Question

Please help im in despirate trouble tomorrow is my exam

Let only class 11th guys and gals answer this

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Angle=cos^{-1}\frac{1}{6}}}$

${\bold{\underline{\underline{Step-by-Step\:explanation:}}}}$

$\underline \bold{given : } \\ \implies A = \hat{i} + 2 \hat{j} + \hat{k} \\ \\ \implies B= 2 \hat{i} - \hat{j} + \hat{k} \\ \\ \underline \bold{to \: find : }\\ \implies Angle \:between \: vector \: A\: and \: B= ?$

• According to given question :

$\implies \cos \theta = \frac{ \vec{A} \times \vec{B}}{ | \vec{S}| | \vec{B}| } \\ \\ \implies \cos \theta = \frac{ (\hat{i} + 2\hat{j} + \hat{k} )\times (2 \hat{i} - \hat{j} + \hat{k})}{ | \sqrt{6} | | \sqrt{6} | } \\ \\ \implies cos \theta = \frac{2( \hat{i }\times \hat{ i}) - 2( \hat{j} \times \hat{j}) + 1( \hat{k} \times \hat{k}}{6} \\ \\ \bold{\hat{i} \times \hat{i} = 1} \\ \\ \bold{\hat{j} \times \hat{j} = 1}\\ \\ \bold{ \hat{k} \times \hat{k} = 1}\\ \\ \implies cos \theta = \frac{2 \times 1 - 2 \times 1 + 1 \times 1}{6} \\ \\ \implies cos \theta = \frac{\cancel2 - \cancel2 + 1}{6} \\ \\ \implies cos \theta = \frac{1}{6} \\ \\ \bold{\implies \theta = {cos}^{ - 1} \frac{1}{6} } \\ \\ \bold{\therefore Angle \: between \: A\: and \: B \: is \: {cos}^{ - 1} \frac{1}{6} }$

Answer 2

Answer:

Explanation:

Hope it helps