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Question: If \({\vec a_1}\) and \({\vec a_2}\) are two non collinear unit vector and if \(\left| {{{\vec a}_1}...

If a1{\vec a_1} and a2{\vec a_2} are two non collinear unit vector and if a1+a2=3\left| {{{\vec a}_1} + {{\vec a}_2}} \right| = \sqrt 3 then find the value of (a1a2)(2a1+a2)({\vec a_1} - {\vec a_2})(2{\vec a_1} + {\vec a_2})
(A) 2
(B) 32\dfrac{3}{2}
(C) 12\dfrac{1}{2}
(D) 1

Explanation

Solution

The relation between the two unit vectors is given by a1+a2=3\left| {{{\vec a}_1} + {{\vec a}_2}} \right| = \sqrt 3 . Squaring on both sides and simplifying the dot product find the value of θ\theta . Now, expand (a1a2)(2a1+a2)({\vec a_1} - {\vec a_2})(2{\vec a_1} + {\vec a_2}) and after simplifying the dot product again substitute the values of both vectors and θ\theta . On evaluating further, we get the required value.

Complete step-by-step solution:
a1{\vec a_1} and a2{\vec a_2} are two non-collinear unit vectors so their magnitude is equal one.
It is given that, a1+a2=3\left| {{{\vec a}_1} + {{\vec a}_2}} \right| = \sqrt 3
Squaring on both sides.
a1+a22=(3)2{\left| {{{\vec a}_1} + {{\vec a}_2}} \right|^2} = {\left( {\sqrt 3 } \right)^2}
a12+a22+2a1.a2=3{\left| {{{\vec a}_1}} \right|^2} + {\left| {{{\vec a}_2}} \right|^2} + 2{\vec a_1}.{\vec a_2} = 3
1+1+2a1a2cosθ=31 + 1 + 2\left| {{{\vec a}_1}} \right|\left| {{{\vec a}_2}} \right|\cos \theta = 3
2×1×1×cosθ=322 \times 1 \times 1 \times \cos \theta = 3 - 2
cosθ=12\cos \theta = \dfrac{1}{2}
We need to find the value of (a1a2)(2a1+a2)({\vec a_1} - {\vec a_2})(2{\vec a_1} + {\vec a_2})
Multiply both to get four individual terms.
=2a122a1.a2+a1.a2a22= 2{\left| {{{\vec a}_1}} \right|^2} - 2{\vec a_1}.{\vec a_2} + {\vec a_1}.{\vec a_2} - {\left| {{{\vec a}_2}} \right|^2}
=2a1.a21= 2 - {\vec a_1}.{\vec a_2} - 1
=1a1a2cosθ= 1 - \left| {{{\vec a}_1}} \right|\left| {{{\vec a}_2}} \right|\cos \theta
Substitute the value of θ\theta .
=11×1×12= 1 - 1 \times 1 \times \dfrac{1}{2}
=12= \dfrac{1}{2}

Hence, the value of (a1a2)(2a1+a2)({\vec a_1} - {\vec a_2})(2{\vec a_1} + {\vec a_2}) is 12\dfrac{1}{2} and the correct option is C.

Note: Collinear vectors are those vectors which act along the same line. So, the angle between them can be zero or 1800{180^0}. Coplanar vectors are where three vectors lie in the same plane. Equal, parallel, anti-parallel, collinear, zero, unit, orthogonal, polar, axial, coplanar and negative are the types of vectors.