If O is the origin and the coordinates of P a | vedclass

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Question

(A) The coordinates of the origin $O$ are $(0, 0, 0)$ and the point $P$ are $(1, 2, -3).$The direction ratios of the line segment $OP$ are $(1-0, 2-0,

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$x+2y-3z-14=0$

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(A) The coordinates of the origin $O$ are $(0, 0, 0)$ and the point $P$ are $(1, 2, -3).$The direction ratios of the line segment $OP$ are $(1-0, 2-0, -3-0) = (1, 2, -3).$Since the plane is perpendicular to $OP,$ the direction ratios of the normal to the plane are the same as the direction ratios of $OP,$ which are $a=1, b=2, c=-3.$The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal direction ratios $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0.$Substituting the point $P(1, 2, -3)$ and the direction ratios $(1, 2, -3)$ into the formula:$1(x-1) + 2(y-2) - 3(z - (-3)) = 0$$1(x-1) + 2(y-2) - 3(z+3) = 0$$x - 1 + 2y - 4 - 3z - 9 = 0$$x + 2y - 3z - 14 = 0.$

If $O$ is the origin and the coordinates of $P$ are $(1, 2, -3),$ find the equation of the plane passing through $P$ and perpendicular to $OP.$

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