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# Show that the points A, B and C with position vectors, and , respectively form the vertices of a right angled triangle.

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.

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