Question

If the points $A(1,-2,2),B(3,1,1),C(-1,p,3)$ are collinear, then find the value of $p$.

Answer 1

Given points are A(1,-2,2) B(3,1,1) and C(-1,p,3)

If three points are collinear, then they lie on a line.

Let first calculate distance between the three points AB,BC and CA.

Hence

$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(3-1{)}^2+(1-(-2){)}^2+(1-2{)}^2}$

$=\sqrt{4+9+1}$

$AB=\sqrt{14}$.

$BC=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}=\sqrt{(-1-3{)}^2+(p-1{)}^2+(3-1{)}^2}$

$=\sqrt{16+(p-1{)}^2+4}$

$=\sqrt{20+(p-1{)}^2}$

$CA=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}=\sqrt{(1-(-1){)}^2+(-2-p{)}^2+(2-3{)}^2}$

$=\sqrt{4+(-2-p{)}^2+1}$

BC=CA

$\sqrt{16+(p-1{)}^2+4}=\sqrt{4+(-2-p{)}^2+1}$

$16+p^2+1-2p+4=4+4+p^2-4p+1$

$2-2p=9-4p$

$4p-2p=21-9$

$2p=12$

$P=12/2$

$\therefore P=6$