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Given the vector v has an initial point at (-1,4) and a terminal point at (0,1), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−1,4) (-1,4) and a terminal point at (0,1) (0,1) , 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 (−1,4) (-1,4) and a terminal point at (0,1) (0,1) , 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. 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=0−(−1)=1\Delta x = x_{\text{terminal}} - x_{\text{initial}} = 0 - (-1) = 1\newlineΔy=yterminal−yinitial=1−4=−3\Delta y = y_{\text{terminal}} - y_{\text{initial}} = 1 - 4 = -3
  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 Δx\Delta x and Δy\Delta y into the formula:\newline∣∣v∣∣=(12+(−3)2)||v|| = \sqrt{(1^2 + (-3)^2)}
  3. Perform Squaring and Addition: Perform the squaring and addition:\newline∣∣v∣∣=1+9||v|| = \sqrt{1 + 9}
  4. Calculate Square Root: Now, we will calculate the square root of the sum:\newline∣∣v∣∣=10||\mathbf{v}|| = \sqrt{10}\newlineSince 10\sqrt{10} is an exact value and cannot be simplified further, this is the exact value of the magnitude of vector v\mathbf{v}.

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