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

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Question

(A) Let the point be $A(2, 1, -3)$ and the origin be $O(0, 0, 0)$.For a plane passing through a fixed point $A$ to have a maximum distance from the or

Vedclass · Accepted Answer

$2x + y - 3z = 14$

Vedclass · Answer

(A) Let the point be $A(2, 1, -3)$ and the origin be $O(0, 0, 0)$.For a plane passing through a fixed point $A$ to have a maximum distance from the origin $O$,the normal vector $\vec{n}$ of the plane must be the vector $\vec{OA}$.Thus,the normal vector is $\vec{n} = \vec{OA} = 2\hat{i} + \hat{j} - 3\hat{k}$.The equation of a plane passing through point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Here,$\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ and $\vec{n} = 2\hat{i} + \hat{j} - 3\hat{k}$.So,the equation is $2x + y - 3z = (2)(2) + (1)(1) + (-3)(-3)$.$2x + y - 3z = 4 + 1 + 9 = 14$.Therefore,the equation of the plane is $2x + y - 3z = 14$.

$A$ 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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