Question

If the subtraction of two unit vectors is a unit vector then the angle between the first two vectors...​

Answer 1

Answer:

Let the two vectors be (ai^+bj^+ck^) & (pi^+qj^+rk^)

Then,

(ai^+bj^+ck^) - (pi^+qj^+rk^)

= (a-p)i^ + (b-q)j^ + (c-k)k^

For this to be an unit vector, EITHER two of (a-p), (b-q) or (c-r) have to be zero. For the sake of simplicity, I take (b-q) and (c-r) to be zero.

Therefore, b=q and c=r, giving us vectors ai^+bj^+ck^ and pi^+bj^+ck^,

Therefore, angle between them is,

∆ = arc(cos[(ap + b^2 + c^2)/{√(a^2 + b^2 + c^2)√(p^2 + b^2 + c^2)}])

[NOTE : I used ∆ instead of the formal theta

arc means inverse i.e. cos inverse]

If b=c=q=r=0,

Angle ∆ = arc(cos(ap/ap))

= arc(cos 1)

= 0°

Explanation:

thank me plz and mark brainliest

Answer 2

Answer:

hhfhnj DHL DHL TTYL week will weep all CNN