the vector a-5b and a-b are perpendiculat. If vector a and b are unit vectors, then determine axb

If this is in just two dimensions:

A = ax i + ay j
B = bx i + by j

C = A-5B = (ax-5bx) i + (ay-5by)j
D = A- B = (ax- bx) i + (ay- by)j

if perpendicular C dot D = 0
(ax-5bx)(ax-bx)+(ay-5by)(ay-by) = 0
ax^2 -6ax bx + 5bx^2 +ay^2 -6ay by +5 by^2 = 0
but unit vectors so ax^2+ay^2 = 1
and
5(bx^2+by^2) = 5
so
6 - 6 ax bx -6 ay by = 0
so
ax bx + ay by = 1

Now we want to know A x B
determinant of
i j k
ax ay 0
bx by 0
= (ax by - ay bx)k
but
ax bx + ay by = 1
ay = (1-ax bx)/bx
so cross product is
(ax by - 1+ax bx)k
= [ax(bx+by)-1] k

Check carefully, I went pretty fast and there may be errors.

To determine the cross product (axb) of vector a and b, we need to find the cross product of their components.

First, let's express vector a - 5b and vector a - b in terms of their components.

Given:
Vector a - 5b = (a₁ - 5b₁, a₂ - 5b₂, a₃ - 5b₃)
Vector a - b = (a₁ - b₁, a₂ - b₂, a₃ - b₃)

Since the two vectors are perpendicular to each other, their dot product is zero.
Therefore, we can write:

(a - 5b) · (a - b) = 0

Expanding the dot product:

(a₁ - 5b₁)(a₁ - b₁) + (a₂ - 5b₂)(a₂ - b₂) + (a₃ - 5b₃)(a₃ - b₃) = 0

Now, substitute the unit vector conditions:

a · a = 1
b · b = 1

Let's simplify the equation:

(a₁² - 6a₁b₁ + 5b₁²) + (a₂² - 6a₂b₂ + 5b₂²) + (a₃² - 6a₃b₃ + 5b₃²) = 0

Expanding further:

a₁² + a₂² + a₃² - 6(a₁b₁ + a₂b₂ + a₃b₃) + 5(b₁² + b₂² + b₃²) = 0

Since a and b are unit vectors, a₁² + a₂² + a₃² = 1 and b₁² + b₂² + b₃² = 1:

1 - 6(a₁b₁ + a₂b₂ + a₃b₃) + 5 = 0

Rearranging the equation:

6(a₁b₁ + a₂b₂ + a₃b₃) = 6

Finally, dividing by 6:

(a₁b₁ + a₂b₂ + a₃b₃) = 1

Therefore, the value of axb is 1.