Question

Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude of 24.5 and is −15.3∘ from the +x‑axis. Resolve ????⃗ and ????⃗ into components, and express in ???????????? unit vector form,

Answer 1

Answer:

For vector u, x component = 10.558 and  y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$