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

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 ${\text{a, b and c}}$ .