Question

The points $(0,0,0,),(0,2,0),(1,0,0),(0,0,4)$ are

• A. coplanar
• B. vertices of a parallelogram
• C. vertices of a rectangle
• D. on a sphere

Answer 1

Answer: (D)

on a sphere

$\left(a\right)$Consider the points

$A(0,0,0),B(0,2,0),C(1,0,0),D(0,0,4)$

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$

$\left|\begin{array}{ccc}x-0& y-0& z-0\\ 0-0& 2-0& 0-0\\ 1-0& 0-0& 0-0\end{array}\right|=0$

$\left|\begin{array}{ccc}x& y& z\\ 0& 2& 0\\ 1& 0& 0\end{array}\right|=0$

$\Rightarrow x(0-0)-y(0-0)+z(0-2)=0$

$\Rightarrow -2z=0$

Substituting the last point $(0,0,4)$ we get

$-2z=-2\times 4\ne 0$

Hence the given points are not co-planar.

$\left(d\right)$The general equation of a sphere is $x^2+y^2+2ux+2vy+2wz+d=0$ ......$\left(1\right)$

It passes through $A(0,0,0),B(0,2,0),C(1,0,0),D(0,0,4)$

Substituting $A(0,0,0)$ in $\left(1\right)$ we get $0+0+0+0+0+0+d=0$ or $d=0$

Substituting $B(0,2,0)$ in $\left(1\right)$ we get

$0+4+0+0+4v+0+d=0$ or $4v+d=-4$

$\Rightarrow 4v=-4$ since $d=0$

$\therefore v=-1$

Substituting $C(1,0,0)$ in $\left(1\right)$ we get $1+0+0+2u+0+0+d=0$

or $2u+d=-1$

$\Rightarrow 2u=-1$ since $d=0$

$\therefore u=-\frac{1}{2}$

Substituting $D(0,0,4)$ in $\left(1\right)$ we get $0+0+16+0+0+8w+d=0$ or $8w+d=-16$

$\Rightarrow 8w=-16$ since $d=0$

$\therefore w=-\frac{16}{8}=-2$

Hence $u=-2,v=-\frac{1}{2},w=-2$

Centre of the sphere is $(-2,-\frac{1}{2},-2)$

Radius$=\sqrt{{(-2)}^2+{(-\frac{1}{2})}^2+{(-2)}^2}=\sqrt{4+\frac{1}{4}+4}=\sqrt{8+\frac{1}{4}}=\sqrt{\frac{32+1}{4}}=\frac{\sqrt{33}}{4}>0$

Hence the given points form a sphere.