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 -

Vedclass · Accepted Answer

$2x + y - 3z = 14$

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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,$|2a + b - 3c| \leq \sqrt{2^2 + 1^2 + (-3)^2} \sqrt{a^2 + b^2 + c^2} = \sqrt{14} \sqrt{a^2 + b^2 + c^2}$.Thus,$d = \frac{|2a + b - 3c|}{\sqrt{a^2 + b^2 + c^2}} \leq \sqrt{14}$.The maximum distance occurs when the normal vector $\vec{n}$ is parallel to the position vector $\vec{OA} = 2\hat{i} + \hat{j} - 3\hat{k}$.Setting $(a, b, c) = (2, 1, -3)$,the equation becomes $2(x - 2) + 1(y - 1) - 3(z + 3) = 0$.$2x - 4 + y - 1 - 3z - 9 = 0 \implies 2x + y - 3z = 14$.

$A$ 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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