Vector A has magnitude 3.7 units; vector B has magnitude 5.9. The angle between vector A and B is 45 degrees. What is the magnitude of vector A+ vector B?

Question

10258 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-07T12:57:04

Magnitude of $(\to A + \to B)$ $(vecA+vecB)$ is $8.91$ $8.91$ units at an angle
of ${27.92}^{0}$ $27.92^0$ from $\to A$ $vecA$

Completing the parallelogram of $\to A and \to B$ $vec A and vec B$ , the

diagonal represents represents resultant of two vectors

$\to A and \to B$ $vecA and vecB$ . The angle between $\to A$ $vecA$ and parallal

$\to B$ $vecB$ is $180 - 45 = 135^{0}$ $180-45=135^0$ By applying cosine law

we get magnitude of ${∣ ∣ ∣ \to A + \to B ∣ ∣ ∣}^{2}$ $|vecA+vecB|^2$

$= {3.7}^{2} + {5.9}^{2} - 2 \cdot 3.7 \cdot 5.9 \cdot cos 135 \approx 79.37$ $= 3.7^2+5.9^2-2*3.7*5.9*cos135~~79.37$ or

$∣ ∣ ∣ \to A + \to B ∣ ∣ ∣ = 8.91$ $|vecA+vecB|= 8.91$ units . Let the resultant vector.

$∣ ∣ ∣ \to A + \to B ∣ ∣ ∣$ $|vecA+vecB|$ makes an angle $θ$ $theta$ with $\to A$ $vecA$ .

By sine law $\frac{5.9}{sin θ} = \frac{8.91}{sin 135}$ $5.9/sintheta=8.91/sin135$ or

$sin θ = \frac{5.9 \cdot sin 135}{8.91} \approx 0.4682$ $sin theta = (5.9*sin135)/8.91~~0.4682$

or $θ = {sin}^{- 1} (0.4682) \approx {27.92}^{0}$ $theta= sin^-1(0.4682)~~27.92^0$ from $\to A$ $vecA$ .

Magnitude of $(\to A + \to B)$ $(vecA+vecB)$ is $8.91$ $8.91$ units at an angle

of ${27.92}^{0}$ $27.92^0$ from $\to A$ $vecA$ [Ans]