) in \((\vec{a} \times \vec{b}) \times \vec{c}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}\):

$$ \begin{aligned} (\vec{a} \times \vec{b}) \times \vec{c} &=0-(1) \vec{a} \\ &=-\hat{i}-\hat{j}-\hat{k} \end{aligned} $$

Hence, \((\vec{a} \times \vec{b}) \times \vec{c}=-\hat{i}-\hat{j}-\hat{k}\)

Though Vector product and Scalar product are quite different in geometric application and their meaning, we still can relate them in a certain way. Recall that the Scalar product of two vectors is defined as

$$\vec{x}.\vec{y}=xy \space cos\theta$$

Squaring both sides, we get

$$(\vec{x}.\vec{y})^{2}=x^{2}y^{2} \space cos^{2}\theta$$

And the vector product between them has the magnitude

$$|\vec{x} \times \vec{y}|=xy \space sin\theta$$

Squaring both sides, we obtain

$$|\vec{x} \times \vec{y}|^{2}=x^{2}y^{2} \space sin^{2}\theta$$

Now adding both the equations involving the vector and scalar product:

$$(\vec{x}.\vec{y})^{2}+|\vec{x} \times \vec{y}|^{2}=x^{2}y^{2} \space (cos^{2}\theta+sin^{2}\theta)$$

and now substituting for the trigonometric identity \(cos^{2}\theta+sin^{2}\theta=1\), we get

$$(\vec{x}.\vec{y})^{2}+|\vec{x} \times \vec{y}|^{2}=x^{2}y^{2}$$

In words, the sum of squares of the magnitudes of the scalar and vector products is equal to the product of the squares of the magnitudes of those vectors. This identity is very useful as it does not involve the angle between the vectors explicitly.

Verify the identity relating dot and vector product: \((\vec{p} \cdot \vec{q})^{2}+|\vec{p} \times \vec{q}|^{2}=p^{2} q^{2}\) for the vectors \(\vec{p}=2 \hat{i}-\hat{k}\) and \(\vec{q}=\hat{j}+\hat{k}\).

**Solution:**

**Step 1: **

Let us evaluate \(\vec{p} \cdot \vec{q}\) first:

$$ \begin{aligned} \vec{p} \cdot \vec{q} &=(2 \hat{i}-\hat{k}) \cdot(\hat{j}+\hat{k}) \\ &=0+0-1 \\ \therefore \vec{p} \cdot \vec{q} &=1 \end{aligned} $$

**Step 2: **

Evaluate \(\vec{p} \times \vec{q}\) :

$$ \begin{aligned} \vec{p} \times \vec{q} &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -1 \\ 0 & 1 & 1 \end{array}\right| \\ \therefore \vec{p} \times \vec{q} &=\hat{i}(0+1)-\hat{j}(2+0)+\hat{k}(2-0) \\ \therefore \vec{p} \times \vec{q} &=\hat{i}-2 \hat{j}+2 \hat{k} \end{aligned} $$

**Step 3: **

Evaluate \(p^{2} q^{2}\) :

$$ \begin{aligned} p=|\vec{p}| &=\sqrt{2^{2}+(-1)^{2}} \\ \therefore p &=\sqrt{5} \end{aligned} $$

And for the magnitude of \(q\) :

$$ \begin{aligned} q=|\vec{q}| &=\sqrt{1^{2}+1^{2}} \\ \therefore q &=\sqrt{2} \end{aligned} $$

Hence, \(p^{2} q^{2}=5 \times 2=10\)

**Ste****p 4:**

Let us verify \( \text{LHS} =(\vec{p} \cdot q)^{2}+|\vec{p} \times q|^{2}\) :

$$ \begin{aligned} \text { LHS } &=1^{2}+\left(1^{2}+(-2)^{2}+2^{2}\right) \\ &=1+1+4+4 \\ \therefore \text{LHS} &=10 \end{aligned} $$

And for \( \text{RHS} =p^{2} q^{2}\), we have already calculated it in **step 3** :

$$ \text { RHS }=10 $$

It can be easily seen that

$$ \text{LHS}= \text{RHS} $$

Thus, the identity has been verified.

