Question

Find the point Q that is a distance 0.1 from the point P=(5,3) in the direction of v=⟨2,1⟩. Give five decimal places in your answer.

Answer 1

To find point Q, normalize the direction vector, multiply it by the given distance, and add the resultant to point P to get Q approximately as (5.08944, 3.04472).

Step-by-step explanation:

To find the point Q that is a distance 0.1 from point P=(5,3) in the direction of vector v=⟨2,1⟩, first normalize the direction vector v. Divide each component of v by its magnitude. The magnitude |v| of v is √(22 + 12) = √5. The normalized direction vector v' is v/|v| = ⟨0.89443, 0.44721⟩. Multiply v' by the distance 0.1 to find the displacement vector: ⟨0.08944, 0.04472⟩. Finally, add this displacement vector to P to get Q: Q = P + 0.1v' = (5 + 0.08944, 3 + 0.04472), giving us Q = (5.08944, 3.04472).