A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,2,1)$. The distance of the plane from the point $(2,3,4)$ is

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Aicharya · Accepted Answer

- A plane perpendicular to two given planes passes through $(1,2,1)$. The distance of the plane from the point $(2,3,4)$. 1. Let the plane equation passing through $(1,2,1)$ be: $a(x-1)+b(y-2)+c(z-1)=0$ 2. The plane is perpendicular to the two given planes: - $2 x-2 y+z=0$ - $x-y+2 z=4$
Thus, the normal vector of the required plane is perpendicular to the normal vectors of these planes. 3. Solve the normal vectors using the cross-product method: - Normal to plane 1: $\overrightarrow{n_1}=(2,-2,1)$ - Normal to plane 2: $\overrightarrow{n_2}=(1,-1,2)$ - Cross product $\overrightarrow{n_1} \times \overrightarrow{n_2}$ gives the required normal. 4. Use the distance formula for the point $(2,3,4)$ to calculate the final distance from the plane. 5. Answer: Option $2 \frac{2 \sqrt{2}}{5}$ units.

A plane which is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, passes through $(1,2,1)$. The…

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