From a point P(a, b, c), perpendiculars PA an | vedclass

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(A) $1$. The coordinates of point $P$ are $(a, b, c)$.$2$. The perpendicular $PA$ is drawn to the $XY$ plane. The $XY$ plane has the equation $z = 0$.

Vedclass · Accepted Answer

$\frac{x}{a} - \frac{y}{b} - \frac{z}{c} = 0$

Vedclass · Answer

(A) $1$. The coordinates of point $P$ are $(a, b, c)$.$2$. The perpendicular $PA$ is drawn to the $XY$ plane. The $XY$ plane has the equation $z = 0$. Thus, the coordinates of $A$ are $(a, b, 0)$.$3$. The perpendicular $PB$ is drawn to the $ZX$ plane. The $ZX$ plane has the equation $y = 0$. Thus, the coordinates of $B$ are $(a, 0, c)$.$4$. The origin $O$ is $(0, 0, 0)$.$5$. The equation of a plane passing through $(0, 0, 0)$, $(a, b, 0)$, and $(a, 0, c)$ is given by the determinant form:$\begin{vmatrix} x & y & z \ a & b & 0 \ a & 0 & c \end{vmatrix} = 0$$6$. Expanding along the first row: $x(bc - 0) - y(ac - 0) + z(0 - ab) = 0$$7$. $xbc - yac - zab = 0$$8$. Dividing the entire equation by $abc$: $\frac{xbc}{abc} - \frac{yac}{abc} - \frac{zab}{abc} = 0$$9$. $\frac{x}{a} - \frac{y}{b} - \frac{z}{c} = 0$.

From a point $P(a, b, c)$, perpendiculars $PA$ and $PB$ are drawn to $XY$ plane and $ZX$ plane respectively. If $O$ is the origin, then the equation of plane $OAB$ is

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