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Q.17   Show that the points A, B and C with position vectors, \vec a = 3 \hat i - 4 \hat j - 4 \hat k , \vec b = 2 \hat i - \hat j + \hat k \: \: and \: \: \: \vec c = \hat i - 3 \hat j - 5 \hat k, respectively form the vertices of a right angled triangle.

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Given 

the position vector of A, B, and C are

\\\vec a = 3 \hat i - 4 \hat j - 4 \hat k ,\\\vec b = 2 \hat i - \hat j + \hat k \: \: and \\\vec c = \hat i - 3 \hat j - 5 \hat k

Now,

\vec {AB}=\vec b-\vec a=-\hat i+3\hat j+5\hat k

\vec {BC}=\vec c-\vec b=-\hat i-2\hat j-6\hat k

\vec {CA}=\vec a-\vec c=2\hat i-\hat j+\hat k

\left | \vec {AB} \right |=\sqrt{(-1)^2+3^2+5^2}=\sqrt{35}

\left | \vec {BC} \right |=\sqrt{(-1)^2+(-2)^2+(-6)^2}=\sqrt{41}

\left | \vec {CA} \right |=\sqrt{(2)^2+(-1)^2+(1)^2}=\sqrt{6}

AS we can see 

\left | \vec {BC} \right |^2=\left | \vec {CA} \right |^2+\left | \vec {AB} \right |^2

Hence ABC is a right angle triangle.

Posted by

Pankaj Sanodiya

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