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Given the vector 
v has an initial point at 
(-2,-3) and a terminal point at 
(-5,-2), find the exact value of 
||v||.
Answer:

Given the vector v \mathbf{v} has an initial point at (2,3) (-2,-3) and a terminal point at (5,2) (-5,-2) , find the exact value of v \|\mathbf{v}\| .\newlineAnswer:

Full solution

Q. Given the vector v \mathbf{v} has an initial point at (2,3) (-2,-3) and a terminal point at (5,2) (-5,-2) , 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=xterminalxinitial=5(2)=5+2=3\Delta x = x_{\text{terminal}} - x_{\text{initial}} = -5 - (-2) = -5 + 2 = -3\newlineΔy=yterminalyinitial=2(3)=2+3=1\Delta y = y_{\text{terminal}} - y_{\text{initial}} = -2 - (-3) = -2 + 3 = 1
  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:\newlinev=(Δx2+Δy2)||v|| = \sqrt{(\Delta x^2 + \Delta y^2)}\newlineSubstitute the values of Δx\Delta x and Δy\Delta y into the formula:\newlinev=((3)2+(1)2)=(9+1)=10||v|| = \sqrt{((-3)^2 + (1)^2)} = \sqrt{(9 + 1)} = \sqrt{10}

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