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Given the vector v has an initial point at (0,2) and a terminal point at (-1,2), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (0,2) (0,2) and a terminal point at (−1,2) (-1,2) , 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 (0,2) (0,2) and a terminal point at (−1,2) (-1,2) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Calculate Distance Formula: To find the magnitude of vector vv, we need to calculate the distance between its initial and terminal points using the distance formula for points in a 22D space: ∣∣v∣∣=((x2−x1)2+(y2−y1)2)||v|| = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}, where (x1,y1)(x_1, y_1) is the initial point and (x2,y2)(x_2, y_2) is the terminal point.
  2. Substitute Coordinates: Substitute the coordinates of the initial point (0,2)(0,2) and the terminal point (−1,2)(-1,2) into the distance formula: ∣∣v∣∣=((−1−0)2+(2−2)2)||v|| = \sqrt{((-1 - 0)^2 + (2 - 2)^2)}.
  3. Calculate Squares: Calculate the squares of the differences: ∣∣v∣∣=(−1)2+(0)2||\mathbf{v}|| = \sqrt{(-1)^2 + (0)^2}.
  4. Simplify Squares: Simplify the squares: ∣∣v∣∣=1+0||\mathbf{v}|| = \sqrt{1 + 0}.
  5. Add Values: Add the values inside the square root: ∣∣v∣∣=1.||v|| = \sqrt{1}.
  6. Take Square Root: Take the square root of 11: ∣∣v∣∣=1||v|| = 1.

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