**Vector Dot Product, **also known simply as **Dot Product, **is another way of multiplying two vectors apart from Vector product. An alternative but equally used name for **Dot product **is **Scalar Product. **

The dot product between two vectors results in a scalar quantity, and not in another vector. The main reason dot product exists is, in order to measure a vector’s magnitude along the direction of another vector.

Let \(\vec{a}\) and \(\vec{b}\) be two non-zero vectors making an acute angle \(\theta\) between them. The formula for the **Dot Product **is given as follows:

$$\vec{a} \cdot \vec{b}=ab \space cos\theta$$

It can be seen that the result of the dot product is a scalar quantity.

Find the dot product of the vectors \(\vec{p}=4\hat{i}+3\hat{j}\) and \(\vec{p}=6\hat{i}-8\hat{j}\) where the acute angle between the vectors is \(\pi/6\).

**Solution:**

Using the formula for dot product:

$$\vec{p} \cdot \vec{q}=pq \space cos\theta$$

Substituting for the appropriate values:

$$\vec{p} \cdot \vec{q}=(\sqrt{4^{2}+3^{2}})(\sqrt{6^{2}+(-8)^{2}}) \space cos(\pi/6)$$

$$\vec{p} \cdot \vec{q}=(5)(10) \space \left( \frac{1}{2} \right)$$

$$\vec{p} \cdot \vec{q}=25$$

Hence, the dot product of the two vectors \(\vec{p}=4\hat{i}+3\hat{j}\) and \(\vec{p}=6\hat{i}-8\hat{j}\) is \(25\) units.

Calculate the vector product of the vectors \(\vec{C}=2\hat{i}+3\hat{j}-\hat{k}\) and \(\vec{D}=-\hat{i}+2\hat{k}\).

**Solution:**

Placing the unit vectors and the components in the determinant form as mentioned earlier, we get

$$\vec{C}\times\vec{D}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\-1 & 0 & 2 \end{vmatrix}$$

Upon expanding the determinant, we get

$$\vec{C} \times \vec{D}=6\hat{i}-3\hat{j}+3\hat{k}$$

which is the resultant vector obtained by the vector product of \(\vec{C}\) and \(\vec{D}\).

For two vectors \(\vec{p}=\hat{i}-2\hat{j}\) and \(\vec{q}=-2\hat{i}+\lambda\hat{j}\) such that their vector product is \(\vec{p} \times \vec{q}=2\hat{k}\). Calculate the value of \(\lambda\).

**Solution:**

Using the earlier derived formula to calculate the vector product of the two vectors, we have

$$\vec{p} \times \vec{q}=\lambda\hat{k}-4\hat{k}=(\lambda-4)\hat{k}$$

But it is also given that the vector product is \(\vec{p} \times \vec{q}=2\hat{k}\), so comparing the coefficient of the unit vector, we get

$$\lambda-4=2$$

which yields \(\lambda=6\).

- The vector product of two vectors is a vector pointing in the direction perpendicular to the plane formed by the two vectors.
- For any two non-zero vectors, \(\vec{x}\) and \(\vec{y}\) making an acute angle \(\theta\) between them, their vector product is given by \(|\vec{x} \times \vec{y}|=xy \space sin\theta \space \hat{n}\).
- The direction of the resulting vector can be obtained using the right-hand screw rule, where the fingers signify the direction of the vectors and the thumb points in the resulting vector.
- The vector product is neither associative nor commutative as opposed to the scalar product which is commutative.
- The sum of the squares of the vector and scalar products is equal to the square of the product of the magnitudes of the vectors, i.e. \((\vec{x}.\vec{y})^{2}+|\vec{x} \times \vec{y}|^{2}=x^{2}y^{2}\).

More about Vector Product

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