ABCDEF is a regular hexagon. Show that vecotr AB+AD+FA+EA=3AB
To show that vector AB + AD + FA + EA = 3AB in a regular hexagon ABCDEF, we'll need to use the properties of a regular hexagon and vector addition.
1. Start by drawing a diagram of the regular hexagon ABCDEF. Label the points as shown:
A
/ \
/ \
F / \ B
\ /
\ /
E \ / C
\ /
D
2. Recall that in a regular hexagon, all the sides are equal in length, and all the angles are equal. Hence, AB = BC = CD = DE = EF = FA = x, where x is the length of a side.
3. Now, let's represent the vectors AB, AD, FA, and EA mathematically:
AB = vector from A to B = (x, 0)
AD = vector from A to D = (-x, 0)
FA = vector from F to A = (-x/2, sqrt(3)x/2)
EA = vector from E to A = (x/2, sqrt(3)x/2)
Notice that AD is the opposite of AB, while FA and EA are diagonals of the hexagon.
4. Now, add up the vectors: AB + AD + FA + EA
(x, 0) + (-x, 0) + (-x/2, sqrt(3)x/2) + (x/2, sqrt(3)x/2)
Simplify the vector addition:
(x + -x + -x/2 + x/2, 0 + 0 + sqrt(3)x/2 + sqrt(3)x/2)
(0, 2sqrt(3)x/2) = (0, sqrt(3)x)
5. We want to show that this sum is equal to 3 times AB. Multiply AB by 3:
3AB = 3(x, 0) = (3x, 0)
6. Finally, compare the sums:
(0, sqrt(3)x) = (3x, 0)
Since the x-component of the sum is equal to the x-component of 3AB (both are 0), and the y-component of the sum is equal to the y-component of 3AB (both are sqrt(3)x), we can conclude that AB + AD + FA + EA = 3AB in a regular hexagon.
So, vector AB + AD + FA + EA = 3AB holds true in a regular hexagon ABCDEF.