Question

If a] and b unit vectors such that \[\left[ {{\text{a b c}}} \right] = \dfrac{1}{4}\] , then angle between \[{\text{a}}\] and \[{\text{b}}\] is?
(A) \[\dfrac{\pi }{3}\]
(B) \[\dfrac{\pi }{4}\]
(C) \[\dfrac{\pi }{6}\]
(D) \[\dfrac{\pi }{2}\]

Answer 1

Hint: In this sum we use formulae related to scalar triple products. And the formula is, \[\left[ {{\text{a b c}}} \right] = ({\text{a}} \times {\text{b)}}{\text{.c}}\], here the product of first two vectors is also a vector and the dot product of this vector and third vector is a scalar. And that is the reason we call this product a scalar triple product.

Complete step-by-step solution:
Here it is given as \[\left[ {{\text{a b c}}} \right] = \dfrac{1}{4}\]
So, we can conclude that, \[({\text{a}} \times {\text{b)}}{\text{.c = }}\dfrac{1}{4}\]
The scalar triple product is also associative. So we can interchange the symbols i.e., we can change the cross with a dot. But here we don’t need that case.
But we also know that the vector \[{\text{c}}\] is the result of cross product of \[{\text{a}}\] and \[{\text{b}}\] .
In a cross product of two vectors, which are in the same plane, gives us a resultant vector, which is perpendicular to both the vectors.
So \[{\text{c = a}} \times {\text{b}}\]
So we get, \[({\text{a}} \times {\text{b)}}{\text{.(a}} \times {\text{b) = }}\dfrac{1}{4}\]
\[ \Rightarrow \]\[|{\text{a}} \times {\text{b}}{{\text{|}}^2} = \dfrac{1}{4}\]
\[ \Rightarrow {\text{a}} \times {\text{b = }}\dfrac{1}{2}\]
And here we have to know that, a vector product or cross product of two vectors is equal to the product of mod of those vectors and the sine value of angle between those vectors.
So, \[|{\text{a||b|sin(a,b) = }}\dfrac{1}{2}\]
As these vectors are unit vectors, it simplifies as,
\[s{\text{in(a,b) = }}\dfrac{1}{2}\]
\[ \Rightarrow \]\[s{\text{in(a,b) = sin}}{30^{\text{o}}}\]
So angle between \[{\text{a}}\] and \[{\text{b}}\] is \[{30^{\text{o}}}\] i.e. \[\dfrac{\pi }{6}\]
So option (C) is the correct option.

Note: In the solution, you can also get a negative value, for which you get the angle between those two vectors as \[{\text{21}}{{\text{0}}^{\text{o}}}\] but that case should not be considered. You should only take the value in the range \[\left[ {0,\pi } \right]\] . And, if the order of three vectors is changed, the scalar triple product that we get will be negative i.e., if order is changed as \[{\text{a, c and b}}\] instead of \[{\text{a, b and c}}\] .