Question

If ${a_1}$ and ${a_2}$ are two non collinear unit vectors and if $\left| {{a_1} + {a_2}} \right| = \sqrt 3$ , then the value of $({a_1} - {a_2}) \cdot (2{a_1} - {a_2})$ is $\dfrac{p}{2}$ . Then find the value of p.

Answer 1

Hint : In this question, we are given two unit vectors namely ${a_1}$ and ${a_2}$ , they both are unit vectors so the magnitude of both of them will be 1. Now, we are given two equations in terms of these two vectors, and one of the equations also involves an unknown variable quantity “p” and we have to find the value of p. On the left-hand side of the second equation, we have an equation involving the dot product of two brackets. So, we will have to find the dot product of ${a_1}$ and ${a_2}$ . The dot product can be found from the first equation by performing some arithmetic operations using the arithmetic identities.

Complete step by step solution:
We are given that $\left| {{a_1} + {a_2}} \right| = \sqrt 3$
We can write $({a_1} + {a_2}) \cdot ({a_1} + {a_2}) = {a_1} \cdot ({a_1} + {a_2}) + {a_2} \cdot ({a_1} + {a_2}) = {a_1} \cdot {a_1} + 2{a_1} \cdot {a_2} + {a_2} \cdot {a_2}$
We know that $\vec a \cdot \vec a = {\left| {\vec a} \right|^2}$ , so we get –
${\left| {{a_1} + {a_2}} \right|^2} = {\left| {{a_1}} \right|^2} + {\left| {{a_2}} \right|^2} + 2{a_1} \cdot {a_2}$
On putting the known values, we get –
$3 = 1 + 1 + 2{a_1} \cdot {a_2} \\\ \Rightarrow {a_1} \cdot {a_2} = \dfrac{{3 - 2}}{2} = \dfrac{1}{2} \\\$
Now, we have to solve $({a_1} - {a_2}) \cdot (2{a_1} - {a_2}) = \dfrac{p}{2}$ for the value of p –
${a_1} \cdot (2{a_1} - {a_2}) - {a_2}(2{a_1} - {a_2}) = \dfrac{p}{2} \\\ \Rightarrow 2{a_1} \cdot {a_1} - {a_1} \cdot {a_2} - 2{a_1} \cdot {a_2} + {a_2} \cdot {a_2} = \dfrac{p}{2} \\\ \Rightarrow 2{\left| {{a_1}} \right|^2} + {\left| {{a_2}} \right|^2} - 3{a_1} \cdot {a_2} = \dfrac{p}{2} \\\ \Rightarrow 2(1) + 1 - 3(\dfrac{1}{2}) = \dfrac{p}{2} \\\ \Rightarrow \dfrac{{6 - 3}}{2} = \dfrac{p}{2} \\\ \Rightarrow p = 3 \;$
Hence if ${a_1}$ and ${a_2}$ are two non collinear unit vectors and if $\left| {{a_1} + {a_2}} \right| = \sqrt 3$ , and the value of $({a_1} - {a_2}) \cdot (2{a_1} - {a_2})$ is $\dfrac{p}{2}$ . Then find the value of p is 3.
So, the correct answer is “3”.

Note : Vectors are those quantities that have both direction and magnitude, the magnitude of a vector $\vec a$ is given as $\left| {\vec a} \right|$ . The dot product of two vectors is given by the formula $\vec a \cdot \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\cos \theta$ where $\theta$ is the angle between the two vectors, that’s why $\vec b \cdot \vec b = \left| {\vec b} \right| \cdot \left| {\vec b} \right|\cos 0 = {\left| {\vec b} \right|^2}$ (angle between two vectors is equal to 0 degrees).