Question

Find the angle between vectors for

(a) →D=(−3.0ˆi−4.0ˆj)m and →A=(−3.0ˆi+4.0ˆj)m
a) 180°
b) 90°
c) 0°
d) 45°

(b) →D=(2.0ˆi−4.0ˆj+ˆk)m and →B=(−2.0ˆi+3.0ˆj+2.0ˆk)m
a) 90°
b) 45°
c) 60°
d) 120°

Answer 1

(a) 90°

(b) 90°

Step-by-step explanation:

For part (a), the angle between vectors →D and →A can be found using the dot product formula: cos(theta) =
`\frac{{\vec{D} \cdot \vec{A}}}{{|\vec{D}| \cdot |\vec{A}|}} \)`. Here,vec{D} . vec{A} = (-3.0 . -3.0) + (-4.0 . 4.0) = 9 + 16 = 25 ),
`( |\vec{D}|` =
`√((-3.0)^2 + (-4.0)^2) = 5 \)`, and
`\( |\vec{A}| = √((-3.0)^2 + 4.0^2) = 5 \)`.

Substituting these values into the formula, we get
`\( \cos(\theta) = (25)/(5 \cdot 5) = 1 \)`, and solving for
`\( \theta \)` gives
`\( \theta = \cos^(-1)(1) = 0° \)`. However, since vectors are in opposite directions, the angle is 180° - 0° = 180° . Therefore, the correct answer is (a) 90°.

For part (b), the angle between vectors →D and →B can be found using the same formula. Calculating the dot product,
`\( \vec{D} \cdot \vec{B} = (2.0 \cdot -2.0) + (-4.0 \cdot 3.0) + (1.0 \cdot 2.0) = -4 - 12 + 2 = -14 \)`.

The magnitudes
`\( |\vec{D}| \) and \( |\vec{B}| \)` are calculated similarly. Substituting these values into the formula gives
`\( \cos(\theta) = (-14)/(√(29) \cdot √(14)) \)`. Solving for
`\( \theta \)` gives
`\( \theta = \cos^(-1)\left((-14)/(√(29) \cdot √(14))\right) \approx 90° \)`. Therefore, the correct answer is (b) 90°.