A variable plane passes through a fixed point | vedclass

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(B) Let the coordinates of points $A, B,$ and $C$ be $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively.The equation of the plane passing through $A

Vedclass · Accepted Answer

$\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1$

Vedclass · Answer

(B) Let the coordinates of points $A, B,$ and $C$ be $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively.The equation of the plane passing through $A, B,$ and $C$ is given by the intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ... $(i)$.Since the plane passes through the fixed point $(\alpha, \beta, \gamma)$,we have: $\frac{\alpha}{a} + \frac{\beta}{b} + \frac{\gamma}{c} = 1$.The plane $P_1$ passes through $A(a, 0, 0)$ and is parallel to the $YZ$-plane,so its equation is $x = a$.The plane $P_2$ passes through $B(0, b, 0)$ and is parallel to the $ZX$-plane,so its equation is $y = b$.The plane $P_3$ passes through $C(0, 0, c)$ and is parallel to the $XY$-plane,so its equation is $z = c$.The point of intersection of these three planes $P_1, P_2,$ and $P_3$ is $(a, b, c)$.Let the coordinates of this point of intersection be $(x, y, z)$. Thus,$x = a, y = b,$ and $z = c$.Substituting these into the equation $\frac{\alpha}{a} + \frac{\beta}{b} + \frac{\gamma}{c} = 1$,we get the locus: $\frac{\alpha}{x} + \frac{\beta}{y} + \frac{\gamma}{z} = 1$.

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