(a) Magnitude of an arbitrary linear combination

We are given that x and y are unit vectors, meaning their magnitudes are both 1:

||x|| = 1

||y|| = 1

Let's find the magnitude (length) of the vector ax + by, denoted by ||ax + by||. We can use the Pythagorean Theorem:

1. Expand the norm (magnitude squared):

||ax + by||^2 = (ax + by) • (ax + by)

2. Use the dot product properties:

Distributive property: (ax + by) • (ax + by) = a^2(x • x) + 2ab(x • y) + b^2(y • y)

Perpendicularity (x • y = 0): a^2(x • x) + 2ab(x • y) + b^2(y • y) = a^2 ||x||^2 + 2ab + b^2 ||y||^2

3. Substitute the fact that ||x|| = 1 and ||y|| = 1:

a^2(1^2) + 2ab + b^2(1^2) = a^2 + 2ab + b^2

4. Take the square root of both sides to find the magnitude:

||ax + by|| = sqrt(a^2 + 2ab + b^2) = a + b

Therefore, the formula for the magnitude of an arbitrary linear combination ax + by of perpendicular unit vectors x and y in terms of a and b is:

||ax + by|| = a + b

(b) Condition for perpendicular linear combinations

We want to find the condition on a, b, c, and d such that ax + by and cx + dy are perpendicular.

Again, we can use the dot product. Two vectors are perpendicular if their dot product is zero:

(ax + by) • (cx + dy) = 0

Expanding the dot product:

a^2(x • x) + ab(x • y) + ac(y • x) + b^2(y • y) + bd(x • y) + cd(x • y) = 0

Then

a^2 ||x||^2 + ab + ac + b^2 ||y||^2 + bd + cd = 0

Substituting ||x|| = 1 and ||y|| = 1:

a^2 + ab + ac + b^2 + bd + cd = 0

Condition for perpendicularity:

a^2 + ab + ac + b^2 + bd + cd =