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  • If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2),

Q15   If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find \angle ABC , [\angle ABC is the angle between the vectors \overline{BA}\: \: and\: \: \overline{BC} ] .

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best_answer

Given points,

A=(1, 2, 3),

B=(–1, 0, 0),

C=(0, 1, 2),

As need to find Angle between  \overline{BA}\: \: and\: \: \overline{BC} ]

\vec {BA}=(1-(-1))\hat i+(2-0)\hat j+(3-0)\hat k=2\hat i+2\hat j+3\hat k

\vec {BC}=(0-(-1))\hat i+(1-0)\hat j+(2-0)\hat k=\hat i+\hat j+2\hat k

Hence angle between them ;

\theta=cos^{-1}(\frac{\vec {BA}.\vec {BC}}{\left | \vec {BA} \right |\left | \vec {BC} \right |})

\theta=cos^{-1}\frac{2+2+6}{\sqrt{17}\sqrt{6}}

\theta=cos^{-1}\frac{10}{\sqrt{102}}

Answer- Angle between the vectors \overline{BA}\: \: and\: \: \overline{BC} is  \theta=cos^{-1}\frac{10}{\sqrt{102}}

Posted by

Pankaj Sanodiya

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