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.
given will be the two vectors of the form:
A = a1 i + b1 j + c1 k
B = a2 i + b2 j + c2 k
where a1, b1, c1 and a2, b2, c2 are the components of A and B.
Step 2 :
Calculate the value of |A| and |B| given as:
|A| = $\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}}$
|B| = $\sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}$
Step 3 :
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.
a1 = 3, b1 = 4, c1 = 5 and
a2 = 4, b2 = 3, c2 =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)
$\theta$ = 12.31o = 0.214 radians.
Answer :The angle between two vectors is $\theta$ = 12.31o = 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 a1 =1, b1 = -4, c1 =5
and a2 = 5, b2 = 3, c2 =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) = 64.92o = 1.137 radians.
Answer :The angle between two vectors is $\theta$ = 64.92o = 1.137 radians.