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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Show that the points A, B and C with position vectors, \( \overrightarrow a = 3\hat i − 4\hat j − 4\hat k, \overrightarrow b = 2\hat i − \hat j + \hat k\) and \( \overrightarrow c = \hat i − 3\hat j − 5\hat k\) , respectively form the vertices of a right angled triangle.

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  • In a $\Delta$ABC, if ${AB}^2+{BC}^2={AC}^2$,then it obeys Pythagoras rule and hence it is a right angled triangle.
Step 1:
Let $\overrightarrow a=3\hat i-4\hat j-4\hat k,\overrightarrow{b}=2\hat i-\hat j+\hat k$ and $\overrightarrow c=\hat i-3\hat j-5\hat k$
Let us check if the three sides of the triangle obeys Pythagoras rule.
To check this,let us find the magnitudes of the three vectors.
Step 2:
Step 3:
It is clear that $b^2+c^2=a^2$
(i.e)$\qquad\quad(\sqrt 6)^2+(\sqrt{35})^2=(\sqrt{41})^2$
$\Rightarrow 6+35=41.$
Hence the vectors form the sides of a right angled triangle.
answered May 17, 2013 by sreemathi.v

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