Question

If the equation of the plane passing through the points (1,2,3) ,(-1,2,0) and perpendicular to the ZX-plane is ax+by+cz+d=0(a>0).

Evaluate for following conditions:
1) a=0 and c=0
2) a+d=0
3) c+d-5 =0

Answer 1

$ax+by+cz+d$

$pts.(1,2,3)$ and $(-1,2,0)$

$a(x-1)+b(y-2)+c(3-3)+d=0------\left(I\right)$

$a(-1,-1)+b(+2-2)+c(0-3)+d=0$

$-2a-3+d=0----\left(II\right)$

$ax+by+cz+d=0$

$a+b+c+d=0-------\left(III\right)$

Equation $\left(II\right)$ to $\left(III\right)$

$\frac{a}{\left|\begin{array}{cc}0& -3\\ 1& 1\end{array}\right|}=\frac{b}{\left|\begin{array}{cc}-2& -3\\ 1& 1\end{array}\right|}=\frac{c}{\left|\begin{array}{cc}-2& 0\\ 1& 1\end{array}\right|}$

$\frac{a}{3}=\frac{b}{1}=\frac{c}{2}=\lambda$

$\Rightarrow a3\lambda ,b=\lambda ,c=2\lambda$

$3\lambda (x-1)+\lambda (y-2)-2\lambda (3-3)+d=0$ (for equation $\left(I\right)$)

$3\lambda x-3\lambda +\lambda y-2\lambda -2\lambda 3+\lambda +d=0$

$3\lambda x+\lambda y-2\lambda 3+\lambda +d=0$

$\Rightarrow 3a+b-2c+\lambda +d=0------\left(IV\right)$

$1)a.$ when $abc=0$

$b+d+\lambda =0$ (for equation $\left(IV\right)$)

$2)a+d=0$ when $a=-d$ or $d=-a$

$2a+b-2c+\lambda =0$ (prove equation $\left(IV\right)$)

$3)$ when $c+d-5=0d=5-c$ there, $3a+b-2c+5-c+\lambda =0$

$3a+b-3c+\lambda =-5$

or $3a+b-3c+\lambda +5=0$ (prove equation $\left(IV\right)$)