If the position vectors of the vertices of $△ A B C$ are $(0, 1, 1), (3, 1, 5)$ and $(0, 3, 3)$ respectively, then show that $∠ C$ is a right angle.

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Solution

$- - \to B C = 3^i - 2^j + 2^b$
$- - \to A C = - 2^j - 2^b$
$- - \to B C \cdot - - \to A C = 4 - 4 = 0$
$∴ - - \to B C$ is perpendicular to $- - \to A C$
$∴ ∠ C = 90^{\circ}$

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If the position vectors of the vertices of △ABC are (0,1,1),(3,1,5) and (0,3,3) respectively, then show that ∠C is a right angle.

−−→BC=3^i−2^j+2^b −−→AC=−2^j−2^b−−→BC⋅−−→AC=4−4=0 ∴−−→BC is perpendicular to −−→AC ∴∠C=90∘

If the position vectors of the vertices of $△ A B C$ are $(0, 1, 1), (3, 1, 5)$ and $(0, 3, 3)$ respectively, then show that $∠ C$ is a right angle.

$- - \to B C = 3^i - 2^j + 2^b$
$- - \to A C = - 2^j - 2^b$
$- - \to B C \cdot - - \to A C = 4 - 4 = 0$
$∴ - - \to B C$ is perpendicular to $- - \to A C$
$∴ ∠ C = 90^{\circ}$