Answer to Question #91546 in Analytic Geometry for Ra

Question #91546
Find the distance of the point of intersection of the line
x-2/1=y+3/-1=z/3
and the plane 2x-3y+4z+4=0 from the origin.
1
Expert's answer
2019-07-15T09:35:29-0400
x21=y+31=z3=t{x=t+2y=t3z=3t\frac{x-2}{1}=\frac{y+3}{-1}=\frac{z}{3}=t\\ \begin{cases} x=t+2\\ y=-t-3\\ z=3t \end{cases}

then


2x3y+4z+4==2t+4+3t+9+12t+4=17t+17=0    t=1{x=1y=2z=32x-3y+4z+4=\\ =2t+4+3t+9+12t+4=17t+17=0\\ \implies t=-1\\ \begin{cases} x=1\\ y=-2\\ z=-3 \end{cases}

and the distance from the origin:

r=(1)2+(2)2+(3)2=14r=\sqrt{(1)^2+(-2)^2+(-3)^2} = \sqrt{14}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment