If the coordinates of point P are (2, 6, 3),w | vedclass

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Question

(D) The coordinates of point $P$ are $(2, 6, 3)$ and the origin $O$ is $(0, 0, 0)$.The vector $\vec{OP}$ is given by $(2 - 0)\hat{i} + (6 - 0)\hat{j}

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

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

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

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