Vector equation of a plane in normal form

[Krushnaraj Pandya]

Homework Statement

A vector n of magnitude 8 units is inclined to x,y and z axis at 45, 60 and 60 degrees resoectively.If the plane passes through (root2, -1, 1) and is normal to n then find its equation.

Homework Equations

(r-a).n=0 where r is position vector of a point on plane, a is a point on the plane and n is a normal vector...(i)
sum of direction cosine^2=1

The Attempt at a Solution

Suppose the normal is ai + bj + ck, its magnitude= sqrt(a^2+b^2+c^2)=8 and direction cosines a/8 , b/8 and c/8- now a/8 = 1/sqrt2 and so on, equating the direction cosines. So the equation of normal is 4√2i + 4j + 4k, also the plane passes through a-(√2,-1,1) so a is √2i-j+k. Putting a and the normal in (i) gives r.(√2i+j+k)=2 but my textbook says its 2i instead of √2i. I have the feeling I'm making a small mistake but can't find it even though I checked- Is there a mistake in my book?

 

[TachyonLord]

Your solution seems to be correct. I can't really seem to find any errors, I mean the fact that it's inclined at 45° has to bring in a √2 term. There might be a mistake in the book.

 

[Krushnaraj Pandya]

Your solution seems to be correct. I can't really seem to find any errors, I mean the fact that it's inclined at 45° has to bring in a √2 term. There might be a mistake in the book.

Alright, thank you very much :D