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Question: The vertices of a triangle are \((2,3,5),( - 1,3,2),(3,5, - 2)\) then the angles are A. \({30^ \ci...

The vertices of a triangle are (2,3,5),(−1,3,2),(3,5,−2)(2,3,5),( - 1,3,2),(3,5, - 2) then the angles are
A. 30∘,30∘,30∘{30^ \circ },{30^ \circ },{30^ \circ }
B. cos⁡−1(15),90∘,cos⁡−1(53){\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 5 }}} \right),{90^ \circ },{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 5 }}{{\sqrt 3 }}} \right)
C. 30∘,60∘,90∘{30^ \circ },{60^ \circ },{90^ \circ }
D. cos⁡−1(13),90∘,cos⁡−1(23){\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right),{90^ \circ },{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{{\sqrt 3 }}} \right)

Explanation

Solution

Hint: Convert whole scenario into vector form, find the position vector then use the information, we can find the angles using the formula cos⁡θ=∣P1→.P2→∣∣P1→∣.∣P2→∣\cos \theta = \dfrac{{\left| {\overrightarrow {{P_1}} .\overrightarrow {{P_2}} } \right|}}{{\left| {\overrightarrow {{P_1}} } \right|.\left| {\overrightarrow {{P_2}} } \right|}}. Here, P stands for position vector, ‘.’ Stands for dot product and ‘||’ is the argument as can be understood by ∣ai^+bj^+ck^∣=a2+b2+c2\left| {a\hat i + b\hat j + c\hat k} \right| = \sqrt {{a^2} + {b^2} + {c^2}} .

Complete step-by-step answer:
We have given three vertices. Let name these vertices as A(2,3,5)A(2,3,5) , B(−1,3,2)B( - 1,3,2) , C(3,5,−2)C(3,5, - 2). Observe that, there are three dimensional points so co-ordinate geometry would not work here. Therefore, we’ll convert the whole scenario into vector form. A will be 2i^+3j^+5k^2\hat i + 3\hat j + 5\hat k , B will be −i^+3j^+2k^ - \hat i + 3\hat j + 2\hat k and C will be 3i^+5j^−2k^3\hat i + 5\hat j - 2\hat k. In order to get the angle, we need to find direction vectors as follows,

AB→=−i^+3j^+2k^−2i^−3j^−5k^=−3i^−3k^ BC→=3i^+5j^−2k^+i^−3j^−2k^=4i^+2j^−4k^ CA→=2i^+3j^+5k^−3i^−5j^+2k^=−i^−2j^+7k^  \overrightarrow {AB} = - \hat i + 3\hat j + 2\hat k - 2\hat i - 3\hat j - 5\hat k = - 3\hat i - 3\hat k \\\ \overrightarrow {BC} = 3\hat i + 5\hat j - 2\hat k + \hat i - 3\hat j - 2\hat k = 4\hat i + 2\hat j - 4\hat k \\\ \overrightarrow {CA} = 2\hat i + 3\hat j + 5\hat k - 3\hat i - 5\hat j + 2\hat k = - \hat i - 2\hat j + 7\hat k \\\

Since, we have got our position vectors, now we can find the angles using the formula cos⁡θ=∣P1→.P2→∣∣P1→∣.∣P2→∣\cos \theta = \dfrac{{\left| {\overrightarrow {{P_1}} .\overrightarrow {{P_2}} } \right|}}{{\left| {\overrightarrow {{P_1}} } \right|.\left| {\overrightarrow {{P_2}} } \right|}}. Here, P stands for position vector, ‘.’ Stands for dot product and ‘||’ is the argument as can be understood by ∣ai^+bj^+ck^∣=a2+b2+c2\left| {a\hat i + b\hat j + c\hat k} \right| = \sqrt {{a^2} + {b^2} + {c^2}} . Now let’s calculate the angles using this information as follows,

