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# given point r=[0 0 -3] and v=[0 2 1]. rotate r around v by 180 degree anticlockwise. find image r?

### 1 Answer

- Anonymous7 years agoFavorite Answer
If r is rotated about v, then the component that is affected is the component of r that is perpendicular to v and the component along v remains the same. So the approach is to express r in terms of v and also the coordinate axis, i, j, k in terms of v. The latter is needed to convert back to the original i j k after we determine the image of r which is easy to do with respect to v, since under the rotation by 180 above v, the perpendicular components change sign while the parallel components remain the same.

So we will write r relative to v where v which we will call r_v.

Then we can decompose r as follows:

r_v = r.v/|v| * (v/|v|) + r x v/|v|

v/|v| = [0 2 1]/sqr( 2^2+1^2) = [0 2 1]/sqrt(5)

r.v = [0 0 -3] . [0 2 1] = -3

v/|v| = [0 2 1]/sqrt(5)

r.v/|v| * v/|v| = -3/5 * [0 2 1]

We will use the orthonormal basis notation i, j, k, where i = [1 0 0], j=[0 1 0], k=[0 0 1]

r x v = determinant of

i....j...k

0..0..-3

0..2...1

=6i=[6 0 0] = 6 * [1 0 0]

so,

r x v/|v| = 6/sqrt(5) * [1 0 0]

r_v = r.v/|v| v/|v| + rxv

r_v = -3/5 * [0 2 1] + [6/sqrt(5) 0 0] = [6/sqrt(5) -6/5 -3/5]

r_v = [6/sqrt(5) -6/5 -3/5]

To recover r in the original system, we need to compute the coordinates-axis in the v system and then use these to compute the components in the old system.

i_v = i.v/|v| * v/|v| + i x v/|v|

i.v/|v| = [1 0 0] . [0 2 1] /sqrt(5) = 0

i x v = determinant of

[i….j…k]

[1..0...0]

[0..2...1]

= -j + 2k =[0 -1 2]

i_v = [0 -1 2]/sqrt(5)

j_v = j.v/|v| * v/|v| + j x v/|v|

j.v = [0 1 0] . [0 2 1] = 2

j x v = determinant of

i j k

0 1 0

0 2 1

= i

j_v = 2/5 * [0 2 1] + [1 0 0]/sqrt(5) = [1/sqrt(5) 4/5 2/5]

k_v = k.v/|v| * v/|v| + k x v/|v|

k.v = [0 0 1] . [0 2 1] = 1

k x v = determinant of

i j k

0 0 1

0 2 1

= -2i = [-2 0 0]

k_v = 1/5 [0 2 1] +[-2 0 0]/sqrt(5) = [-2/sqrt(5), 2/5, 1/5]

To check that I can recover the original r

r = r_v . i_v * i + r_v . j_v * j + r_v . k_v * k

r_v . i_v = [6/sqrt(5) -6/5 -3/5] . [0 -1 2]

= -1 * -6/5 + -3/5 * 2

= 6/5 - 6/5 = 0

r_v . j_v = [6/sqrt(5) -6/5 -3/5] . [1/sqrt(5) 4/5 2/5]

= 6/5 -24/25 -6/25 = 6/5 -30/25 = 6/5 -6/5 = 0

k_v = [-2/sqrt(5), 2/5, 1/5]

r_v . k_v = [6/sqrt(5) -6/5 -3/5] . [-2/sqrt(5), 2/5, 1/5]

=-12/5 -12/25 -3/25 =-12/5 - 15/25 = -12/5 - 3/5 = -15/5 = -3

r = [0 0 -3] which is the original r.

Okay so we now have the machinery setup. Let find image of r under the 180 rotation of v.

First we established that we can write:

r_v = r.v/|v| v/|v| + r x v

Now rotate the space about v by 180 degrees

Then as we argued earlier, the components perpendicular to v change signs under 180 rotation.

rxv -> -rxv

so the image of r in v-space r_v'

r_v' = r.v/|v| v/|v| - r x v

r_v' = -3/5 * [0 2 1] - [6/sqrt(5) 0 0] = [-6/sqrt(5) -6/5 -3/5]

i_v = [0 -1 2]/sqrt(5)

j_v = [1/sqrt(5) 4/5 2/5]

k_v = [-2/sqrt(5), 2/5, 1/5]

Now to get the components of the rotated r vector back into the original system,

r_v' . i_v = [-6/sqrt(5) -6/5 -3/5] . [0 -1 2]/sqrt(5) = (6/5-6/5)/sqrt(5) = 0

r_v' . j_v = [-6/sqrt(5) -6/5 -3/5] . [1/sqrt(5) 4/5 2/5] = -6/5 -24/25 -6/25 = -6/5 -30/25=-12/5

r_v' . k_v = [-6/sqrt(5) -6/5 -3/5] . [-2/sqrt(5), 2/5, 1/5] = 12/5 -12/25 - 3/25 = 12/5 -15/25 = 12/5-3/5=9/5

r' = [0 -12/5 9/5] = 3/5 [0 -4 3] (FINAL ANSWER)

check:

since it is a reflection,

r . v = r' . v

r' . v = 3/5 [0 -4 3] . [0 2 1] = 3/5 * (-8 + 3) = 3/5 * (-5) = -3 = r . v