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Given the vector v has an initial point at (-5,6) and a terminal point at (-6,1), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−5,6) (-5,6) and a terminal point at (−6,1) (-6,1) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:

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Transformations of absolute value functions: translations and reflectionsright chevron icon

Full solution

Q. Given the vector v \mathbf{v} has an initial point at (−5,6) (-5,6) and a terminal point at (−6,1) (-6,1) , 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. Then, we will use the Pythagorean theorem to find the magnitude.
  2. Find Δx\Delta x: The difference in the x-coordinates (Δx\Delta x) is the x-coordinate of the terminal point minus the x-coordinate of the initial point: Δx=−6−(−5)=−6+5=−1\Delta x = -6 - (-5) = -6 + 5 = -1.
  3. Find Δy\Delta y: The difference in the y-coordinates (Δy\Delta y) is the y-coordinate of the terminal point minus the y-coordinate of the initial point: Δy=1−6=−5\Delta y = 1 - 6 = -5.
  4. Use Pythagorean Theorem: Now, we use the Pythagorean theorem to find the magnitude of vector vv, which is the square root of the sum of the squares of Δx\Delta x and Δy\Delta y: ∣∣v∣∣=Δx2+Δy2||v|| = \sqrt{\Delta x^2 + \Delta y^2}.
  5. Substitute and Solve: Substitute the values of Δx\Delta x and Δy\Delta y into the formula: ∥v∥=(−1)2+(−5)2=1+25=26.\|v\| = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}.

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