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Given the vector v has an initial point at (-2,-6) and a terminal point at (-1,-4), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−2,−6) (-2,-6) and a terminal point at (−1,−4) (-1,-4) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:

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Full solution

Q. Given the vector v \mathbf{v} has an initial point at (−2,−6) (-2,-6) and a terminal point at (−1,−4) (-1,-4) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Calculate Differences: 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. The magnitude of vector vv, denoted as ∣∣v∣∣||v||, is the square root of the sum of the squares of these differences.\newlineLet's calculate the differences:\newlineΔx=xterminal−xinitial=(−1)−(−2)=−1+2=1\Delta x = x_{\text{terminal}} - x_{\text{initial}} = (-1) - (-2) = -1 + 2 = 1\newlineΔy=yterminal−yinitial=(−4)−(−6)=−4+6=2\Delta y = y_{\text{terminal}} - y_{\text{initial}} = (-4) - (-6) = -4 + 6 = 2
  2. Use Pythagorean Theorem: Now, we use the Pythagorean theorem to find the magnitude of vector vv:
    ∣∣v∣∣=(Δx2+Δy2)||v|| = \sqrt{(\Delta x^2 + \Delta y^2)}
    Substitute Δx\Delta x and Δy\Delta y with the values we found:
    ∣∣v∣∣=(12+22)||v|| = \sqrt{(1^2 + 2^2)}
    ∣∣v∣∣=(1+4)||v|| = \sqrt{(1 + 4)}
    ∣∣v∣∣=5||v|| = \sqrt{5}

More problems from Transformations of absolute value functions: translations and reflections

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