cos⁡A=∣AB→.CA→∣∣AB→∣.∣CA→∣ =∣(−3i^−3k^).(−i^−2j^+7k^)∣∣−3i^−3k^∣.∣−i^−2j^+7k^∣ =∣3+0−21∣(−3)2+(−3)2.(−1)2+(−2)2+(7)2 =1818.54 =1854 =13 ∴cos⁡A=13 A=cos⁡−113  \cos A = \dfrac{{\left| {\overrightarrow {AB} .\overrightarrow {CA} } \right|}}{{\left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {CA} } \right|}} \\\ = \dfrac{{\left| {\left( { - 3\hat i - 3\hat k} \right).\left( { - \hat i - 2\hat j + 7\hat k} \right)} \right|}}{{\left| { - 3\hat i - 3\hat k} \right|.\left| { - \hat i - 2\hat j + 7\hat k} \right|}} \\\ = \dfrac{{\left| {3 + 0 - 21} \right|}}{{\sqrt {{{( - 3)}^2} + {{( - 3)}^2}} .\sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{(7)}^2}} }} \\\ = \dfrac{{18}}{{\sqrt {18} .\sqrt {54} }} \\\ = \dfrac{{\sqrt {18} }}{{\sqrt {54} }} \\\ = \dfrac{1}{{\sqrt 3 }} \\\ \therefore \cos A = \dfrac{1}{{\sqrt 3 }} \\\ A = {\cos ^{ - 1}}\dfrac{1}{{\sqrt 3 }} \\\

Similarly,

cos⁡B=∣AB→.BC→∣∣AB→∣.∣BC→∣ =∣(−3i^−3k^).(4i^+2j^−4k^)∣∣−3i^−3k^∣.∣4i^+2j^−4k^∣ =∣−12+12∣(−3)2+(−3)2.(4)2+(2)2+(−4)2 =0 ∴cos⁡B=0 B=90∘  \cos B = \dfrac{{\left| {\overrightarrow {AB} .\overrightarrow {BC} } \right|}}{{\left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {BC} } \right|}} \\\ = \dfrac{{\left| {\left( { - 3\hat i - 3\hat k} \right).\left( {4\hat i + 2\hat j - 4\hat k} \right)} \right|}}{{\left| { - 3\hat i - 3\hat k} \right|.\left| {4\hat i + 2\hat j - 4\hat k} \right|}} \\\ = \dfrac{{\left| { - 12 + 12} \right|}}{{\sqrt {{{( - 3)}^2} + {{( - 3)}^2}} .\sqrt {{{(4)}^2} + {{(2)}^2} + {{( - 4)}^2}} }} \\\ = 0 \\\ \therefore \cos B = 0 \\\ B = {90^ \circ } \\\

And finally,

cos⁡C=∣CA→.BC→∣∣CA→∣.∣BC→∣ =∣(−i^−2j^+7k^).(4i^+2j^−4k^)∣∣−i^−2j^+7k^∣.∣4i^+2j^−4k^∣ =∣−4−4−28∣(−1)2+(−2)2+(7)2.(4)2+(2)2+(−4)2 =3654.36 =3654 =23 ∴cos⁡C=23 C=cos⁡−1(23)  \cos C = \dfrac{{\left| {\overrightarrow {CA} .\overrightarrow {BC} } \right|}}{{\left| {\overrightarrow {CA} } \right|.\left| {\overrightarrow {BC} } \right|}} \\\ = \dfrac{{\left| {\left( { - \hat i - 2\hat j + 7\hat k} \right).\left( {4\hat i + 2\hat j - 4\hat k} \right)} \right|}}{{\left| { - \hat i - 2\hat j + 7\hat k} \right|.\left| {4\hat i + 2\hat j - 4\hat k} \right|}} \\\ = \dfrac{{\left| { - 4 - 4 - 28} \right|}}{{\sqrt {{{( - 1)}^2} + {{( - 2)}^2} + {{(7)}^2}} .\sqrt {{{(4)}^2} + {{(2)}^2} + {{( - 4)}^2}} }} \\\ = \dfrac{{36}}{{\sqrt {54} .\sqrt {36} }} \\\ = \dfrac{{\sqrt {36} }}{{\sqrt {54} }} \\\ = \dfrac{{\sqrt 2 }}{{\sqrt 3 }} \\\ \therefore \cos C = \dfrac{{\sqrt 2 }}{{\sqrt 3 }} \\\ C = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{{\sqrt 3 }}} \right) \\\

Hence the required angles are cos⁡−1(13),90∘,cos⁡−1(23){\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 3 }}} \right),{90^ \circ },{\cos ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{{\sqrt 3 }}} \right) and the correct option is D.

Note: There is one more method to solve this question in which we’ll find these lines as in our solution we have calculated position vectors. Then we use the angle formula for three-dimensional geometry and get the result.