Question

Find $x$ if $△ABC$ is right angled at $A$, where $A\equiv (4,2,3),B\equiv (3,1,8),C\equiv (x,-1,2)$.

Answer 1

Given $A(4,2,3)B(3,1,8)C(x,-1,2)$

is right angled at -A so $B{C}^2=A{B}^2+A{C}^2$

$A{B}^2=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}=\sqrt{(3-4{)}^2+(1-2{)}^2+\left)\right(8-3{)}^2}$

$A{B}^2=\sqrt{1+1+25}\Rightarrow A{B}^2=27$

$B{C}^2=(x-3{)}^2+(-1-1{)}^2+(2-8{)}^2=(x-3{)}^2+4+36=(x-3{)}^2+40$

$A{C}^2=(x-4{)}^2+(-1-2{)}^2+(2-3{)}^2=(x-4{)}^2+9+1=(x-4{)}^2+10.$

$B{C}^2=A{B}^2+A{C}^2$

$(x-3{)}^2+40=27+(x-4{)}^2+10$

$x^2+9-6x+40=37+x^2+16-8x$

$-6x+49+8x-53=0$

$2x-4=0$

$x=2$