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Q. The vertices of a triangle are $i+2 j+4 k,-2 i+2 j+k$ and $2 i+4 j-3 k$. The triangle is

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Solution:

$\overrightarrow{O A}=i+2 j+4 k ; \overrightarrow{O B}=-2 i+2 j+k, \overrightarrow{O C}=2 i+4 j-3 k$
$\overrightarrow{A B}=-3 i-3 k ; \overrightarrow{B C}=4 i+2 j-4 k ; \overrightarrow{A C}=i+2 j-7 k$
$|\overrightarrow{A B}|=\sqrt{18} ;|\overrightarrow{B C}|=6 ;|\overrightarrow{A C}|=\sqrt{54} $
$ \therefore|\overrightarrow{A C}|^{2}=|\overrightarrow{A B}|^{2}+|\overrightarrow{B C}|^{2}$
Hence it is a right angled triangle