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Given the vector v has an initial point at (-1,3) and a terminal point at (-2,4), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−1,3) (-1,3) and a terminal point at (−2,4) (-2,4) , 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 (−1,3) (-1,3) and a terminal point at (−2,4) (-2,4) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Use Distance Formula: To find the magnitude (∥v∥\|\mathbf{v}\|) of the vector v\mathbf{v}, 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)\|\mathbf{v}\| = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)}
  2. Substitute Given Points: Substitute the given points into the distance formula:\newlineInitial point: (−1,3)(-1, 3)\newlineTerminal point: (−2,4)(-2, 4)\newline∣∣v∣∣=((−2−(−1))2+(4−3)2)||v|| = \sqrt{((-2 - (-1))^2 + (4 - 3)^2)}
  3. Simplify Expression: Simplify the expression inside the square root:\newline||v|| = \sqrt{((-2 + 1)^2 + (4 - 3)^2)}\(\newline||v|| = \sqrt{((-1)^2 + (1)^2)}\)
  4. Calculate Squares: Calculate the squares and sum them up:\newline∣∣v∣∣=1+1||v|| = \sqrt{1 + 1}
  5. Take Square Root: Add the values inside the square root and then take the square root: ∣∣v∣∣=2||v|| = \sqrt{2}

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