# normal plane to line in space ?

what is normal plane to line in space ?

### 5 Answers

- RaymondLv 78 years agoFavorite Answer
In vector space, normal = perpendicular (in all directions simultaneously)

In a 3-dimension vector space, the space will be spanned by three independent vectors

let's call them u, v and w for now

they would look like this (for example)

u = (3, 1, -2)

v = (1, 0, 4)

w = (0, -1, -1)

Any vector in the space could be represented by a linear combination of these

G = 3u + 4v - 7w

(for example)

Imagine that you have a line. It can be represented by any vector parallel to the line (any segment of the line)

L = (2, 2, 2)

is a vector that could represent a line.

Your task, in a question like "finding a plane that is normal to a line" (a normal plane to that line) is to find a plane

(represented by two independent vectors that would span it, for example)

(it could also be represented by some equation)

on which any vector (including the vectors of the spanning set) will be perpendicular to L

How do you tell that a vector is parallel to another?

With the "dot product", the sum of the products of corresponding elements.

For example,

a = (1, 2, 3)

b = (4, 5, 6)

a "dot" b =

(1, 2, 3) "dot" (4, 5, 6) =

(1)(4) + (2)(5) + (3)(6) =

4 + 10 + 18 = 32

It two vectors are perpendicular, then the dot product must be exactly zero.

If the dot product is exactly zero, the two vectors must be perpendicular.

(it is an "if and only if" situation)

In the example, where the dot product gives 32, the vectors a and b are not perpendicular.

A random vector can be written as

R = (x, y, z)

We would like to find one that is perpendicular to L = (2, 2, 2)

We need to find

R "dot" L = 0

(x, y, z) "dot" (2, 2, 2) = 0

2x + 2y + 2x = 0

2x = -2y - 2z (for example)

x = -y - z

so, if we set y = z = 1

then we would have to use x = -2

The vector R = (-2, 1, 1) is perpendicular to L.

To find another (independent) vector, we isolate another variable

2y = -2x - 2z

y = -x - z

set x=0 and z = 3

then y = -3

S = (0, -3, 3) is also perpendicular to L

The plane spanned by these two vectors, R and S, is normal to the line L.

- campbelp2002Lv 78 years ago
Normal just means perpendicular. So a normal plane to a line is a plane (flat surface like a table top) with a line standing straight out from it (like a pencil balanced on end on the table).

- crabillLv 43 years ago
[ a million ] Line popular to the airplane 1x + 4y - 1z = -2 has path d? = < a million, 4, -a million >. If it passes by making use of making use of the factor (3,4,-a million), then its symmetric cartesian equations are ( x-3 ) / a million = ( y- 4 ) / 4 = ( z + a million ) / -a million= t, and, subsequently, its parametric equations are x = t+3, y = 4t+4, z = -t-a million ............. (a million) If x = 0, then t+3 = 0 and t = -3. yet neither y nor z is 0 at t = -3. next, if y = 0, then 4t+4 = 0 and t = -a million. additionally, at t = -a million, z = -(-a million)-a million = a million-a million = 0. subsequently, at t = -a million, y = 0 and z = 0 and the factor lies on the X-axis. At t = -a million, x = t+3 = -a million+3 = 2. subsequently, the line in question intersects X-axis on the factor ( 2, 0, 0 ). .................................... Ans. ........................................... [ 2 ] A airplane parallel to X-axis don't have an x-term in its equation. subsequently, its equation is of the form by making use of technique of + cz = a million ..................... (a million) If it passes by making use of making use of the factors A(a million,2,3) and B(0,-a million,2), then 2b + 3c = a million ..................... (2) - b + 2c = a million ..................... (3) Multiplying (2) by making use of technique of two, -2b + 4c = 2 ...................... (4) which incorporate (2) and (4), 7c = 3 ? c = 3/7 ? from (2), ... b = 2c - a million = (6/7) - a million = -a million/7 ? from (a million), the req'd eq of airplane is ... (-a million/7)y + (3/7)z = a million ? -y + 3z = 7 ? y - 3z = -7 ..................... Ans. ........................................... happy to help ! ...........................................

- Anonymous8 years ago
Haven't got a clue.

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- Anonymous8 years ago
No, theres nothing to push through so it would not work

Source(s): The guy above me ... SMART! idk equations like his :P