If P is the point (2, 6, 3),then the equation | vedclass

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(D) The coordinates of point $P$ are $(2, 6, 3)$ and the origin $O$ is $(0, 0, 0)$.The vector $\vec{OP}$ acts as the normal vector to the plane,so $\v

Vedclass · Accepted Answer

$2x + 6y + 3z = 49$

Vedclass · Answer

(D) The coordinates of point $P$ are $(2, 6, 3)$ and the origin $O$ is $(0, 0, 0)$.The vector $\vec{OP}$ acts as the normal vector to the plane,so $\vec{n} = \vec{OP} = 2\hat{i} + 6\hat{j} + 3\hat{k}$.The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.Substituting the values $a=2, b=6, c=3$ and $(x_1, y_1, z_1) = (2, 6, 3)$:$2(x - 2) + 6(y - 6) + 3(z - 3) = 0$Expanding the equation:$2x - 4 + 6y - 36 + 3z - 9 = 0$$2x + 6y + 3z - 49 = 0$$2x + 6y + 3z = 49$.Thus,the correct option is $D$.

If $P$ is the point $(2, 6, 3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$,where $O$ is the origin,is:

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