Question

If we have unit vectors as $\hat{a}$ and $\hat{b}$ inclined at an angle $\theta$ then prove that $\cos \dfrac{\theta }{2}=\dfrac{1}{2}\left| \hat{a}+\hat{b} \right|$

Answer 1

We solve this problem simple by adding the given unit vectors and squaring to get the required result. Here, we get the dot product after squaring the sum of vectors. We use the formula of dot product as
$\hat{a}.\hat{b}=\left| {\hat{a}} \right|\left| {\hat{b}} \right|\cos \theta$
Where $\theta$ is the angle between $\hat{a}$ and $\hat{b}$

Complete step-by-step solution:
We are given that $\hat{a}$ and $\hat{b}$ are unit vectors
We know that the modulus of unit vector is 1 that is
$\Rightarrow \left| {\hat{a}} \right|=\left| {\hat{b}} \right|=1$
Now, let us assume that the sum of given two vectors as
$\Rightarrow S=\hat{a}+\hat{b}$
Now, by squaring on both sides we get
$\Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}={{S}^{2}}$
Now, by expanding the square we get
$\Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}={{\left| {\hat{a}} \right|}^{2}}+{{\left| {\hat{b}} \right|}^{2}}+2\left( \hat{a}.\hat{b} \right)$
We know that the modulus of unit vectors as1 that is
$\Rightarrow \left| {\hat{a}} \right|=\left| {\hat{b}} \right|=1$
By substituting this value in above equation we get

& \Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=1+1+2\left( \hat{a}.\hat{b} \right) \\\ & \Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=2+2\left( \hat{a}.\hat{b} \right) \\\ \end{aligned}$$ We know that the formula of dot product is given as $$\hat{a}.\hat{b}=\left| {\hat{a}} \right|\left| {\hat{b}} \right|\cos \theta $$ By substituting this formula in above equation we get $$\begin{aligned} & \Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=2+2\left( 1.1.\cos \theta \right) \\\ & \Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=2\left( 1+\cos \theta \right) \\\ \end{aligned}$$ We know that the formula of half angles for the trigonometric ratios is given as $$\Rightarrow 1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$$ By substituting this formula in above equation we get $$\begin{aligned} & \Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}=2\left( 2{{\cos }^{2}}\dfrac{\theta }{2} \right) \\\ & \Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}={{\left( 2\cos \dfrac{\theta }{2} \right)}^{2}} \\\ \end{aligned}$$ By applying square root on both sides we get $$\begin{aligned} & \Rightarrow \left| \hat{a}+\hat{b} \right|=2\cos \dfrac{\theta }{2} \\\ & \Rightarrow \cos \dfrac{\theta }{2}=\dfrac{1}{2}\left| \hat{a}+\hat{b} \right| \\\ \end{aligned}$$ Hence the required result has been proved. **Note:** Students may make mistakes in squaring the sum of two vectors. While squaring the sum of two vectors we get the dot product of those vectors that is $$\Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}={{\left| {\hat{a}} \right|}^{2}}+{{\left| {\hat{b}} \right|}^{2}}+2\left( \hat{a}.\hat{b} \right)$$ But students may miss this point of having the dot product and take the formula as $$\Rightarrow {{\left| \hat{a}+\hat{b} \right|}^{2}}={{\left| {\hat{a}} \right|}^{2}}+{{\left| {\hat{b}} \right|}^{2}}+2\left( \hat{a}\hat{b} \right)$$ This gives the wrong result. Also while applying the square root we need to give modulus to either of side that is we have $$\Rightarrow \left| \hat{a}+\hat{b} \right|=2\cos \dfrac{\theta }{2}$$ But students will take the equation as $$\Rightarrow \left( \hat{a}+\hat{b} \right)=2\cos \dfrac{\theta }{2}$$ This is wrong because we get $$\pm $$ condition while applying the square root. So, modulus function is important.