Show that the points A ( 2 i - j + k ),B ( i - 3 j - 5 k ),C ( 3 i - 4 j - 4 k ) are the vertices of a right angled triangle.

AB = - i - 2 j - 6 k BC=2 i- j+ k CA = - i + 3 j + 5 k | AB |^2 = 1^2+2^2+6^2 = 41 | BC |^2 = 2^2+1^2+1^2=6 | CA |^2 = 1^2+3^2+5^2=35 | AB |^2 = | BC |^2 + | CA |^2 Hence, the is a right angled triangle.

Show that the points A ( 2 i - j + k ),B ( i - 3 j - 5 k ),C ( 3 …

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Question

Show that the points $A\left( {2\hat i - \hat j + \hat k} \right),B\left( {\hat i - 3\hat j - 5\hat k} \right),C\left( {3\hat i - 4\hat j - 4\hat k} \right)$ are the vertices of a right angled triangle.

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Answer

$\overrightarrow {AB} = - \hat i - 2\hat j - 6\hat k$
$\vec{BC}=2\vec i-\vec j+\vec k$
$\overrightarrow {CA} = - \hat i + 3\hat j + 5\hat k$
${\left| {\overrightarrow {AB} } \right|^2} ={1^2+2^2+6^2}= 41$
$\left| {\overrightarrow {BC} } \right|^2 = 2^2+1^2+1^2=6$
$\left| {\overrightarrow {CA} } \right|^2 = 1^2+3^2+5^2=35$
${\left| {\overrightarrow {AB} } \right|^2} = {\left| {\overrightarrow {BC} } \right|^2} + {\left| {\overrightarrow {CA} } \right|^2}$
Hence, the $\Delta $ is a right angled triangle.

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