Question

If $ {{\mathbf{a}}_{\mathbf{1}}} $ and $ {{\mathbf{a}}_{\mathbf{2}}} $ are two non collinear unit vectors and if $ \left| {{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}} \right| = \sqrt 3 $ , then value of $ {\mathbf{(}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ - }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}} \cdot {\mathbf{(2}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ - }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}} $ is $ \dfrac{p}{2} $ Then the value of $ p $ is _______.

Answer 1

Hint: Use the formula for the dot product of two vectors. Formula for dot product of two vectors is, $ {\mathbf{a}}{\mathbf{.b = }}\left| {\mathbf{a}} \right|\left| {\mathbf{b}} \right|\cos \theta $ where $ {\text{a}} $ and $ {\text{b}} $ are two vectors with $ \theta $ being the angle between them.

Complete answer:
We know, for two non linear vectors, $ {\text{a}} $ and $ {\text{b}} $ , the dot product or scalar product of them is, $ {\mathbf{a}}{\mathbf{.b = }}\left| {\mathbf{a}} \right|\left| {\mathbf{b}} \right|\cos \theta $ where, $ \theta $ is the angle between the vectors.
Here we have , $ \left| {{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}} \right| = \sqrt 3 $ …..1 , where $ {a_1} $ and $ {a_2} $ are non collinear ( having some angle between them) unit vectors , meaning, magnitude of them is one, i.e. $ \left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|{\mathbf{ = }}\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|{\mathbf{ = }}1 $ .
Let’s say, the angle between the unit vectors $ {a_1} $ and $ {a_2} $ is $ \theta $ . Then we have, $ {{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}} = \left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|\cos \theta $ and
Squaring both sides of the equation (1) we get,
 $ {\left| {{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}} \right|^2} = {(\sqrt 3 )^2} $
Now, putting the value $ {\mathbf{a = b = (}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}} $ in the dot product rule, we can have,
 $ {\mathbf{(}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}}{\mathbf{.(}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}}{\text{ = }}\left| {{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}} \right|\left| {{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}} \right|\cos {0^ \circ } $ [ Since, the angle between the same vectors is zero . Hence $ \cos \theta = 1 $ for dot product of the same vector]
On simplifying further using the distributive property we get,
  $ \Rightarrow {{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{ + 2}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}} = {\left| {{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}} \right|^2} = 3 $ since the value of $ \cos {0^ \circ } = 1 $
Now, using the formula for dot product and simplifying we get,
 $ \Rightarrow {\left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|^{\mathbf{2}}}{\mathbf{ + }}{\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|^{\mathbf{2}}}{\mathbf{ + }}2\left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|\cos \theta = 3 $
Now we have, $ \left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|{\mathbf{ = }}\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|{\mathbf{ = }}1 $
 $ \therefore \Rightarrow 1 + 1 + 2 \cdot 1 \cdot 1 \cdot \cos \theta = 3 $
 $ \Rightarrow 2 \cdot \cos \theta = 3 - 2 = 1 $
Then we get,
 $ \Rightarrow \cos \theta = \dfrac{1}{2} $ ……(2)
Now, we have to find the value of $ {\mathbf{(}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}}{\mathbf{.(}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}} $
Therefore on simplifying using the rule for dot product we get,
 $ {\mathbf{ = }}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ - }}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{ - 2}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}} $
Since, the scalar product of two vectors follows commutative relation. Therefore,
 $ {\mathbf{ = }}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ - 3}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{ + }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{.}}{{\mathbf{a}}_{\mathbf{2}}} $
 $ {\mathbf{ = }}2\left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|\left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|{\mathbf{ - }}3\left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|\cos \theta {\mathbf{ + }}\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right| $
Therefore that becomes,
 $ = 2{\left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|^{\mathbf{2}}} - 3\left| {{{\mathbf{a}}_{\mathbf{1}}}} \right|\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|\cos \theta + {\left| {{{\mathbf{a}}_{\mathbf{2}}}} \right|^{\mathbf{2}}} $
Putting, $ \cos \theta = \dfrac{1}{2} $ and $ \left| {{a_1}} \right| = \left| {{a_2}} \right| = 1 $
 $ = {2.1^2} - 3.1.1.\dfrac{1}{2} + 1 $
 $ = 2 + 1 - \dfrac{3}{2} $
Hence,
 $ {\mathbf{(}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ - }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}} \cdot {\mathbf{(2}}{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{ - }}{{\mathbf{a}}_{\mathbf{2}}}{\mathbf{)}} = \dfrac{3}{2} $
Hence, the value of $ p $ is $ 3 $ , which is our required answer.

Note:
The vectors are unit vectors here, if vectors of same magnitudes were given then also we could solve this problem but if vectors of different magnitudes were given then we have to find the values of their magnitude first, in this problem which is not possible.