Find the equation of the plane in the xyz-space through the point P=(3,4,2) and perpendicular to the vector n = (-5,-5,5)

z = ________

Question

Find the equation of the plane in the xyz-space through the point P=(3,4,2) and perpendicular to the vector n = (-5,-5,5)

z = ________

amistre64 · Answer

yay were back lol

anonymous · Answer

The equation of the plane is the dot product of P and n.

amistre64 · Answer

they give you the normal; and a point, so all we do is attach it all

anonymous · Answer

ya...i'm aware...i'm getting marked wrong though

anonymous · Answer

Could you post some of your workings?

anonymous · Answer

i get it as -5(x-3) -5(y-4) + 5(z-2)

amistre64 · Answer

-5<1,1,-1>

(x-Px)+(y-Py)-(z-Pz)=0

amistre64 · Answer

your right, but reduce it

anonymous · Answer

right...distribute i get -5x -5y + 5z = -25

amistre64 · Answer

and they might not like a -leading coeff

anonymous · Answer

Looks good to me. You can make it simpler though.

anonymous · Answer

well i keep getting it marked wrong and i cant figure out why

amistre64 · Answer

format, thats all

anonymous · Answer

it wants the z = obv.....so if i plug in 2 arbitrary constants for x and y

amistre64 · Answer

x+y-z-5=0 is what i would put for it

anonymous · Answer

hang on lemme try that

amistre64 · Answer

err.... z = x+y-5  :)

anonymous · Answer

I would say to rearrange it to be z = x + y -5.

anonymous · Answer

yes that worked

anonymous · Answer

u guys are awesome thanks a ton

amistre64 · Answer

yep :)