Have a few vector questions I cant seem to get my head around.

[knowlewj01]

1. Homework Statement

Ok, so here are the questions, I literally have no idea how to begin on them.

1.Show that the vector 2i+3j+3k is normal to the plane containing the points (1,1,-2), (4,2,-5) and (1,-1,0). Hence find the equation of the plane in the form
(r-a).n(hat)

what is the perpendicular distance from the origin to the plane?


There are a couple more questions but i'll take them one at a time.


3. The Attempt at a Solution

Been looking through my notes for ages and on th internet, can't find anything that makes sense, anyone good with vectors? thanks :)
If anyone knows of a good vectors tutorial then i would be happy to see it.

 

[anathema]

2. Response

Hi there! I can definitely help you with these questions. Let's start with the first one.

To show that the vector 2i+3j+3k is normal to the plane containing the given points, we can use the fact that the dot product of two perpendicular vectors is equal to 0. So, we can set up the following equation:

(2i+3j+3k) • (r-a) = 0

Where (r-a) is a vector on the plane and the dot product represents the angle between the two vectors. Now, we can substitute in any of the given points for (r-a), let's use (1,1,-2):

(2i+3j+3k) • (1,1,-2) = 0

Simplifying this, we get:

2+3-6 = 0

Therefore, the vector 2i+3j+3k is indeed normal to the plane containing the given points.

To find the equation of the plane in the form (r-a).n(hat), we can use the cross product of two vectors on the plane to find the normal vector (n(hat)). Let's use the vectors (1,1,-2) and (4,2,-5) as our two vectors. The cross product is given by:

(n(hat)) = (1,1,-2) x (4,2,-5)

= (1*2 - (-2)*2, (-2)*4 - 1*(-5), 1*2 - 1*4)

= (4, -3, -2)

Now, we can use this normal vector and one of the given points (let's use (1,-1,0)) to write the equation of the plane as:

(r- (1,-1,0)) • (4,-3,-2) = 0

Simplifying this, we get:

4x - 3y - 2z = 7

So, the equation of the plane is 4x - 3y - 2z = 7.

To find the perpendicular distance from the origin to the plane, we can use the formula:

d = |ax + by + cz + d| / √(a^2 + b^2 + c^2)

Where a, b, and c are the coefficients of x,

 

[kerimek]

Hello, it seems like you are struggling with vector questions. Vector problems can be tricky, but with the right approach, they can be solved easily. Let me try to help you understand these questions.

1. To show that a vector is normal to a plane, we need to show that it is perpendicular to all the vectors that lie on the plane. In this case, we have three points that lie on the plane, so we need to show that the vector 2i+3j+3k is perpendicular to the vectors formed by these points. To do this, we can use the dot product. If the dot product of two vectors is 0, then they are perpendicular. So, we can find the vectors formed by the points (1,1,-2) and (4,2,-5) and take their dot product with 2i+3j+3k. If the result is 0, then we have shown that the given vector is normal to the plane.

Next, to find the equation of the plane in the form (r-a).n(hat), we can use the cross product. The cross product of two vectors gives us a vector that is perpendicular to both of them. So, we can take the cross product of the two vectors formed by the points (1,1,-2) and (4,2,-5) and use that as the normal vector (n(hat)) in the equation. The point (1,-1,0) can be used as the point (a) in the equation. This will give us the equation of the plane in the form (r-a).n(hat).

To find the perpendicular distance from the origin to the plane, we can use the formula d = |(a.b)|/|n|, where a is the point (0,0,0) and b is any point on the plane. We already have the normal vector (n) from the previous step, so we can use any point on the plane to find the distance.

I hope this helps you understand how to approach vector questions. If you need more help, I suggest looking for online tutorials or asking your teacher for clarification. Best of luck!