Question

The angle between the vectors $\overline{u}=<3,0>and\overline{v}=<5,5>$ is _____

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Answer 1

We know that for two vectors $\overrightarrow{a}and\overrightarrow{b}$ their dot product

$\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|cos\theta Socos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$

Now here

$Cos\theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{|\overrightarrow{v}||u|}$ ...(1)

$|\overrightarrow{v}|=\sqrt{3^2+0^2}=3|\overrightarrow{v}|=\sqrt{5^2+5^2}=\sqrt{50}Also\overrightarrow{u}.\overrightarrow{v}=<3,0>.<5,5>$

$=(3\times 5)+(0\times 5)\left[\right(a_1\hat{i}+a_2\hat{j}).(b_1\hat{i}+b_2\hat{j})=(a_1{b}_2+a_2{b}_2\left)\right]$

So substituting the values in (1)

We get

$Cos\theta =\frac{15}{3\times \sqrt{50}}=\frac{15}{3\times 5\times \sqrt{2}}=\frac{1}{\sqrt{2}}Cos\theta \frac{1}{\sqrt{2}}$ $\Rightarrow \theta ={45}^{\circ }$, which is the correct answer.