A plane passes through the point A(2, 1, -3). | vedclass

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(A) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}.$ The equation of the plane passing through $A(2, 1, -3)$ is $a(x-

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$2x+y-3z=14$

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(A) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}.$ The equation of the plane passing through $A(2, 1, -3)$ is $a(x-2) + b(y-1) + c(z+3) = 0,$ which simplifies to $ax + by + cz = 2a + b - 3c.$The distance $d$ of this plane from the origin $(0, 0, 0)$ is given by $d = \frac{|2a + b - 3c|}{\sqrt{a^2 + b^2 + c^2}}.$By the Cauchy-Schwarz inequality,for any vectors $\vec{u} = (2, 1, -3)$ and $\vec{n} = (a, b, c),$ we have $|\vec{u} \cdot \vec{n}| \leq |\vec{u}| |\vec{n}|.$ Thus,$d = \frac{|\vec{u} \cdot \vec{n}|}{|\vec{n}|} \leq |\vec{u}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}.$The distance is maximum when the normal vector $\vec{n}$ is parallel to the position vector $\vec{OA} = 2\hat{i} + \hat{j} - 3\hat{k}.$Setting $\vec{n} = (2, 1, -3),$ the equation of the plane becomes $2(x-2) + 1(y-1) - 3(z+3) = 0,$$2x - 4 + y - 1 - 3z - 9 = 0,$$2x + y - 3z = 14.$

$A$ plane passes through the point $A(2, 1, -3).$ If the distance of this plane from the origin is maximum,then its equation is

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