Closest Point Plane 9727E7

1. **State the problem:** We need to find the point $P$ on the plane given by the equation $$2x + 3y - z = 1$$ that is closest to the point $(2, 1, 0)$, and then find the distance between $P$ and the plane. 2. **Formula and concept:** The closest point on a plane to a given point is the orthogonal projection of that point onto the plane. The distance from a point $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is given by: $$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$ 3. **Rewrite the plane equation:** The plane equation is: $$2x + 3y - z = 1$$ Rewrite as: $$2x + 3y - z - 1 = 0$$ So, $A=2$, $B=3$, $C=-1$, and $D=-1$. 4. **Calculate the distance from point $(2,1,0)$ to the plane:** $$d = \frac{|2(2) + 3(1) - 1(0) - 1|}{\sqrt{2^2 + 3^2 + (-1)^2}} = \frac{|4 + 3 - 0 - 1|}{\sqrt{4 + 9 + 1}} = \frac{|6|}{\sqrt{14}} = \frac{6}{\sqrt{14}}$$ 5. **Find the closest point $P$ on the plane:** The vector normal to the plane is: $$\vec{n} = (2, 3, -1)$$ The point $P$ is the projection of $(2,1,0)$ onto the plane along the direction of $\vec{n}$: Let $P = (x, y, z) = (2, 1, 0) - t(2, 3, -1)$ for some scalar $t$. 6. **Substitute $P$ into the plane equation:** $$2x + 3y - z = 1$$ Substitute $x = 2 - 2t$, $y = 1 - 3t$, $z = 0 + t$: $$2(2 - 2t) + 3(1 - 3t) - (t) = 1$$ Simplify: $$4 - 4t + 3 - 9t - t = 1$$ $$7 - 14t = 1$$ 7. **Solve for $t$:** $$7 - 14t = 1 \implies -14t = 1 - 7 = -6 \implies t = \frac{6}{14} = \frac{3}{7}$$ 8. **Find coordinates of $P$:** $$x = 2 - 2 \times \frac{3}{7} = 2 - \frac{6}{7} = \frac{14}{7} - \frac{6}{7} = \frac{8}{7}$$ $$y = 1 - 3 \times \frac{3}{7} = 1 - \frac{9}{7} = \frac{7}{7} - \frac{9}{7} = -\frac{2}{7}$$ $$z = 0 + \frac{3}{7} = \frac{3}{7}$$ So, the closest point is: $$P = \left( \frac{8}{7}, -\frac{2}{7}, \frac{3}{7} \right)$$ 9. **Distance between $P$ and the plane:** Since $P$ lies on the plane, the distance is zero. 10. **Distance between original point and plane:** As calculated in step 4, the distance from $(2,1,0)$ to the plane is: $$\frac{6}{\sqrt{14}}$$ **Final answers:** - Closest point $P$ on the plane: $$\left( \frac{8}{7}, -\frac{2}{7}, \frac{3}{7} \right)$$ - Distance from $(2,1,0)$ to the plane: $$\frac{6}{\sqrt{14}}$$