Vector equation of a plane in normal form

In summary, the conversation discussed finding the equation of a plane passing through a given point and being normal to a vector with specific inclinations to the x, y, and z axes. The solution involved finding the direction cosines and using them to determine the normal vector, and then plugging in the given point and normal vector to the formula (r-a).n=0. There was a discrepancy with the solution in the textbook, suggesting a possible error.
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Krushnaraj Pandya
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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?
 
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  • #2
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.
 
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  • #3
TachyonLord said:
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
 

Related to Vector equation of a plane in normal form

1. What is a vector equation of a plane in normal form?

A vector equation of a plane in normal form is an equation that represents a plane in three-dimensional space using a point on the plane and a vector perpendicular to the plane. It is written in the form r · n = p · n, where r is the position vector of any point on the plane, p is a known point on the plane, and n is the normal vector to the plane.

2. How is a normal vector determined in a vector equation of a plane in normal form?

The normal vector in a vector equation of a plane in normal form is determined by finding the cross product of two non-parallel vectors in the plane. The resulting vector will be perpendicular to both of the original vectors and therefore will be a normal vector to the plane.

3. Can a vector equation of a plane in normal form be converted to standard form?

Yes, a vector equation of a plane in normal form can be converted to standard form. To do so, the equation must be rearranged to have the form Ax + By + Cz = D, where A, B, and C are the coefficients of x, y, and z, and D is a constant term. The normal vector can then be used to determine the values of A, B, and C.

4. How many points are needed to define a plane in a vector equation of a plane in normal form?

Only one point is needed to define a plane in a vector equation of a plane in normal form. This point is represented by the vector p in the equation r · n = p · n. The normal vector n provides the orientation of the plane, and the position vector r determines the location of the plane in relation to the origin.

5. How can a vector equation of a plane in normal form be used to determine the distance from a point to the plane?

The distance from a point to a plane can be determined by using the formula d = |(r - p) · n| / ||n||, where r is the position vector of the given point, p is a known point on the plane, and n is the normal vector to the plane. This formula calculates the perpendicular distance from the point to the plane.

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