# Thread: Two short questions on vectors

1. ## Two short questions on vectors

Hi
The following question i am having problems with:

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

$2x+y-3z = d$

$|d|=5$
$d= 5$ or $-5$

so

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

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

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

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

$A1,6,1), B3,5,2) and C3,2,1)" alt="A1,6,1), B3,5,2) and C3,2,1)" />

$n=AB*AC$
$n=(OB-OA) * (OC-OA)$
$n=((3i+5j+2k)-(i+6j+k)) * ((3i+2j+k)-(i+6j+k))$
$n=(2i-j+k)*(2i-4j)$

$
\begin{bmatrix}
i & j & k \\
2 & -1 & 1 \\
2 & -4 & 0 \\
\end{bmatrix}$

$n=i(0-4)-j(0-2)+k(-8+2)$
$n=4i+2j+6k$

At plane (1,6,1)
4(1)+2(6)-6(1)=d
4+12-6=d
d=10

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

however book's answer is $2x+y-3z=5$
P.S

2. Originally Posted by Paymemoney
Hi
The following question i am having problems with:

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

$2x+y-3z = d$

$|d|=5$
$d= 5$ or $-5$

so

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

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

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

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

$A1,6,1), B3,5,2) and C3,2,1)" alt="A1,6,1), B3,5,2) and C3,2,1)" />

$n=AB*AC$
$n=(OB-OA) * (OC-OA)$
$n=((3i+5j+2k)-(i+6j+k)) * ((3i+2j+k)-(i+6j+k))$
$n=(2i-j+k)*(2i-4j)$

$
\begin{bmatrix}
i & j & k \\
2 & -1 & 1 \\
2 & -4 & 0 \\
\end{bmatrix}$

$n=i(0-4)-j(0-2)+k(-8+2)$
$n=4i+2j+6k$

At plane (1,6,1)
4(1)+2(6)-6(1)=d
4+12-6=d
d=10

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

however book's answer is $2x+y-3z=5$
P.S
There is a common factor of 2 on the LHS of your equation.

$4x + 2y + 6z = 10$

$2(2x + y + 3z) = 10$

$2x + y + 3z = 5$.

So you are correct.

3. oh ok thanks, any ideas about question 1)

4. The first question sounds strange to me. All equations of the form

$2x+y-3z=d$

are equations for planes with $2i+j-3k$ as a normal vector, regardless of what $d$ is. Is there some extra information, like a point the plane must pass through?

Where did the expression $|d|=5$ come from? Was $P$ meant to be $d$?

As for question 2, notice that you have a slight disagreement with the book's answer. While you have $2x+y+3z=5$, the book has $2x+y-3z=5$ (so notice the $-3$ on the $z$).

5. Originally Posted by HappyJoe
Is there some extra information, like a point the plane must pass through?

Where did the expression $|d|=5$ come from? Was $P$ meant to be $d$?
well the full description of the question is:

Find the equation of the plane normal to the vector 2i+j-3k and a distance 5 from the origin.

This equation of the plane is given by $P = \frac{|d|}{\sqrt{a^2+b^2+c^2}}$

Originally Posted by HappyJoe
As for question 2, notice that you have a slight disagreement with the book's answer. While you have $2x+y+3z=5$, the book has $2x+y-3z=5$ (so notice the $-3$ on the $z$).

6. Ah, I see.

You write that the equation of the plane is given by $P = \frac{|d|}{\sqrt{a^2+b^2+c^2}}$. You probably didn't mean this. This expression is the distance from the plane to the origin.

You are being told that this distance is 5, so that $P=5$. You already have $a$, $b$, and $c$:

$a = 2,$
$b = 1,$
$c = -3.$

So insert these pieces of information into the formula for the distance from the plane to the origin:

$5 = \frac{|d|}{\sqrt{2^2+1^2+(-3)^2}},$

or

$|d| = 5\sqrt{14},$

giving you the two required solutions.

7. yes, but isn't it $\frac{5}{\sqrt{14}}$, how did you get $5\sqrt{14}$ that is what i don't understand

8. Check that last bit of algebra carefully,

$5 = \frac{|d|}{\sqrt{14}}$

$5\sqrt{14} = |d|$

$|d| = 5\sqrt{14}$

9. oh =__+