Equation for Plane through (-5, 1, 2) Perpendicular to n = 3 i - 5 j + 2 k

[Square1]

Find an equation for the plane through
(−5, 1, 2) and perpendicular to n = 3 i − 5 j + 2 k.



They say that :
The plane has an equation
3(x + 5) − 5(y − 1) + 2(z − 2) = 0

What confuses me is why they switched the add/subtract signs and the order of everything when in my notes they tell us the following:

vector P'knot'·P = (x − x'knot', y − y'knot', z − z'knot').
Thus, n ·P'knot'·P = 0 is equivalent to a(x − x'knot') + b(y − y'knot') + c(z − z'knot') = 0


Thanks in advance

 

[tiny-tim]

Hi Square!

Find an equation for the plane through
(−5, 1, 2) and perpendicular to n = 3 i − 5 j + 2 k.



They say that :
The plane has an equation
3(x + 5) − 5(y − 1) + 2(z − 2) = 0

Never mind the formula (which btw i can't read ) …

isn't it obvious that the LHS will be zero if x = -5, y = 1, z = 2, and not for the "minus" values?

 

[Mark44]

P₀ sounds like P "knot" but it's really P "nought". A knot is formed by joining two pieces of rope; nought means "nothing" or "zero".

You can make subscripts and exponents by clicking Go Advanced, which opens the advanced menu across the top of the input window. Use the X² button to make exponents, and use the X₂ to make subscripts.

 

[Square1]

Yes tim, yes mark and thanks mark. :)

Ok I got it though I misread the question. Lesson learned - don't do homework at stupid hours.