Yacht Moving

Vectop problem plz help?

1) Yacht A moves according to x= 4 =t, y= 5-2t where the distance units are kilometers and the time units are hours. Yacht B moves according to x = 1 + 2t, y = -8+t, t > or equal to 0.
a. Find the initial position of each yacht
b.Find the initial velocity vector of each yacht
c. State the speeds of each Yacht
d. Prove the paths of the Yachts are at right angles to each other

I think you meant “x = 4 + t” for yacht A so I’ll work the problem under that assumption.

(a.) The initial positions will be the positions of the yachts when t = 0.

For yacht a: x = 4, y = 5. For yacht b: x = 1, y = -8.

(b.) Differentiate the position vector to get the velocity vector, then find the initial velocity by plugging in 0 for t.

Velocity of yacht a: (no matter what the value of t is)
Velocity of yacht b: (no matter what the value of t is)

(c.) Speed is given by the absolute value of the magnitude (norm) of the velocity vector.

Speed of yacht a: absolute value [sqrt(1^2 + (-2)^2)] = sqrt(3) (kilometers per hour)

Speed of yacht b: absolute value [sqrt(2^2 + 1^2)] = sqrt(5) (kilometers per hour)

(d.) The direction vectors of yacht A and yacht B are just the position vectors without those constant terms.

Yacht A:
Yact B:

Two vectors are orthogonal (meet at right angles) if their dot product = 0. The dot product of the two direction vectors is t(2t) + (-2t)(t) = 0, so the two yachts are at right angles.

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