Question

the angle between the two vectors to ICAP + 3 J cap + K cap and minus 3 ICAP + 6 k cap is​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow \theta= \cos^{-1}\left(\frac{-3}{275}\right)$

Explanation:

Let say :

$\displaystyle \vec{\text{A}} =\hat{i}+3\hat{j}+\hat{k}$

$\displaystyle \vec{\text{B}} =-3\hat{i}+6\hat{k}$

We are asked to find angle between them :

Using dot product :

$\displaystyle \sf A.B=AB\cos \theta$

$\displaystyle \sf \longrightarrow \cos \theta=\frac{AB}{|A|.|B|}$

Substituting values here we get :

$\displaystyle \sf \longrightarrow \cos \theta=\frac{(\hat{i}+3\hat{j}+\hat{k}).(-3\hat{i}+6\hat{k} )}{\left|\left(\sqrt{1^2+3^2+1^2}\right)\right|.\left|\left(\sqrt{(-3)^2+6^2}\right)\right|}$

$\displaystyle \sf \longrightarrow \cos \theta=\frac{(-9+6)}{\left|\left(\sqrt{11}\right)\right|.\left|\left(\sqrt{25}\right)\right|}$

$\displaystyle \sf \longrightarrow \cos \theta=\frac{-3}{275}$

$\displaystyle \sf \longrightarrow \theta= \cos^{-1}\left(\frac{-3}{275}\right)$

Hence we get required answer!