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Given the vector v has an initial point at (5,1) and a terminal point at (4,-4), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (5,1) (5,1) and a terminal point at (4,−4) (4,-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 (5,1) (5,1) and a terminal point at (4,−4) (4,-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=4−5=−1\Delta x = x_{\text{terminal}} - x_{\text{initial}} = 4 - 5 = -1\newlineΔy=yterminal−yinitial=−4−1=−5\Delta y = y_{\text{terminal}} - y_{\text{initial}} = -4 - 1 = -5
  2. Use Pythagorean Theorem: Now, we will use the Pythagorean theorem to find the magnitude of vector vv. The magnitude ∣∣v∣∣||v|| is given by the formula:\newline∣∣v∣∣=(Δx2+Δy2)||v|| = \sqrt{(\Delta x^2 + \Delta y^2)}\newlineSubstitute the values of Δx\Delta x and Δy\Delta y into the formula:\newline∣∣v∣∣=((−1)2+(−5)2)||v|| = \sqrt{((-1)^2 + (-5)^2)}
  3. Perform Squaring and Summation: Perform the squaring of Δx\Delta x and Δy\Delta y and sum them up: ∣∣v∣∣=1+25||v|| = \sqrt{1 + 25}
  4. Add Squared Values: Now, add the squared values to find the value under the square root: ∣∣v∣∣=26||v|| = \sqrt{26}
  5. Final Magnitude Calculation: Since 2626 is not a perfect square, we leave the square root as it is. The exact value of ∣∣v∣∣||v|| is 26\sqrt{26}.

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