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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the equation of the plane passing through the point $A(2, 1, -3)$ be $a(x - 2) + b(y - 1) + c(z + 3) = 0$,where $\vec{n} = (a, b, c)$ is the n

Vedclass · Accepted Answer

$2x + y - 3z = 14$

Vedclass · Answer

(A) Let the equation of the plane passing through the point $A(2, 1, -3)$ be $a(x - 2) + b(y - 1) + c(z + 3) = 0$,where $\vec{n} = (a, b, c)$ is the normal vector to the plane.The distance $d$ of the plane from the origin $(0, 0, 0)$ is given by $d = \frac{|-2a - b + 3c|}{\sqrt{a^2 + b^2 + c^2}}$.To maximize the distance $d$,the normal vector $\vec{n}$ must be parallel to the position vector $\vec{OA} = (2, 1, -3)$.Thus,we set $(a, b, c) = (2, 1, -3)$.The equation of the plane becomes $2(x - 2) + 1(y - 1) - 3(z + 3) = 0$.Expanding this,we get $2x - 4 + y - 1 - 3z - 9 = 0$,which simplifies to $2x + y - 3z = 14$.

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

Similar Questions

The angle between the planes $3x - 4y + 5z = 0$ and $2x - y - 2z = 5$ is

The Cartesian equation of the plane passing through the point $(1, -2, 3)$ and perpendicular to the vector $-\hat{i} + 2\hat{j} - 3\hat{k}$ is:

The equations $x + y + z + 2 = 0$ and $x + y + z + 3 = 0$ together represent what in space?

$A$ plane passing through the points $A(2, 3, 5)$ and $B(-3, -5, -7)$ is perpendicular to the plane $x - y + z = 1$. Which of the following points lies on this plane?

Find the equation of the plane that passes through the three points $(1,1,0), (1,2,1),$ and $(-2,2,-1)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module