Question

If $\overrightarrow a$ and $\overrightarrow b$ are two non collinear unit vectors and $\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt 3$ then $\left( {2\overrightarrow a - 5\overrightarrow b } \right) \cdot \left( {3\overrightarrow a + \overrightarrow b } \right) =$

Answer 1

Hint : Here in this question, we have to find the value of the given vector equation i.e., $\left( {2\overrightarrow a - 5\overrightarrow b } \right) \cdot \left( {3\overrightarrow a + \overrightarrow b } \right)$ using a vector condition $\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt 3$. This can solve by substituting the value of $\overrightarrow a$ and $\overrightarrow b$ (given that two vectors are unit vectors then it values are $\overrightarrow a = 1$ and $\overrightarrow b = 1$) and find the value of $\overrightarrow a \cdot \overrightarrow b$ by simplifying the condition $\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt 3$ and by substituting all these values in a simplified vector equation we get the required solution.

Complete step-by-step answer :
Vector is a quantity that has both magnitude and direction. It is typically represented by an arrow whose direction is the same as that of the quantity and whose length is proportional to the quantity’s magnitude. Although a vector has magnitude and direction, it does not have position.
Let us consider the condition,
$\Rightarrow \left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt 3$
Squaring on both sides, we get
$\Rightarrow {\left( {\left| {\overrightarrow a + \overrightarrow b } \right|} \right)^2} = {\left( {\sqrt 3 } \right)^2}$
Apply a algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, then
$\Rightarrow {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + 2 \cdot \overrightarrow a \cdot \overrightarrow b = {\left( {\sqrt 3 } \right)^2}$
Given $\overrightarrow a$ and $\overrightarrow b$ are unit vectors which have a length 1 unit.
i.e., $\left| {\overrightarrow a } \right| = 1$ and $\left| {\overrightarrow b } \right| = 1$, on substituting we have
$\Rightarrow {1^2} + {1^2} + 2 \cdot \overrightarrow a \cdot \overrightarrow b = {\left( {\sqrt 3 } \right)^2}$
On simplification, we have
$\Rightarrow 1 + 1 + 2 \cdot \overrightarrow a \cdot \overrightarrow b = 3$
$\Rightarrow 2 + 2 \cdot \overrightarrow a \cdot \overrightarrow b = 3$
Subtract 2 on both sides, then
$\Rightarrow 2 \cdot \overrightarrow a \cdot \overrightarrow b = 3 - 2$
$\Rightarrow 2 \cdot \overrightarrow a \cdot \overrightarrow b = 1$
Divide both side by 2, then
$\Rightarrow \overrightarrow a \cdot \overrightarrow b = \dfrac{1}{2}$
Now consider,
$\Rightarrow \left( {2\overrightarrow a - 5\overrightarrow b } \right) \cdot \left( {3\overrightarrow a + \overrightarrow b } \right) =$
Remove parenthesis by multiplying a binomial
$\Rightarrow 6{\left| {\overrightarrow a } \right|^2} + 2\overrightarrow a \cdot \overrightarrow b - 15\overrightarrow b \cdot \overrightarrow a - 5{\left| {\overrightarrow b } \right|^2}$
Where, $\left| {\overrightarrow a } \right| = 1$, $\left| {\overrightarrow b } \right| = 1$ and $\overrightarrow a \cdot \overrightarrow b = \dfrac{1}{2}$, on substituting we have
$\Rightarrow 6{\left( 1 \right)^2} + 2\left( {\dfrac{1}{2}} \right) - 15\left( {\dfrac{1}{2}} \right) - 5{\left( 1 \right)^2}$
On simplification, we have
$\Rightarrow 6 + 1 - \dfrac{{15}}{2} - 5$
$\Rightarrow 7 - \dfrac{{15}}{2} - 5$
$\Rightarrow 2 - \dfrac{{15}}{2}$
Take 2 as LCM, then
$\Rightarrow \dfrac{{4 - 15}}{2}$
On simplification, we get
$\Rightarrow - \dfrac{{11}}{2}$
Hence, it’s a required solution.
So, the correct answer is “$\Rightarrow - \dfrac{{11}}{2}$”.

Note : Remember, a vector is a quantity that has both magnitude, as well as direction. A vector that has a magnitude of 1 is a unit vector. It is also known as Direction Vector. When two or more vectors which are parallel to the same line irrespective of their magnitudes and direction are known as Collinear vectors otherwise its non collinear.