Answer:
d=10u
Q(5/3,5/3,-19/3)
Step-by-step explanation:
The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane [tex]n=(-2,-2,1)[/tex], then r will have the next parametric equations:
[tex]x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda[/tex]
To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T
[tex]-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3[/tex]
Substitute the value of [tex]\lambda[/tex] in the parametric equations:
[tex]x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\[/tex]
Those values are the coordinates of Q
Q(5/3,5/3,-19/3)
The distance from Po to the plane
[tex]d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u[/tex]
