Let be the plane passing through the intersection of the planes r( i+3 j- k)=5 and r(2 i- …

Question

Let $\pi$ be the plane passing through the intersection of the planes $\vec r\cdot(\hat i+3\hat j-\hat k)=5$ and $\vec r\cdot(2\hat i-\hat j+\hat k)=3$, and the point $(2,1,-2)$. Let the position vectors of the points $X$ and $Y$ be $\hat i-2\hat j+4\hat k$ and $5\hat i-\hat j+2\hat k$ respectively. Then the points:

Accepted Answer

Let the equation of required plane

$$\pi :(x + 3y - z - 5) + \lambda (2x - y + z - 3) = 0$$

$$\because$$ (2, 1, -2) lies on it so, $$2 + \lambda ( - 2) = 0$$

$$ \Rightarrow \lambda  = 1$$

Hence, $$\pi :3x + 2y - 8 = 0$$

$$\because$$ $${\pi _x} =  - 9$$, $${\pi _y} = 5$$, $${\pi _{x + y}} = 4$$

$${\pi _{x - y}} =  - 22$$ and $${\pi _{y - x}} = 6$$

Clearly X and Y are on opposite sides of plane $$\pi$$

Solutions · 1