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Given the vector v has an initial point at (0,-3) and a terminal point at (1,1), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (0,−3) (0,-3) and a terminal point at (1,1) (1,1) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:

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Q. Given the vector v \mathbf{v} has an initial point at (0,−3) (0,-3) and a terminal point at (1,1) (1,1) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Define Initial and Terminal Points: To find the magnitude of vector vv, we need to calculate the difference in the xx-coordinates and the difference in the yy-coordinates between the terminal point and the initial point. Then, we will use the Pythagorean theorem to find the magnitude.\newlineLet's denote the initial point as (x1,y1)=(0,−3)(x_1, y_1) = (0, -3) and the terminal point as (x2,y2)=(1,1)(x_2, y_2) = (1, 1).
  2. Calculate Differences: Calculate the difference in the x-coordinates (Δx\Delta x) and the difference in the y-coordinates (Δy\Delta y).\newlineΔx=x2−x1=1−0=1\Delta x = x_2 - x_1 = 1 - 0 = 1\newlineΔy=y2−y1=1−(−3)=1+3=4\Delta y = y_2 - y_1 = 1 - (-3) = 1 + 3 = 4
  3. Use Pythagorean Theorem: Now, we use the Pythagorean theorem to find the magnitude of vector vv, which is denoted as ∣∣v∣∣||v||.∣∣v∣∣=(Δx2+Δy2)=(12+42)=(1+16)=17||v|| = \sqrt{(\Delta x^2 + \Delta y^2)} = \sqrt{(1^2 + 4^2)} = \sqrt{(1 + 16)} = \sqrt{17}

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