Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z.
we can measur ethe perpendicular dist d ten find out angle theta between normal and line given and then puut our ans=d/costheta
is this ok????
Moderation Team
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![[tex] \\ \mbox{pt is} \ (3,4,5) \\ 2x=y=z \ \mbox{corresponds to} \ \frac{x-0}{\frac{1}{2}} = \frac{y-0}{1} = \frac{z-0}{1} \\ \\ \mbox{The equation of line through } \ (3,4,5) \ \mbox{and parallel to } \ 2x=y=z \ \mbox{is} \\ \\ \frac{x-3}{\frac{1}{2}} = \frac{y-4}{1} = \frac{z-5}{1} = \alpha \\ \\ x = \frac{\alpha}{2}+3 , y = \alpha+4 , z = \alpha+5 \\ \\ \mbox{Now this pt lies on the plane } \ x+y+z = 2 \\ \\ \mbox{Thus} \ \frac{\alpha}{2}+3+\alpha+4+\alpha+5 = 2 \\ \\ \mbox{Solving} \ \alpha = -4 \\ \mbox{and} \ x=1,y=0,z=1 \\ \\ \mbox{Now distance between} \ (1,0,1) \ \mbox{and} \ (3,4,5) \ \mbox{is what we have to find which is } \ \ 6 \ \mbox{units} [\tex]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/0/5/305b17f077f9b375ac0fc7c7be351264e2a54c39.gif)
