In mathematics, there are several important topics that can help students to score good marks in their final exam. One such topic is the cross product of the same vector or the cross product of two parallel vectors. And if you have always wanted to learn more about these topics, then you are in the right place. We will also focus on concepts like right-hand rule cross product and the cross product is zero.
According to experts, cross product can be defined as the binary operation on two vectors in a three-dimensional space. This further results in a sector that is perpendicular to both of the vectors. It should be noted by students that the cross product of two vectors can be calculated by using the right-hand rule.
For students who don’t know what the right-hand rule is, it is defined as nothing but the resultant of any two vectors. These two vectors should be perpendicular to the other two vectors. By using the cross product, one can also find the magnitude of the final resulting vector.
Students should remember that the cross product of two vectors, which is also known as the vector product, A and B is denoted by A × B. also, the resultant vector will be perpendicular to both the vectors named A and B. You can also refer to the image that is attached below to get a good visual representation of this concept.
(Image will be uploaded soon)
When a student is working with vectors, there are certain key points that he or she should remember. We have created a list of those key points and that list is mentioned below.
The cross product of any two vectors will always result in a vector quantity.
In the concept of vector product, the resulting vector will contain a negative sign if the order of the vectors is changed by the student.
The direction of both A and B will always be perpendicular to the plan that contains A and B.
The null vector is the cross product of any two linear vectors.
There are several formulas that are related to the chapter on vectors. In this section, we will look at some of the most important vector formulas. Let’s start with the formula of the cross product. If we assume that θ is the angle that exists between any two given vectors, then the formula can be given by:
A × B = AB sin θ
The same formula can also be written as
A × B = ab sin θ n̂
Here, n̂ is the unit vector.
Students should also be familiar with the concept of direction of the cross product. It should be noted that the direction of the cross product of any two non zero parallel vectors, a and b, can be given by using the right-hand thumb rule.
To apply the right-hand rule, simply use your right hand and point your index finger along the vector a. After that, point your middle finger along the vector b. Finally, use the thumb to ascertain the direction of the cross product.
Let’s look at the cross product of two vectors. Students should remember that the cross product of two vectors can be indicated in the following manner:
X × Y = |X| . |Y| sin θ
Now, take any two vectors like × = xi + yj + z and Y = ai + bj + ck
We know that the cross product of these two vectors can be explained in the matrix form, which is also known as the determinant form. This expression is given below.
X × Y = i (yc - zb) - j (xc - za) + k (xb - ya)
After the cross product of two vectors, the next important topic is the triple cross product. As you might have guessed, the product of three vectors is also known as the triple product. It can also be explained as the cross product of a vector with the cross product of any other two vectors.
To arrive at the formula for the triple cross product, we must assume that there are three vectors, which are represented by A, B, and C. These three vectors can be denoted in the following manner.
A × (B × C) = (A . C) B - (A . B) C
(A × B) × C = -C × (A × B) = -(C . B) A + (C . A) B
The last major topic we need to look at is the cross product in spherical coordinates. Students should remember that the resultant vector of the cross product of any two vectors is perpendicular to both vectors. It is also normal to the plane in which the vectors are lying. The same can be represented by using spherical coordinates in a 3-dimensional system or space.
It is also possible to define a vector in a 3-dimensional system or space. This is done as the first radical distance, also known as r. It can also be explained as the distance of a fixed point to the origin, the second point to the polar angle θ, and the third point to the azimuth angle ϕ.
All this information can also be explained in a form of a formula as we know about the transformation from Cartesian to a spherical form. This representation is given below.
X = r sin θ cos ϕ
Y = r sin θ sin ϕ
Z = r cos θ
Do you know that cross product has several applications in different contexts? For example, the understanding of cross products is used in computational geometry, engineering, and physics. These are very interesting fields and applications that students can pursue further if they are interested in cross products and the field of science in general.
Question 1. It is Given that A = i + 2j + 3k, B = 2i + 3j - 2k, and C = i + j + k. Use this Information to Find the Solution to A × (B × C) if A × (B × C) = (A . C) B - (A . B) C.
Answer: A . C = (i + 2j + 3k) . (i + j + k) = 6
A . B = (i + 2j + 3k) . (2i + 3j - 2k) = 2
After that, A × (B × C) = 6 (2i + 3j - 2k) - 2 (i + j + k)
= 10i - 16j - 14k
Question 2. If a + b + c = 0, then Find out the Relation that is Correct from the Following Options:
a = b = c = 0
a . b = b . c = c . a
c × a = a × b = b × c
None of these
Answer: We know that,
a + b + c = 0
= a × (a + b + c) = 0
= a × a × a × b + a × c = 0
= a × b = a × c = c × a ----(1)
Similarly, b × (a + b + c) = 0
= b × c = a × b ----(2)
Now, if we use both (1) and (2), then we will get
a × b = b × c = c × a
Question 3. FInd the cross products of the following vectors.
a = [3, 4, 7] and b = [4, 9, 2]
Answer: We know that the given vectors are:
a = [3, 4, 7] and b = [4, 9, 2]
This means that the cross product is give by:
a × b = |i j k|
|a1 a2 a3|
|b1 b2 b3|
a × b = |i j k|
|3 4 7|
|4 9 2|
a × b = i (4 × 2 - 9 × 7) - j (3 × 2 - 4 × 7) + k (3 × 9 - 4 × 4)
a × b = i (8 - 63) - j (6 - 28) + k (27 - 16)
a × b = -55i + 22j + 11k