Question

If $A(0,4,1),B(a,b,c),C(4,5,0),D(2,6,2)$ are the consecutive vertices of a square, then the distance $BD$ is:

• A. $\sqrt{(a+2{)}^2+(b+6{)}^2+(c+2{)}^2}$
• B. $\sqrt{34}$
• C. $18$
• D. $3\sqrt{2}$

Answer 1

The correct option is D $3\sqrt{2}$

Side of square $CD=\sqrt{(4-2{)}^2+(5-6{)}^2+(0-2{)}^2}=3$
So, the distance $BD=$the length of diagonal $=\sqrt{2}\left(CD\right)=3\sqrt{2}$