Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

What is the angle between following planes $2x-3y+4z=1$ and $-x+y=4$?

Can you answer this question?

1 Answer

0 votes
  • The angle between two planes is the angle between the normals to the two planes. $ \theta=\cos^{-1}\bigg( \large\frac{\overrightarrow n_1.\overrightarrow n_2}{|\overrightarrow n_1||\overrightarrow n_2|} \bigg).$ If $ \overrightarrow n_1.\overrightarrow n_2=0$ the planes are at right angles.
$ 2x-3y+4z=1$ (i) and $-x+y=4$ (ii)
$ \overrightarrow n_1 = 2\overrightarrow i-3\overrightarrow j+4\overrightarrow k \: \: \overrightarrow n_2=-\overrightarrow i+\overrightarrow j$
$ \theta = \cos^{-1} \large\frac{[\overrightarrow n_1.\overrightarrow n_2]}{|\overrightarrow n_1||\overrightarrow n_2|}$$=\cos^{-1} \bigg[ \large\frac{(2)(-1)+(-3)(1)+0}{\sqrt{4+9+16}\sqrt{1+1}} \bigg]$
$ = \cos^{-1} \large\frac{-5}{\sqrt{58}}$
answered Jan 30, 2014 by balaji

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App