How To Find Parallel Vectors

In vector algebra, different types of vectors are defined and various operations can be performed on these vectors such as improver, subtraction, production and and so on. In this commodity, the cantankerous product of two vectors, formulas, properties, and examples is explained.
Table of Contents:

Definition
Cross production of two vectors
Formula
Cross product matrix
Right hand rule
Properties
Perpendicular vectors
Parallel vectors
Magnitude
Example

What is a Cross Product?

Cantankerous product is a binary performance on two vectors in three-dimensional infinite. It results in a vector that is perpendicular to both vectors. The Vector product of 2 vectors, a and b, is denoted by a × b. Its resultant vector is perpendicular to a and b. Vector products are as well called cross products. Cross production of ii vectors will give the resultant a vector and calculated using the Correct-hand Dominion.

Cross Product of Two Vectors

The vector product or cross product of two vectors A and B is denoted by A × B, and its resultant vector is perpendicular to the vectors A and B. The cross product is mostly used to determine the vector, which is perpendicular to the plane surface spanned by ii vectors, whereas the dot product is used to discover the angle between ii vectors or the length of the vector. The cross product of two vectors, say A × B, is equal to another vector at right angles to both, and information technology happens in the three dimensions.

Cross Product of Two Vectors

Cross Product Formula

If θ is the angle betwixt the given two vectors A and B, then the formula for the cross product of vectors is given past:

A × B =|A| |B| sin θOr,

\(\begin{array}{l}\vec{A}\times \vec{B}=||\vec{A}|| \ ||\vec{B}|| sin\theta \ \hat{n} \end{array} \)

Here,

\(\begin{array}{50}\vec{A},\vec{B} \ are\ the\ two\ vectors.\end{array} \)

\(\begin{assortment}{50}||\vec{A}||, \ ||\vec{B}|| \ are\ the\ magnitudes\ of\ given\ vectors.\end{array} \)

θ is the angle between two vectors

\(\begin{array}{50}\hat{due north} \ \text {is the unit of measurement vector perpendicular to the airplane containing the given ii vectors, in the direction given by the right-hand rule.}\end{array} \)

Cross product of two vectors Formula

Consider two vectors,

A = ai + bj + ck

B = xi + yj + zk

We know that the standard footing vectors i, j, and k satisfy the below-given equalities.

i × j = thousand and j × i = -one thousand j × k = i and k × j = -i thou × i = j and i × k = -j

Also, the anti-commutativity of the cross product and the distinct absence of linear independence of these vectors signifies that:

i × i = j × j = chiliad × k = 0

Now,

A × B = (ai + bj + ck) × (11 + yj + zk)

= ax(i × i) + ay(i × j) + az(i × g) + bx(j × i) + by(j × j) + bz(j × k) + cx(one thousand × i) + cy(k × j) + cz(k × k)

Past applying the above-mentioned equalities,

A × B = ax(0) + ay(k) + az(-j) + bx(-k) + by(0) + bz(i) + cx(j) + cy(-i) + cz(0) = (bz – cy)i + (cx – az)j + (ay – bx)k

Cantankerous Product Matrix

We can also derive the formula for the cross product of 2 vectors using the determinant of the matrix as given beneath.

A = ai + bj + ck B = xi + yj + zk

Thus,

\(\begin{array}{l}\mathbf{A}\times \mathbf{B} = \brainstorm{vmatrix} \boldsymbol{i} & \mathbf{j} & \mathbf{g}\\ a & b & c\\ ten & y & z \end{vmatrix}\cease{assortment} \)

A × B = (bz – cy)i – (az – cx)j + (ay – bx)k = (bz – cy)i + (cx – az)j + (ay – bx)k

Correct-mitt Dominion Cantankerous Product

Nosotros can discover the management of the unit vector with the aid of the right-hand rule.

Right-hand Rule Cross Product

In this dominion, we tin can stretch our right paw so that the alphabetize finger of the right hand is in the direction of the showtime vector and the middle finger is in the direction of the second vector. And so, the thumb of the right hand indicates the direction or unit of measurement vector n. With the help of the correct-hand rule, we can easily show that vectors' cross product is non commutative. If nosotros have two vectors A and B, then the diagram for the right-hand rule is as follows:

Cantankerous Production Properties

To find the cross product of two vectors, we can use properties. The properties such as anti-commutative property, zero vector property plays an essential role in finding the cantankerous production of two vectors. Apart from these properties, some other properties include Jacobi belongings, distributive property. The properties of cross-production are given below:

Cross Product Properties

Cross Product of Perpendicular Vectors

Cross product of two vectors is equal to the product of their magnitude, which represents the area of a rectangle with sides X and Y. If 2 vectors are perpendicular to each other, then the cantankerous product formula becomes:θ = ninety degreesWe know that, sin 90° = 1.

So,

Cross Product of Perpendicular Vectors

Cantankerous Production of Parallel vectors

The cross product of two vectors are zippo vectors if both the vectors are parallel or opposite to each other. Conversely, if two vectors are parallel or opposite to each other, and so their product is a cipher vector. Two vectors have the aforementioned sense of management.θ = 90 degreesAs we know, sin 0° = 0 and sin ninety° = 1.

And then,

Cross Product of Parallel vectors

Magnitude of Cantankerous Product

Allow united states of america presume two vectors,

\(\begin{array}{50}\vec{A}= A_{x}+ A_{y}+ A_{z}\end{assortment} \)

\(\begin{array}{l}\vec{B}= B_{ten}+ B_{y}+ B_{z}\end{array} \)

Then the magnitude of 2 vectors is given by the formula,

\(\begin{array}{fifty}|\vec{A}| = \sqrt{A_{x}^{2} + A_{y}^{2}+ A_{z}^{two}}\cease{array} \)

\(\brainstorm{array}{50}|\vec{B}|| = \sqrt{B_{x}^{2} + B_{y}^{2}+ B_{z}^{2}}\finish{assortment} \)

Hence, the magnitude of the cross product of ii vectors is given by the formula,

\(\begin{array}{50}|\vec{A}\times \vec{B}|= |\vec{A}| |\vec{B}| |sin\theta|\terminate{assortment} \)

Cross Product Example

Example:

Find the cross product of the given two vectors:

\(\begin{array}{l}\vec{X}= 5\vec{i} + six\vec{j} + ii\vec{chiliad} \ and\ \vec{Y}= \vec{i} + \vec{j} + \vec{yard}\end{array} \)

Solution:

Given:

\(\begin{array}{l}\vec{Ten}= 5\vec{i} + 6\vec{j} + ii\vec{grand} \\ \vec{Y}= \vec{i} + \vec{j} + \vec{g}\end{assortment} \)

To find the cantankerous production of two vectors, we have to write the given vectors in determinant form. Using the determinant course, we tin can find the cross product of two vectors as:

\(\brainstorm{array}{l}\vec{10}\times \vec{Y} = \brainstorm{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 5 & half-dozen & 2\\ i & 1 & ane \end{vmatrix}\end{assortment} \)

By expanding,

\(\begin{array}{l}\vec{10}\times \vec{Y}= (half-dozen-2)\vec{i}-(5-ii)\vec{j}+ (5-half dozen)\vec{k}\end{array} \)

Therefore,

\(\begin{array}{l}\vec{X}\times \vec{Y}= four\vec{i}-three\vec{j}- \vec{k}\finish{array} \)

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