Solveeit Logo
Login

Question

Mathematics Question on introduction to three dimensional geometry

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

form a rhombus which is not a square

Explanation

Solution

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