Question

The position-vectors of $A,B,C$ are $(6,0,1),(8,-3,7)$ and $(2,-5,10)$ respectively. Prove that there exists a point $D$ such that $ABCD$ is a rhombus. Find the position vector of $D.$

Answer 1

Given $A(6,0,1)$, $B(8,-3,7)$, $C(2,-5,10)$

Let $D=(x,y,z)$

Vertices $A,B,C$ & $D$ have position vectors

Now, for $ABCD$ to be rhombus,

$AD\perp BC$

i.e, $AD.BC=0$

i.e, $6(6-x)+2(-y)+(-3)(1-z)=0$

i.e, $36-6x-2y-3+3z=0$

or, $6x+2y-3z-33=0⟶\left(i\right)$

$AB\parallel CD$

i.e, $AB\times CD=0$

$\Rightarrow$ $|ijk-23-62-x-5-y10-z|=0$

$\Rightarrow 3(10-z)+6(-5-y)=0$

$2(3-10)+6(2-x)=0$

$2(5+y)-3(2-x)=0$

i.e, $30-3z-30-6y=0\Rightarrow z+3y=0⟶\left(ii\right)$

$2z-20+12-6x=0\Rightarrow 3x-z+4=0⟶\left(iii\right)$

$10+2y-6+3x=0\Rightarrow 3x+2y+4=0⟶\left(iv\right)$

$\Rightarrow x=-4/3,y=0,z=0$

$\therefore$ Position vector of $D=(-4/3,0,0)$

i.e, $-4/3i+0j+ok$