Question

Find equation of plane passing through the point (1,0,1) (1,-1,1) (4,-3,2)

Answer 1

The equation of plane passing through three points is x - 3z + 2 = 0.

Step-by-step explanation:

Let,

$P_{1} = (x_{1} ,y_{1} ,z_{1} ) = (1,0,1)\\P_{2} = (x_{2},y_{2},z_{2}) = (1,-1,1)\\P_{3} = (x_{3},y_{3},z_{3}) = (4,-3,2)$

Equation of plane passing through three points is,

$\left[\begin{array}{ccc}(x-x_{1})&(y - y_{1})&(z-z_{1})\\(x_{2}-x_{1})&(y_{2}-y_{1})&(z_{2}-z_{1})\\{(x_{3}-x_{1})&(y_{3}-y_{1})&(z_{3}-z_{1})\end{array}\right] = 0$........................(i)

Plug the values in equation (i),

$\left[\begin{array}{ccc}(x-1)&(y-0)&(z-1)\\(1-1)&(-1-0)&(1-1)\\(4-1)&(-3-0)&(2-1)\end{array}\right] =0$

$\left[\begin{array}{ccc}(x-1)&y&(z-1)\\0&-1&0\\3&-3&1\end{array}\right] =0$

=> (x - 1)(-1 - 0) - y(0 - 0) + (z - 1)(0 + 3) = 0

=> -x + 1 + 3z - 3 = 0

=> -x + 3z - 2 = 0

=> x - 3z + 2 = 0

Hence, the required equation of plane is x - 3z + 2 = 0.