Question

(a) Write down the angle between the vector i + j and the vector i- j.

(6) Calculate the angle between the vector 2i + 5j and the vector 4i.
(2)
(c) Two forces, A = (14i +2j) N and B = (2i - 6j) N, act on a ball.
Show that the magnitude of A is Root k times larger than the magnitude of B, where k is
an integer
(3)

Answer 1

(a) Let θ be the angle between i + j and i - j. Then
(i + j) • (i - j) = ||i + j|| ||i -j|| cos(θ)
We also recall that for any vector x,
||x||² = x • x
We have
(i + j) • (i - j) = (i • i) + (j • i) + (i • (-j)) + (j • (-j))
= (i • i) + (j • i) + (i • (-j)) + (j • (-j))
= ||i||² + (i • j) - (i • j) - ||j||²
= 1 + 0 - 1
= 0
and
||i + j|| = √(1² + 1²) = √2
||i - j|| = √(1² + (-1)²) = √2
So,
0 = 2 cos(θ)
cos(θ) = 0
θ = arccos(0) = 180°

(b) By the same process, we find
(2i + 5j) • 4i = 8
||2i + 5j|| = √(2² + 5²) = √29
||4i|| = 4
Then if θ is the angle between 2i + 5j and 4i, we have
8 = 4√29 cos(θ)
cos(θ) = 2/√29
θ = arccos(2/√29) ≈ 68.1986°

(c) Compute the magnitudes:
||A|| = √(14² + 2²) = √200 = 10√2
||B|| = √(2² + (-6)²) = √40 = 2√10
Now we can write
||B|| = 2√10 = 2 √(5 × 2) = 2√5 × √2 = 1/√5 × 10 √2 = 1/√5 ||A||
or
||A|| = √5 ||B||
so that k = 5.