The problem concern the position vectors r1=2.43mi-4.74mj and r2=-3.02mi+0.85mj.
Find the angle between the two vectors. Hint: Find the angle each makes with the +x-axis.
r1 = 2.43-i4.74.
tanA1 = -4.74/2.43 = -1.95062,
A1 = -62.86deg, CW = 297.1deg,CCW.
r2 = -3.02+i0.85.
tanA2 = 0.85/-3.02 = -0.281457,
A2 = -15.72deg,CW = 344.3deg,CCW.
A = 344.3 - 297.1 = 47.2deg. = Angle
between r1 and r2.
Or
A = -15.7 - (-62.9) = -15.7 + 62.9
= 47.2deg.
To find the angle between two vectors, we can use the dot product formula. The dot product of two vectors is given by the equation:
A · B = |A| |B| cos(θ)
Where A and B are vectors, |A| and |B| are the magnitudes of the vectors, and θ is the angle between the vectors.
In this case, we need to find the angles each vector makes with the +x-axis. The angle that vector A makes with the +x-axis, θ1, can be found using the formula:
θ1 = atan2(Ay, Ax)
Where Ay is the y-component of vector A and Ax is the x-component of vector A. Similarly, the angle that vector B makes with the +x-axis, θ2, can be found using the formula:
θ2 = atan2(By, Bx)
Where By is the y-component of vector B and Bx is the x-component of vector B.
Let's calculate θ1 and θ2 for the given vectors r1 = 2.43mi - 4.74mj and r2 = -3.02mi + 0.85mj.
For r1:
Ax = 2.43mi
Ay = -4.74mj
θ1 = atan2(-4.74, 2.43) = -62.1° (approx.)
For r2:
Bx = -3.02mi
By = 0.85mj
θ2 = atan2(0.85, -3.02) = 164.9° (approx.)
Now that we have the angles θ1 = -62.1° and θ2 = 164.9°, we can find the angle between the two vectors using the dot product formula:
A · B = |A| |B| cos(θ)
|A| = sqrt(Ax^2 + Ay^2)
= sqrt((2.43)^2 + (-4.74)^2)
= sqrt(5.9049 + 22.5076)
= sqrt(28.4125)
= 5.33 (approx.)
|B| = sqrt(Bx^2 + By^2)
= sqrt((-3.02)^2 + (0.85)^2)
= sqrt(9.1204 + 0.7225)
= sqrt(9.8429)
= 3.13 (approx.)
A · B = (2.43)(-3.02) + (-4.74)(0.85)
= -7.3566 - 4.034
= -11.39 (approx.)
Now we can solve for cos(θ):
-11.39 = (5.33)(3.13) cos(θ)
cos(θ) = -11.39 / (5.33)(3.13)
= -0.718 (approx.)
Since cos(θ) = -0.718, we can find θ using the inverse cosine function:
θ = acos(-0.718)
= 133.82° (approx.)
Therefore, the angle between the two vectors r1 and r2 is approximately 133.82°.