Find the equation of the plane that passes through $P(2,-1,3)$ and is perpendicular to the line $\frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-0}{5}$.

Step-by-step solution

1. Identify direction vector of given line: $\mathbf{d}=(4,,-2,,5)$.
2. Since the plane is perpendicular to the line, its normal vector $\mathbf{n}$ is parallel to $\mathbf{d}$, so $\mathbf{n}=(4,,-2,,5)$.
3. Use point‑normal form: $\mathbf{n}cdot(\mathbf{r}-{\mathbf{r}}_0)=0$, where ${\mathbf{r}}_0=(2,-1,3)$.
4. Compute dot product: $4(x-2)-2(y+1)+5(z-3)=0$.
5. Expand: $4x-8-2y-2+5z-15=0$.
6. Collect like terms: $4x-2y+5z-25=0$.
7. Rearrange: $4x-2y+5z=25$.
8. Hence the required plane equation is $4x-2y+5z=25$.

Concept tested: Plane perpendicular to line

Published 31 August 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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