Question

Show that the four points vectors $4\hat{i}+8\hat{j}+12\hat{k},2\hat{i}+4\hat{j}+6\hat{k},3\hat{i}+5\hat{j}+4\hat{k}$ and $5\hat{i}+8\hat{j}+5\hat{k}$ are co-planar.

Answer 1

Let

$\overrightarrow{a}=4\hat{i}+8\hat{j}+12\hat{k}$

$\overrightarrow{b}=2\hat{i}+4\hat{j}+6\hat{k}$

$\overrightarrow{c}=3\hat{i}+5\hat{j}+4\hat{k}$

$\overrightarrow{d}=5\hat{i}+8\hat{j}+5\hat{k}$

For four points to be coplanar, they must satisfy the condition $xa+yb+zc+td=0$.

Since, $x,y,z,t$ are scalars, so the condition should be $x+y+z+t=0$.

Let $t=1$, then,

$xa+yb+zc+d=0$

$x\left(4\hat{i}+8\hat{j}+12\hat{k}\right)+y\left(2\hat{i}+4\hat{j}+6\hat{k}\right)+z\left(3\hat{i}+5\hat{j}+4\hat{k}\right)+5\hat{i}+8\hat{j}+5\hat{k}=0$

$\left(4x+2y+3z+5\right)\hat{i}+\left(8x+4y+5z+8\right)\hat{j}+\left(12x+6y+4z+5\right)\hat{k}=0$

This gives,

$4x+2y+3z+5=0$

$8x+4y+5z+8=0$

$12x+6y+4z+5=0$

$x+y+z+1=0$

Solving these four equations, we get,

$x=-\frac{1}{2},y=\frac{3}{2},z=-2,t=1$.

The sum of the scalars $x=-\frac{1}{2},y=\frac{3}{2},z=-2,t=1$ is 0, therefore,

$-\frac{1}{2}a+\frac{3}{2}b-2c+1d=0$

Hence, the given four points are coplanar.