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  • Show that the vectors 2i − j + k, i− 3 j − 5k and 3i − 4 j − 4k form the vertices of a right angled triangle.

Q17  Show that the vectors 2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k  form the vertices of a right angled triangle.

 

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Given the position vector of A, B , and C are 

2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k

To show that the vectors 2 \hat i - \hat j + \hat k , \hat i - 3 \hat j - 5 \hat k \: \: and \: \: 3 \hat i - 4 \hat j - 4 \hat k  form the vertices of a right angled triangle

\vec {AB}=(1-2)\hat i + (-3-(-1))\hat j+(-5-1)\hat k=-1\hat i -2\hat j-6\hat k

\vec {BC}=(3-1)\hat i + (-4-(-3))\hat j+(-4-(-5))\hat k=-2\hat i -\hat j+\hat k

\vec {AC}=(3-2)\hat i + (-4-(-1))\hat j+(-4-(1))\hat k=\hat i -3\hat j-5\hat k

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

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

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

Here we see that 

|\vec {AC}|^2+|\vec {BC}|^2=|\vec {AB}|^2

Hence A,B, and C are the vertices of a right angle triangle.

Posted by

Pankaj Sanodiya

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