Angle between Two Vectors Calculator

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**Step 1 :**

**Step 2 :**

**Step 3 :**

Vector can be defined as a quantity which has both magnitude and direction. The magnitude expresses the distance of the vector whereas the direction shows the direction in which it travels.

When two vectors are sharing a common vertex then angle is formed between them is called angle between two vectors.

**Angle between two vectors calculator** calculates the angle between any two vectors which are sharing the common vertex.

When two vectors are sharing a common vertex then angle is formed between them is called angle between two vectors.

given will be the two vectors of the form:

A = a_{1} i + b_{1} j + c_{1} k

B = a_{2} i + b_{2} j + c_{2} k

where a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are the components of A and B.

Calculate the value of |A| and |B| given as:

|A| = $\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}$

|B| = $\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}$

Then using the below formula, calculate the value of cos $\theta$

cos $\theta$ = $\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{|A||B|}$

and from that get the value of $\theta$ using formula:

$\theta$ = cos^{-1} $\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{|A||B|}$

which gives the answer.

Calculate the angle between the the two vectors A and B where A = 3i + 4j + 5k and B = 4i + 3j + 6k.

**Step 1 :**given: The two vectors are A = 3i + 4j + 5k and B = 4i + 3j + 6k.

a

_{1}= 3, b_{1}= 4, c_{1}= 5 and

a

_{2}= 4, b_{2}= 3, c_{2}=6.**Step 2 :**The magnitude of two vectors |A| and |B| are given as

|A| = $\sqrt{3^{2} + 4^{2} + 5^{2}}$ = $\sqrt{50}$

|B| = $\sqrt{4^{2} + 3^{2} + 6^{2}}$ = $\sqrt{61}$**Step 3 :**Then using the formula for cos $\theta$, we get

cos $\theta$ = $\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{|A||B|}$

= $\frac{(3)(4) + (4)(3) + (5)(6)}{\sqrt{50} \sqrt{61}}$

cos $\theta$ = $\frac{12 + 12 + 30}{\sqrt{50} \sqrt{61}}$

cos $\theta$ = $\frac{54}{\sqrt{50} \sqrt{61}}$

= $\frac{54}{55.27}$

= 0.977

$\theta$ = cos

^{-1}(0.977)

$\theta$ = 12.31

^{o}= 0.214 radians.

**Answer :**The angle between two vectors is $\theta$ = 12.31

^{o}= 0.214 radians.Calculate the angle between the the two vectors A and B where A = i - 4j + 5k and B = 5i + 3j + 6k.

**Step 1 :**given: A = i - 4j + 5k and B = 5i + 3j + 6k.

where a

_{1}=1, b_{1}= -4, c_{1}=5

and a

_{2}= 5, b_{2}= 3, c_{2}=6.**Step 2 :**The magnitude of two vectors |A| and |B| are given as

|A| = $\sqrt{1^{2} + (-4)^{2} + 5^{2}}$ = $\sqrt{42}$

|B| = $\sqrt{5^{2} + 3^{2} + 6^{2}}$ = $\sqrt{70}$**Step 3 :**Then using the formula:

cos $\theta$ = $\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{|A||B|}$

= $\frac{(1)(5) + (-4)(3) + (5)(6)}{\sqrt{42} \sqrt{70}}$

cos $\theta$ = $\frac{5 - 12 + 30}{\sqrt{42} \sqrt{70}}$

cos $\theta$ = $\frac{23}{\sqrt{42} \sqrt{70}}$

= $\frac{23}{54.27}$

= 0.42

$\theta$ = cos^{-1}(0.42) = 64.92^{o}= 1.137 radians.**Answer :**The angle between two vectors is $\theta$ = 64.92

^{o}= 1.137 radians.