If A and B are the feet of the perpendiculars | vedclass

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(A) The coordinates of point $Q$ are $(a, b, c)$.The foot of the perpendicular from $Q(a, b, c)$ to the $yz$-plane $(x=0)$ is $A(0, b, c)$.The foot of

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The coordinates of point $Q$ are $(a, b, c)$.The foot of the perpendicular from $Q(a, b, c)$ to the $yz$-plane $(x=0)$ is $A(0, b, c)$.The foot of the perpendicular from $Q(a, b, c)$ to the $zx$-plane $(y=0)$ is $B(a, 0, c)$.The origin is $O(0, 0, 0)$.The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant:$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}ight|=0$Substituting $O(0, 0, 0)$,$A(0, b, c)$,and $B(a, 0, c)$:$\left|\begin{array}{ccc} x & y & z \ 0 & b & c \ a & 0 & c \end{array}ight|=0$Expanding the determinant along the first row:$x(bc - 0) - y(0 - ac) + z(0 - ab) = 0$$bcx + acy - abz = 0$Dividing the entire equation by $abc$ $(abc 
eq 0)$:$\frac{bcx}{abc} + \frac{acy}{abc} - \frac{abz}{abc} = 0$$\frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$

If $A$ and $B$ are the feet of the perpendiculars drawn from point $Q(a, b, c)$ to the planes $yz$ and $zx$ respectively,then the equation of the plane passing through the points $A, B$ and the origin $O(0, 0, 0)$ is $.........$

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