Question

The points $ (1,3,4),\,(-1,6,10),\,(-7,4,7) $ and $ (-5,1,1) $

• A. form a rectangle which is not a square
• B. form a rhombus which is not a square
• C. form a parallelogram which is not a rhombus
• D. are collinear

Answer 1

The Correct Option is B

Given points are $ A(1,3,4),B(-1,6,10),C(-7,4,7) $ and $ D(-5,1,1). $ $ AB=\sqrt{{{(-1-1)}^{2}}+{{(6-3)}^{2}}+{{(10-4)}^{2}}} $
$=\sqrt{{{(-2)}^{2}}+{{(3)}^{2}}+{{(6)}^{2}}} $
$=\sqrt{4+9+36}=\sqrt{49}=7 $ $ BC=\sqrt{{{(-7+1)}^{2}}+{{(4-6)}^{2}}+{{(7-10)}^{2}}} $
$=\sqrt{{{(-6)}^{2}}+{{(-2)}^{2}}+{{(-3)}^{2}}} $
$=\sqrt{36+4+9}=\sqrt{49}=7 $ $ CD=\sqrt{{{(-5+7)}^{2}}+{{(1-4)}^{2}}+{{(1-7)}^{2}}} $
$=\sqrt{{{(2)}^{2}}+{{(-3)}^{2}}+{{(-6)}^{2}}} $
$=\sqrt{4+9+36}=7 $ $ AD=\sqrt{{{(-5-1)}^{2}}+{{(1-3)}^{2}}+{{(1-4)}^{2}}} $
$=\sqrt{{{(-6)}^{2}}+{{(-2)}^{2}}+{{(-3)}^{2}}} $
$=\sqrt{36+4+9}=7 $ $ AC=\sqrt{{{(-7-1)}^{2}}+{{(4-3)}^{2}}+{{(7-4)}^{2}}} $
$=\sqrt{{{(-8)}^{2}}+{{(1)}^{2}}+{{(3)}^{2}}} $
$=\sqrt{64+1+9}=\sqrt{74} $ $ BD=\sqrt{{{(-5+1)}^{2}}+{{(1-6)}^{2}}+{{(1-10)}^{2}}} $
$=\sqrt{{{(-4)}^{2}}+{{(-5)}^{2}}+{{(-9)}^{2}}} $
$=\sqrt{16+25+81}=\sqrt{112} $ Since, sides $ AB=BC=CD=AD=7 $ and diagonals, $ AC\ne BD $ So, given points form a rhombus which is not a square.