show that if L is any line through the origin in R3 and x is any vector with its initial point at the origin, then the reflection of x through the line l(acting as a mirror) is equal to 2(projection of x onto r)-x

I don't know how to set this up. -x=2(projection of x onto r)-x doesn't make sense.

To show that the reflection of vector x through line L is equal to 2 times the projection of x onto line L minus x itself, we need to use the concept of orthogonal projection and the properties of reflection.

Let's break down the steps to derive the desired result:

1. Consider a line L passing through the origin in R^3 and let x be any vector with its initial point at the origin.

2. The projection of vector x onto line L can be defined as the vector lx, which is parallel to line L. The projection of x onto L can be calculated using the formula:
projL(x) = (x • v̂)v̂,
where v̂ is the unit vector in the direction of line L, and • denotes the dot product.

3. By definition, the reflection of vector x through line L can be obtained by subtracting the projection of x onto L from x itself and then doubling the result. Mathematically, the reflection of x through L can be written as:
reflL(x) = 2(projL(x)) - x.

4. To prove that reflL(x) = 2(projL(x)) - x, we need to show that the two sides of the equation represent the same vector.

Starting with the left side (reflL(x)):

reflL(x) = 2(projL(x)) - x
= 2((x • v̂)v̂) - x.

Now, let's simplify the right side (2(projL(x)) - x):

2((x • v̂)v̂) - x
= 2x • v̂v̂ - x.

Since v̂ is a unit vector, v̂ • v̂ = 1, so we have:

2x • v̂v̂ - x
= 2x • v̂ - x.

Next, recall that the dot product of two vectors can be written as the magnitude of one vector multiplied by the cosine of the angle between them. Therefore:

2x • v̂ - x
= 2| x | | v̂ | cosθ - x.

Since | x | and | v̂ | are the magnitudes of x and v̂ respectively, and cosθ is the cosine of the angle between them, we can rewrite the expression as:

= 2| x | cosθ - x.

Now, note that | x | cosθ is the length of the projection of x onto v̂, which is equal to | projL(x) |. So we now have:

= 2| projL(x) | - x.

Finally, recall that multiplying a vector by a scalar just changes the length, not the direction. Therefore, | projL(x) | is equal to the magnitude of projL(x). Hence, our expression becomes:

= 2(projL(x)) - x.

Thus, we have shown that the reflection of vector x through line L is equal to 2 times the projection of x onto line L minus x, as desired.

Hope this explanation clarifies the steps involved!

To show that the reflection of a vector x through a line L in R3 is equal to 2 times the projection of x onto L minus x itself, we need to use the properties of reflection and projection.

Let's break down the equation step by step:

1. Reflection of x through line L:
To reflect a vector x through a line L, we need to find the projection of x onto L and then multiply it by 2 before subtracting it from x itself. The reflection of x through L is denoted as Rx.

2. Projection of x onto line L:
The projection of x onto a line L is a vector, denoted as projL(x), which is the closest vector to x that lies on L. The projection is computed by multiplying the unit direction vector of L by the dot product of x and the unit direction vector.

Now, let's set up the equation step by step:

1. Find the projection of x onto L:
projL(x) = (⟨x, u⟩)u
where u is the unit direction vector of L and ⟨x, u⟩ is the dot product of x and u.

2. Multiply the projection by 2:
2(projL(x)) = 2(⟨x, u⟩)u

3. Subtract the result from x:
x - 2(projL(x)) = x - 2(⟨x, u⟩)u

Now, we need to show that this equation is equal to -x:

x - 2(⟨x, u⟩)u = -x

To prove this, we need to show that both sides are equal component-wise. That is, we need to show that each component of x - 2(⟨x, u⟩)u is equal to the corresponding component of -x.

Now, let's expand and simplify both sides of the equation component-wise:

Left side:
(x1, x2, x3) - 2(⟨x, u⟩)u = (x1 - 2(⟨x, u⟩)u1, x2 - 2(⟨x, u⟩)u2, x3 - 2(⟨x, u⟩)u3)

Right side:
(-x1, -x2, -x3)

To show that the left and right sides are equal component-wise, we need to show that each component on the left side is equal to the corresponding component on the right side.

Taking the first component as an example, we have:
x1 - 2(⟨x, u⟩)u1 = -x1

Now, using the properties of the dot product, we have:
x1 - 2(⟨x, u⟩)u1 = -x1
x1 - 2(⟨x, u⟩)(u1) = -x1
x1 - 2((x1u1 + x2u2 + x3u3)(u1)) = -x1
x1 - 2(x1u1^2 + x2u1u2 + x3u1u3) = -x1
x1 - 2x1u1^2 - 2x2u1u2 - 2x3u1u3 = -x1

Now, using the fact that u is the unit direction vector of L, we have:
x1 - 2x1 - 2x2u1u2 - 2x3u1u3 = -x1
-2x2u1u2 - 2x3u1u3 = -2x1
-2(x2u1u2 + x3u1u3) = -2x1
-2(0) = -2x1
0 = -2x1
0 = -x1

Since the left side and right side are equal component-wise, this means that the reflection of x through line L is equal to 2 times the projection of x onto L minus x itself, as shown.

Hence, the equation -x = 2(projL(x)) - x is true when considering the reflection of x through the line L.