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Q.
If the coordinate of the vertices of a triangle $ ABC $ be $ A(-1, 3, 2), $ $ B(2, 3, 5) $ and $ C(3, 5, -2) $ then $ \angle A $ is equal to:
Jharkhand CECEJharkhand CECE 2005
Solution:
Given that coordinates of the vertices of a triangle $ ABC $ are
$ A(-1, 3, 2),\,B(2, 3, 5) $ and $ C(3,\,5,\,-2) $.
$ DR's $ of a line $ AB $ is $ (3,\,0,\,3) $
and $ CA $ is $ (-4,\,-2,\,4) $ .
$ \therefore \cos \theta =\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}}} $
$ =\frac{3(-4)+0(-2)+3(4)}{\sqrt{9+0+9}\sqrt{16+4+16}}=0 $
Alternative Solution:
Given vertices of a triangle are
$ A(-1,\,3,\,2),\,B(2,\,3,\,5) $ and $ C(3,\,5,\,-2) $.
$ \therefore AB=\sqrt{(2+1)^{2}+(3-3)^{2}+(5-2)^{2}} $
$ =\sqrt{9+0+9} =3\sqrt{2} $
$ BC=\sqrt{(3-2)^{2}+(5-3)^{2}+(-2-5)^{2}} $
$ =\sqrt{1+4+49} =3\sqrt{6} $
$ CA=\sqrt{(3+1)^{2}+(5-3)^{2}+(-2-2)^{2}} $
$ =\sqrt{16+4+16} =6 $
Now, $ AB^{2}+CA^{2}=(3\sqrt{2})^{2}+(6)^{2} $
$ =18+36=54 $
$ =BC^{2} $
$ \therefore \Delta ABC $ is right angled triangle, right angled at $ A $ .
$ \therefore \angle A=90^{\circ} $