Question

Find the angle between the vectors 2i+3j-k and i+2j+3k​

Answer 1

$\large\underline{\sf{Solution-}}$

Given vectors are

$\rm :\longmapsto\:2\hat{i} + 3\hat{j} - \hat{k}$

and

$\rm :\longmapsto\:\hat{i} + 2\hat{j} + 3\hat{k}$

Let assume that

$\rm :\longmapsto\:\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$

and

$\rm :\longmapsto\:\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$

So,

$\rm :\longmapsto\:\vec{a}.\vec{b}$

$\rm \:  =  \: (2\hat{i} + 3\hat{j} - \hat{k}).(\hat{i} + 2\hat{j} + 3\hat{k})$

$\rm \:  =  \: 2 + 6 - 3$

$\rm \:  =  \: 5$

$\purple{\rm\implies \:\boxed{\tt{ \vec{a}.\vec{b} = 5}}}$

Now, Consider

$\rm :\longmapsto\: |\vec{a}|$

$\rm \:  =  \: |2\hat{i} + 3\hat{j} - \hat{k}|$

$\rm \:  =  \: \sqrt{ {(2)}^{2} + {(3)}^{2} + {( - 1)}^{2} }$

$\rm \:  =  \: \sqrt{4 + 9 + 1}$

$\rm \:  =  \: \sqrt{14}$

$\purple{\rm\implies \:\boxed{\tt{ |\vec{a}| = \sqrt{14}}}}$

Now, Consider

$\rm :\longmapsto\: |\vec{b}|$

$\rm \:  =  \: |\hat{i} + 2\hat{j} + 3\hat{k}|$

$\rm \:  =  \: \sqrt{ {(1)}^{2} + {(2)}^{2} + {(3)}^{2} }$

$\rm \:  =  \: \sqrt{1 + 4 + 9}$

$\rm \:  =  \: \sqrt{14}$

$\purple{\rm\implies \:\boxed{\tt{ |\vec{b}| = \sqrt{14}}}}$

Now, the required angle is given by

$\rm :\longmapsto\:cos \theta \: = \: \bigg |\dfrac{\vec{a}.\vec{b}}{ |\vec{a}| |\vec{b}| } \bigg|$

$\rm :\longmapsto\:cos \theta \: = \: \bigg |\dfrac{5}{ \sqrt{14} \times \sqrt{14} } \bigg|$

$\rm :\longmapsto\:cos \theta \: = \: \bigg |\dfrac{5}{ 14} \bigg|$

$\rm :\longmapsto\: \theta \: = {cos}^{ - 1}\bigg [\dfrac{5}{ 14} \bigg]$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

$\rm :\longmapsto\:\boxed{\tt{ \vec{a}.\vec{b} = \vec{b}.\vec{a}}}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}}}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{a} \times \vec{b} =0\rm\implies \:\vec{a} \: \parallel \: \vec{b}}}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{a}.\vec{b} = 0\rm\implies \:\vec{a} \: \perp \: \vec{b}}}$

$\rm :\longmapsto\:\boxed{\tt{ \vec{a}.\vec{b} = |\vec{a}| |\vec{b}| cos \theta}}$

$\rm :\longmapsto\:\boxed{\tt{ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| sin \theta}}$

Answer 2

Let the two vectors be $\overrightarrow{A}$ and $\overrightarrow{B}$ respectively.

We've been provided two vectors and we've been asked to find out the angle between both vectors.

Whenever we have been provided two different vectors and asked to find angle between them then the concept by which we can solve this problem is Dot product of both vectors.

We know that, the angle between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by,

$\implies\boxed{\cos(\theta) = \dfrac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}}$

Where θ refers angle between both vectors.

$\implies \cos(\theta) = \dfrac{ \big(2 \hat{i} +3\hat{j} - \hat{k} \big) \cdot \big(\hat{i} + 2\hat{j} + 3\hat{k} \big)} { \big( \sqrt{(2)^{2} +(3)^{2} + ( - 1)^{2}} \big) \big( \sqrt{(1)^{2} + (2)^{2} + (3)^{2}} \big)} \\ \\ \implies \cos(\theta) = \dfrac{(2\hat{i})(1\hat{i}) + (3\hat{j})(2\hat{j}) - (1\hat{k})(3\hat{k})}{ \big(\sqrt{4 + 9 + 1} \big) \big( \sqrt{1 + 4 + 9} \big)} \\ \\ \implies \cos(\theta) = \dfrac{2 + 6 - 3}{ \big( \sqrt{14} \big) + \big( \sqrt{14} \big)} \\ \\ \implies \cos(\theta) = \dfrac{8 - 3}{7 + 7} \\ \\ \implies \cos(\theta) = \dfrac{5}{14} \\ \\ \implies \boxed{\theta = \cos^{ - 1} \bigg[ \dfrac{5}{14} \bigg]}$