Question

If $ \theta $ is the angle between the planes $ 2x-y+z-1=0 $ and $ x-2y+z+2=0, $ then $ \cos \,\theta $ is equal to

• A. $ 2/3 $
• B. $ 3/4 $
• C. $ 4/5 $
• D. $ 5/6 $ s

Answer 1

Answer: (D)

$ \cos \theta =\left| \frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \right| $
Here, $ {{a}_{1}}=2,\,{{b}_{1}}=-1,\,{{c}_{1}}=1,\,{{d}_{1}}=-1 $
and $ {{a}_{2}}=1,{{b}_{2}}=-2,\,{{c}_{2}}=1,\,{{d}_{2}}=2 $
$ \therefore $ $ \cos \theta =\left| \frac{(2\times 1)+(-1\times -2)+(1\times 1)}{\sqrt{4+1+1}\sqrt{1+4+1}} \right| $
$ =\left| \frac{2+2+1}{\sqrt{6}\sqrt{6}} \right| $
$ \cos \theta =\left| \frac{5}{6} \right| $ s