Question

Find the angle between two vectors $\overrightarrow a$ and $\overrightarrow b$ with magnitude $\sqrt 3$ and $2,$ respectively having $\overrightarrow a .\overrightarrow b = \sqrt 6$

Answer 1

Hint : An algebraic operation which takes two equal-length sequences of numbers generally coordinate vectors and gives the resultant of the single number is known as the dot product or the scalar product. It is the sum of the product of the corresponding entries of the sequences of the two numbers. Use formula - $\overrightarrow a \cdot \overrightarrow b = \left| a \right|\left| b \right|\cos \theta$. Place the value and simplify the equation.

Complete step-by-step answer :
The magnitude of vector $\overrightarrow a = \left| a \right| = \sqrt 3$
The magnitude of vector $\overrightarrow b = \left| b \right| = 2$
$\overrightarrow a .\overrightarrow b = \sqrt 6$
To find the angle between the two given vectors, we will use the formula for the dot product.
$\Rightarrow \overrightarrow a \cdot \overrightarrow b = \left| a \right|\left| b \right|\cos \theta$
Place the given values in the above equation –
$\Rightarrow \sqrt 6 = \sqrt 3 \times 2 \times \cos \theta$
When the term multiplicative at one side when changes its side goes to the division and vice-versa.
$\Rightarrow \dfrac{{\sqrt 6 }}{{\sqrt 3 \times 2}} = \cos \theta$
Simplify the above equation -
$\Rightarrow \cos \theta = \dfrac{{\sqrt 6 }}{{\sqrt 3 \times 2}}$
$\Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 }}$
Make the required angle the subject –
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$
We know that, $\cos 45^\circ = \left( {\dfrac{1}{{\sqrt 2 }}} \right)$, so place the value in the above equation –
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\cos 45^\circ } \right)$
Cosine inverse and cosine functions cancel each other on the right hand side of the equation.
$\Rightarrow \theta = 45^\circ$
Or
$\Rightarrow \theta = \dfrac{\pi }{4}$

Note : Remember the trigonometric values for all the functions for the angles $0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ$ for direct substitutions. Also, Remember the All STC rule, it is also known as ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ($0^\circ \;{\text{to 90}}^\circ$ ) are positive, sine and cosec are positive in the second quadrant ($90^\circ {\text{ to 180}}^\circ$ ), tan and cot are positive in the third quadrant ($180^\circ \;{\text{to 270}}^\circ$ ) and sin and cosec are positive in the fourth quadrant ($270^\circ {\text{ to 360}}^\circ$ ).