A) \[{{45}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{30}^{o}}\]
Correct Answer: C
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, \[A{{B}^{2}}+C{{A}^{2}}={{(3\sqrt{2})}^{2}}+{{(6)}^{2}}\] \[=18+36=54\] \[=B{{C}^{2}}\] \[\therefore \]\[\Delta ABC\]is right angled triangle, right angled at\[A\]. \[\therefore \] \[\angle A={{90}^{o}}\]
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