
If the magnitude of two vectors is \[\;3\] and \[4\] their scalar product is \[6\], then find the angle between the two vectors.
Answer
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Hint: There are two types of multiplication of vectors one of which is a scalar product and another one is the vector product. The scalar product is also called the dot product and the vector product is also called the cross product. The result of the two scalar multiplication will be scalar and the result of the two vector multiplication will be vector.
Complete answer:
Given that the magnitude of two vectors is \[\;3\] and \[4\]. The scalar product of the two vectors is given as \[6\].
We know that the scalar products of any two vectors can be obtained by multiplying the magnitude of the same vectors with the cosine of the angle between them.
Using this we can find the angle between the two vectors.
\[\vec A.\vec B = AB\cos \theta \]
Substituting \[\vec A.\vec B\]=\[6\].
Given that the magnitude of \[\vec A\]=\[3\] and the magnitude of \[\vec B\]=\[6\]. Therefore substituting all these values in the above equation, we get,
\[6 = 3 \times 4\cos \theta \]
\[ \Rightarrow \dfrac{6}{{12}} = \cos \theta \]
\[ \Rightarrow \dfrac{1}{2} = \cos \theta \]
Therefore taking the inverse of the cosine theta, we get,
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
\[ \Rightarrow \theta = {60^0}\]
Therefore the angle between the two vectors \[\vec A\] and \[\vec B\] is \[{60^0}\].
Note:
The scalar product and the vector product find many applications in physics and astronomy. Both scalar and vector products have the distributive property in common, while the commutative property can be found only in scalar products. The vector product has the anti-commutative property. This means that if we change the order of the vectors in which they have multiplied then the product will acquire a negative sign. Also, the scalar products of the orthogonal vectors (vectors that are perpendicular to each other) vanish. And the scalar products of the anti-parallel vectors (two vectors that are parallel to each other and have opposite direction) is negative.
Complete answer:
Given that the magnitude of two vectors is \[\;3\] and \[4\]. The scalar product of the two vectors is given as \[6\].
We know that the scalar products of any two vectors can be obtained by multiplying the magnitude of the same vectors with the cosine of the angle between them.
Using this we can find the angle between the two vectors.
\[\vec A.\vec B = AB\cos \theta \]
Substituting \[\vec A.\vec B\]=\[6\].
Given that the magnitude of \[\vec A\]=\[3\] and the magnitude of \[\vec B\]=\[6\]. Therefore substituting all these values in the above equation, we get,
\[6 = 3 \times 4\cos \theta \]
\[ \Rightarrow \dfrac{6}{{12}} = \cos \theta \]
\[ \Rightarrow \dfrac{1}{2} = \cos \theta \]
Therefore taking the inverse of the cosine theta, we get,
\[\theta = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
\[ \Rightarrow \theta = {60^0}\]
Therefore the angle between the two vectors \[\vec A\] and \[\vec B\] is \[{60^0}\].
Note:
The scalar product and the vector product find many applications in physics and astronomy. Both scalar and vector products have the distributive property in common, while the commutative property can be found only in scalar products. The vector product has the anti-commutative property. This means that if we change the order of the vectors in which they have multiplied then the product will acquire a negative sign. Also, the scalar products of the orthogonal vectors (vectors that are perpendicular to each other) vanish. And the scalar products of the anti-parallel vectors (two vectors that are parallel to each other and have opposite direction) is negative.
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