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Given the vector v has an initial point at (-2,0) and a terminal point at (-4,-3), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−2,0) (-2,0) and a terminal point at (−4,−3) (-4,-3) , 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 (−2,0) (-2,0) and a terminal point at (−4,−3) (-4,-3) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Calculate Components: To find the magnitude of vector vv, we need to use the distance formula, which is derived from the Pythagorean theorem. The distance formula for a vector with initial point (x1,y1)(x_1, y_1) and terminal point (x2,y2)(x_2, y_2) is:\newline∣∣v∣∣=((x2−x1)2+(y2−y1)2)||v|| = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}\newlineLet's calculate the components of the vector vv.
  2. Find Change in Coordinates: The change in the x-coordinate (Δx\Delta x) is the difference between the x-coordinates of the terminal and initial points: Δx=−4−(−2)=−4+2=−2\Delta x = -4 - (-2) = -4 + 2 = -2 The change in the y-coordinate (Δy\Delta y) is the difference between the y-coordinates of the terminal and initial points: Δy=−3−0=−3\Delta y = -3 - 0 = -3 Now we can substitute Δx\Delta x and Δy\Delta y into the distance formula.
  3. Substitute into Distance Formula: Substitute Δx\Delta x and Δy\Delta y into the distance formula to find the magnitude of vector vv: \newline∣∣v∣∣=((−2)2+(−3)2)||v|| = \sqrt{((-2)^2 + (-3)^2)}\newline∣∣v∣∣=(4+9)||v|| = \sqrt{(4 + 9)}\newline∣∣v∣∣=13||v|| = \sqrt{13}\newlineThe exact value of the magnitude of vector vv is 13\sqrt{13}.

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