how to find angle between two vectors

The vector quantities possess magnitude as well equally direction, whereas scalar quantities take magnitude only, but not direction. A vector may be represented in the following grade:

In two dimensions:

\(\begin{array}{l}\vec{A} = A_{x}i+A_{y}j\end{array} \)

Case: 2i+ 3j

In 3 dimensions:

\(\begin{array}{l}\vec{A} = A_{x}i+A_{y}j+A_{z}k\finish{array} \)

Example: 2i+ 3j -7k

Ii vectors may exist inclined at an angle from each other equally shown in the following figure:

Also, read:

• Vectors
• Types of Vector
• Vector Space
• Vectors Joining Ii Points
• Dot Product of Two Vectors
• Multiplication of Vectors with Scalar

Angle Between Two Vectors Formula

Before understanding the formula of the bending between two vectors, permit u.s.a. understand how to observe a scalar production or dot production of ii vectors.

Let united states of america presume that ii vectors are given such that:

\(\begin{array}{l}\vec{A} = A_{x}i+A_{y}j+A_{z}k\finish{array} \)

and

\(\begin{array}{l}\vec{B} = B_{x}i+B_{y}j+B_{z}k\end{array} \)

So dot product or scalar production of

\(\begin{array}{fifty}\vec{A}\terminate{assortment} \)

and

\(\brainstorm{assortment}{l}\vec{B}\cease{array} \)

can be calculated equally:

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

The angle θ betwixt the 2 vectors

\(\begin{array}{l}\vec{A}\terminate{array} \)

  and

\(\begin{assortment}{l}\vec{B}\cease{assortment} \)

can be calculated using the post-obit formula:

\(\brainstorm{array}{fifty}\theta=cos^{-1}\frac{\vec{A}.\vec{B}}{\left | \vec{A} \correct |\left | \vec{B} \correct |}\end{assortment} \)

Where

\(\brainstorm{assortment}{fifty}\vec{A}.\vec{B}\end{assortment} \)

= Dot product of

\(\begin{array}{l}\vec{A}\end{array} \)

  and

\(\brainstorm{array}{l}\vec{B}\end{array} \)

\(\begin{array}{fifty}\left | \vec{A} \right |\cease{array} \)

= Magnitude of vector A

\(\brainstorm{array}{50}\left | \vec{A} \right |=\sqrt{A_x^two+A_y^2+A_z^two}\terminate{assortment} \)

\(\begin{array}{l}\left | \vec{B} \right |\end{array} \)

= Magnitude of vector B

\(\begin{array}{50}\left | \vec{B} \right |=\sqrt{B_x^2+B_y^2+B_z^2}\finish{array} \)

Cosine of Bending Between 2 Vectors

The formula for finding cosine of angle between two vectors can be deduced by the formula of bending between two vectors

\(\begin{array}{50}\vec{A}\terminate{array} \)

  and

\(\begin{assortment}{50}\vec{B}\end{assortment} \)

is

\(\brainstorm{array}{l}\theta=cos^{-1}\frac{\vec{A}.\vec{B}}{\left | \vec{A} \right |\left | \vec{B} \right |}\end{array} \)

Therefore, the cosine angle between two vectors is given by

\(\begin{array}{l}\theta=cos^{-i}\frac{\vec{A}.\vec{B}}{\left | \vec{A} \right |\left | \vec{B} \right |}\end{assortment} \)

And then, the cosine of the angle betwixt two vectors can be calculated by dividing the dot product of the vectors by-product of their magnitudes.

Distance Between Two Vectors

Permit u.s. suppose that two vectors that are defined in two-dimensional infinite exist:

\(\begin{assortment}{l}\vec{A} = A_{x}i+A_{y}j\end{array} \)

and

\(\brainstorm{array}{l}\vec{B} = B_{x}i+B_{y}j\end{array} \)

Therefore, the altitude between two vectors such as vector A and vector B is given as

\(\brainstorm{assortment}{fifty}d =\sqrt{(A_{x}-B_{x})^{ii}+ (A_{y}-B_{y})^{2}}\cease{array} \)

Similarly, consider 2 vectors defined in three-dimensional space be:

\(\begin{assortment}{l}\vec{A} = A_{ten}i+A_{y}j+A_{z}g\stop{assortment} \)

and

\(\brainstorm{assortment}{fifty}\vec{B} = B_{x}i+B_{y}j+B_{z}chiliad\end{array} \)

Therefore, the distance between two vectors such as vector A and vector B is given as

\(\begin{array}{fifty}d =\sqrt{(A_{x}-B_{x})^{2}+ (A_{y}-B_{y})^{2}+ (A_{z}-B_{z})^{2}}\end{assortment} \)

How to Detect the Angle Between Two Vectors?

To find the bending between ii vectors, ane needs to follow the steps given below:

Pace 1: Calculate the dot product of two given vectors by using the formula :

\(\begin{array}{50}\vec{A}.\vec{B} = A_{x}B_{x}+ A_{y}B_{y}+A_{z}B_{z}\finish{array} \)

Step 2: Calculate the magnitude of both the vectors separately. Magnitude can be calculated by squaring all the components of vectors and adding them together and finding the square roots of the upshot.

Step 3: Substitute the values of dot production and magnitudes of both vectors in the following formula for finding the angle between 2 vectors, i.e.

\(\brainstorm{array}{50}\theta=cos^{-1}\frac{\vec{A}.\vec{B}}{\left | \vec{A} \correct |\left | \vec{B} \correct |}\end{assortment} \)

Bending Betwixt 2 vectors Instance

Question: Summate the angle between vectors 2i + 3j – yard and i – 3j + 5k

Solution:

Let

\(\brainstorm{array}{l}\vec{A}\end{assortment} \)

=2i + 3j – k and

\(\begin{assortment}{l}\vec{B}\stop{assortment} \)

=i – 3j + 5k

Scalar product of

\(\begin{array}{l}\vec{A}\end{array} \)

and

\(\brainstorm{array}{fifty}\vec{B}\end{array} \)

is

\(\begin{assortment}{l}\vec{A}. \vec{B}\end{array} \)

= 2 + (-nine) + (-5) = – 12

Magnitude of

\(\brainstorm{array}{fifty}\vec{A}\terminate{assortment} \)

is:

\(\begin{array}{l}\left | \vec{A} \right |=\sqrt{iv+9+i}=\sqrt{14}\end{assortment} \)

Magnitude of

\(\begin{array}{l}\vec{B}\finish{array} \)

is:

\(\begin{array}{l}\left | \vec{B} \correct |=\sqrt{1+9+25}=\sqrt{35}\finish{assortment} \)

The formula to notice the angle betwixt two vectors is

\(\begin{array}{l}\theta=cos^{-one}\frac{\vec{A}.\vec{B}}{\left | \vec{A} \right |\left | \vec{B} \correct |}\end{array} \)

\(\begin{array}{fifty}\theta = cos^{-1}\frac{-12}{\sqrt{xiv}\sqrt{35}}\stop{array} \)

\(\begin{array}{l}\theta = cos^{-1}\frac{-12}{3.64 \times five.92}\end{array} \)

\(\brainstorm{assortment}{l}\theta = cos^{-ane}(-0.542)\end{array} \)

\(\begin{array}{l}\theta = 122. 82^{\circ}\terminate{assortment} \)

Therefore, the angle betwixt two vectors 2i + 3j – one thousand and i – 3j + 5k is

\(\begin{assortment}{50}122. 82^{\circ}\end{assortment} \)

.

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