Vectors: Find dist. from P(2,5,5) to line x=0, y=5+5t, z=5+t

[chucknorrisfish]

Find the distance from P(2,5,5) to the line x=0,y=5+5t,z=5+t
Can anybody help me with this problem. I know the formula:

d=QPxQR / |QR|

Not sure how to get point Q and R tho.

Also i have something similar that i have tried many ways but still isn't right:
Find the distance from the point (1,3,-3) to the plane -3x-y+2z=1.

Here's what i have done:

P1: (1,3,-3)
P0(point on the plane)-3.-1,5/2)

P0P1(disp)4,4,-5)

so then i took -3(4)-1(4)+2(-5) / sqrt(-3,-1,2)

gave me: -26/sqrt(14) which isnt correct, i tried a couple other things as well but they don't work either.

 

[pka]

First of all the correct formula is \(\displaystyle \L d = \frac{{\left\| {\vec{QP} \times \vec{QR} } \right\|}}{{\left\| {\vec{QR} } \right\|}}\).

Both Q & R are points on the line. They an be any two points.

 

[galactus]

Re: Vectors: Find dist. from P(2,5,5) to line x=0, y=5+5t, z

Find the distance from the point (1,3,-3) to the plane -3x-y+2z=1.

Here's what i have done:

P1: (1,3,-3)
P0(point on the plane)-3.-1,5/2)

P0P1(disp)4,4,-5)

so then i took -3(4)-1(4)+2(-5) / sqrt(-3,-1,2)

gave me: -26/sqrt(14) which isnt correct, i tried a couple other things as well but they don't work either.

\(\displaystyle \L\\D=\frac{|-3(1)-1(3)+2(-3)-1|}{\sqrt{(-3)^{2}+(-1)^{2}+(2)^{2}}}=\frac{13}{\sqrt{14}}\)

Remember the absolute value. You can not have a negative distance.