Hi

The following question i am having problems with:

1) Find the equation of the plane normal to the vector 2i+j-3k

$\displaystyle 2x+y-3z = d$

$\displaystyle |d|=5$

$\displaystyle d= 5 $ or $\displaystyle -5$

so

$\displaystyle P = \frac{5}{\sqrt{14}}$ or $\displaystyle \frac{-5}{\sqrt{14}}$

so $\displaystyle 2x+y-3z = \frac{5}{\sqrt{14}}$ or $\displaystyle 2x+y-3z = \frac{-5}{\sqrt{14}}$

book's answer says $\displaystyle 2x+y-3z =5\sqrt{14}$ or $\displaystyle 2x+y-3z =-5\sqrt{14}$

2) Find the equation of the plane which passes through the points (1,6,1), (3,5,2) and (3,2,1)

$\displaystyle A

1,6,1), B

3,5,2) and C

3,2,1)$

$\displaystyle n=AB*AC$

$\displaystyle n=(OB-OA) * (OC-OA)$

$\displaystyle n=((3i+5j+2k)-(i+6j+k)) * ((3i+2j+k)-(i+6j+k))$

$\displaystyle n=(2i-j+k)*(2i-4j)$

$\displaystyle

\begin{bmatrix}

i & j & k \\

2 & -1 & 1 \\

2 & -4 & 0 \\

\end{bmatrix}$

$\displaystyle n=i(0-4)-j(0-2)+k(-8+2)$

$\displaystyle n=4i+2j+6k$

At plane (1,6,1)

4(1)+2(6)-6(1)=d

4+12-6=d

d=10

therefore equation is $\displaystyle 4x+2y+6z=10

$

however book's answer is $\displaystyle 2x+y-3z=5$

P.S