Question

Suppose A=(−3.39 m)i+(4.84 m)j,B=(2.37 m)i+(−3.02 m)j+(1.57 m)k, and D=(−3.75 m)i+(−3.75 m)j 0% Part (a) What is the angle, in degrees, between D and A ?

Answer 1

To find the angle between vectors D and A, we can use the dot product formula. By substituting the given values into the formula and solving for θ, we find that the angle between vectors D and A is approximately 103.3°.

Step-by-step explanation:

To find the angle between vectors D and A, we can use the dot product formula:

D · A = |D||A|cos(θ)

Where D · A represents the dot product of D and A, |D| and |A| represent the magnitudes of D and A respectively, and θ represents the angle between them.

Using the given values for D and A, we can calculate the dot product and the magnitudes:

D · A = (-3.75 * -3.39) + (-3.75 * 4.84) = -3.75 * 1.45 = -5.4375

|D| = sqrt((-3.75)^2 + (-3.75)^2) = sqrt(28.125) ≈ 5.31

|A| = sqrt((-3.39)^2 + (4.84)^2) = sqrt(35.2327) ≈ 5.93

By substituting the values into the formula, we get:

-5.4375 = 5.31 * 5.93 * cos(θ)

Solving for θ:

cos(θ) = -5.4375 / (5.31 * 5.93) ≈ -0.205

θ ≈ arccos(-0.205) ≈ 103.3°

So, the angle between vectors D and A is approximately 103.3°.