Question

lf $OABC$ is a tetrahedron such that the $O{A}^2+B{C}^2=O{B}^2+C{A}^2=O{C}^2+A{B}^2$, then which of the following is/are correct

• A. $AB\perp OC$
• B. $OB\ne CA$
• C. $OC=AB$
• D. $AB\perp BC$

Answer 1

Answer: (A)

$AB\perp OC$

$OABC$ is a tetrahedron such that $O{A}^2+B{C}^2=O{B}^2+C{A}^2=O{C}^2+A{B}^2$
Let $O(0,0,0),A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3)$ are coordinates of vertices.
$O{A}^2+B{C}^2$ = $(x_1^2+y_1^2+z_1^2)+\left(\right(x_3-x_2{)}^2+(y_3-y_2{)}^2+(z_3-z_2{)}^2)$
= $(x_1^2+y_1^2+z_1^2)+(x_2^2+y_2^2+z_2^2)+(x_3^2+y_3^2+z_3^2)-2(x_2{x}_3+y_2{y}_3+z_2{z}_3)$
Similarly,
$O{B}^2+C{A}^2$ = $(x_1^2+y_1^2+z_1^2)+(x_2^2+y_2^2+z_2^2)+(x_3^2+y_3^2+z_3^2)-2(x_1{x}_3+y_1{y}_3+z_1{z}_3)$
$O{C}^2+A{B}^2$ = $(x_1^2+y_1^2+z_1^2)+(x_2^2+y_2^2+z_2^2)+(x_3^2+y_3^2+z_3^2)-2(x_1{x}_2+y_1{y}_2+z_1{z}_2)$
$\therefore$ $O{A}^2+B{C}^2=O{B}^2+C{A}^2\Rightarrow x_2{x}_3+y_2{y}_3+z_2{z}_3=x_1{x}_3+y_1{y}_3+z_1{z}_3$
$\Rightarrow (x_1-x_2)x_3+(y_1-y_2)y_3+(z_1-z_2)z_3=0$

$\Rightarrow (x_1-x_2)(x_3-0)+(y_1-y_2)(y_3-0)+(z_1-z_2)(z_3-0)=0$

$\Rightarrow \overrightarrow{AB}\cdot \overrightarrow{OC}=0$
Hence, $OC$ is perpendicular to $AB.$