MHB Unit vector perpendicular to a plane.

[Drain Brain]

Find in rectangular coordinates a unit vector which is: A. in the direction of E at P(2,3,-4)if $\overline{E}=(x^2+y^2+z^2)\left(\frac{xa_{x}}{\sqrt{y^2+z^3}}+\frac{ya_{y}}{\sqrt{x^2+z^2}}+\frac{za_{z}}{\sqrt{x^2+y^2}}\right)$; B. Perpendicular to the plane passing through M(1,-5,5), N(-2,4,0) and Q(2,3,4) and having a positive $x$ component.

I managed to solve for a.

$E=11.6a_{x}+19.46a_{y}-32.161a_{z}$ at P(2,3,-4)

then, $\overline{a_{E}}=\frac{E}{|E|}=0.295a_{x}+0.495a_{y}-0.818a_{z}$

what I did for prob B was I find all the possible cross products of the vectors defined by those 3 points given above. I get different results and one of them matched the key answer in my book. Are those results that I get from taking all the cross products of the vectors on the plane defined by the given points valid?

$\overline{MQ}\times\overline{NQ}$ by the way, this is the cross product that matched the answer in my book.

can you help me with prob B. TIA!

 

[Prove It]

Find in rectangular coordinates a unit vector which is: A. in the direction of E at P(2,3,-4)if $\overline{E}=(x^2+y^2+z^2)\left(\frac{xa_{x}}{\sqrt{y^2+z^3}}+\frac{ya_{y}}{\sqrt{x^2+z^2}}+\frac{za_{z}}{\sqrt{x^2+y^2}}\right)$; B. Perpendicular to the plane passing through M(1,-5,5), N(-2,4,0) and Q(2,3,4) and having a positive $x$ component.

I managed to solve for a.

$E=11.6a_{x}+19.46a_{y}-32.161a_{z}$ at P(2,3,-4)

then, $\overline{a_{E}}=\frac{E}{|E|}=0.295a_{x}+0.495a_{y}-0.818a_{z}$

what I did for prob B was I find all the possible cross products of the vectors defined by those 3 points given above. I get different results and one of them matched the key answer in my book. Are those results that I get from taking all the cross products of the vectors on the plane defined by the given points valid?

$\overline{MQ}\times\overline{NQ}$ by the way, this is the cross product that matched the answer in my book.

can you help me with prob B. TIA!

You know that the points (1, -5, 5), (-2, 4, 0) and (2, 3, 4) lie in the plane which has equation $\displaystyle \begin{align*} a\,x + b\,y+ c\,z = d \end{align*}$, where a,b,c,d are constants. So substituting these values in gives the system of equations

$\displaystyle \begin{align*} \phantom{-}1a - 5b + 5c &= d \\ -2a + 4b + 0c &= d \\ \phantom{-}2a + 3b + 4c &= d \end{align*}$

Once you solve the system for a,b,c in terms of d, you will be able to write down a vector that is normal to the plane. Dividing by d and its length will give you a unit vector in that direction.

 

[Drain Brain]

You know that the points (1, -5, 5), (-2, 4, 0) and (2, 3, 4) lie in the plane which has equation $\displaystyle \begin{align*} a\,x + b\,y+ c\,z = d \end{align*}$, where a,b,c,d are constants. So substituting these values in gives the system of equations

$\displaystyle \begin{align*} \phantom{-}1a - 5b + 5c &= d \\ -2a + 4b + 0c &= d \\ \phantom{-}2a + 3b + 4c &= d \end{align*}$

Once you solve the system for a,b,c in terms of d, you will be able to write down a vector that is normal to the plane. Dividing by d and its length will give you a unit vector in that direction.

Is there other quick method than this?
what can you say about the cross product that I mentiond above? It gave me the right answer though.

 

[Fallen Angel]

Hi,

Your method was right, and even simpler, given three not collinear points in the space, you can choose two vectors between them that will generate the whole plane containing that three points, and the crossproduct will give you a perpendicular vector to the plane, and if you flip the order of the vectors into the cross product you will get the same answer but with the sign changed.

 

[Drain Brain]

Hi, Fallen Angel! thanks!