If A(-1,3,2), B(2,3,5) and C(3,5,-2) are vertices of a A B C, then angles of A B C are

Correct Answer is : None of the above

Correct Answer is : None of the above

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CGPET 2014 Solved Paper

Section: Mathematics

Question : 144 of 150

Marks: +1, -0

A(-1,3,2), B(2,3,5)

C(3,5,-2)

\triangle A B C

\triangle A B C

Solution:

Given vertices of a

\triangle A B C

are

A(-1,3,2), B(2,3,5)

and

C(3,5,-2)

Now DR's of

AB=(2+1,3-3,5-2)

=(3,0,3)

DR's of

BC=(3-2,5-3,-2-5)

=(1,2,-7)

and DR's of

CA=(-1-3,3-5,2+2)

=(-4,-2,4)

Now, the angle between

AB

and

BC

\cos B = \frac{\left|3 \times 1 + 0 \times 2 + 3 \times (-7)\right|}{\sqrt{3^2+0^2+3^2} \sqrt{1^2+2^2+(-7)^2}}

= \frac{\left|3+0-21\right|}{\sqrt{9+0+9} \sqrt{1+4+49}}

= \frac{18}{3\sqrt{2} \times 3\sqrt{6}}

= \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}

angle between

BC

and

CA

\cos C = \frac{\left|1 \times (-4) + 2(-2) + (-7)(4)\right|}{\sqrt{1^2+2^2+(-7)^2} \sqrt{(-4)^2+(-2)^2+4^2}}

= \frac{\left|-4-4-28\right|}{\sqrt{1+4+49} \sqrt{16+4+16}}

= \frac{36}{\sqrt{54} \sqrt{36}} = \frac{36}{3\sqrt{6} \times 6}

= \frac{2}{\sqrt{2}\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}}

and angle between

AC

and

AB

\cos A = \frac{\left|-4 \times 3 + (-2) \times 0 + 4 \times 3\right|}{\sqrt{(-4)^2+(-2)^2+4^2} \sqrt{3^2+0^2+3^2}}

= \left|0\right|

\Rightarrow \quad A=90^{\circ}

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