Question

we have two vectors a→ and b→ with magnitudes a and b, respectively. suppose c→=a→ b→ is perpendicular to b→ and has a magnitude of 2b . what is the ratio of a / b ?

Answer 1

Since c→ is perpendicular to b→, the dot product of c→ and b→ is zero:

c→ · b→ = 0

Taking the dot product of c→ and b→, we get:

c→ · b→ = a b cos(90°) = 0

Since cos(90°) = 0, we have:

a b = 0

Therefore, either a = 0 or b = 0. However, since c→ has a magnitude of 2b, we must have b ≠ 0. Hence, we have a = 0.

Now, since c→ = a→ b→, we have:

|c→| = |a→| |b→| = 2b

Substituting a = 0, we get:

|b→| = 2b

Dividing both sides by b, we get:

|b→| / b = 2

Since |b→| / b = |b→| / |b| = 1 + a / b, we have:

1 + a / b = 2

Subtracting 1 from both sides, we get:

a / b = 1

Therefore, the ratio of a / b is 1.