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Given the vector v has an initial point at (-3,0) and a terminal point at (-5,-5), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−3,0) (-3,0) and a terminal point at (−5,−5) (-5,-5) , 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 (−3,0) (-3,0) and a terminal point at (−5,−5) (-5,-5) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Use Distance Formula: 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)}
  2. Substitute Given Points: Substitute the given points into the distance formula:\newlineInitial point (x1,y1)=(−3,0)(x_1, y_1) = (-3, 0)\newlineTerminal point (x2,y2)=(−5,−5)(x_2, y_2) = (-5, -5)\newline∣∣v∣∣=((−5−(−3))2+(−5−0)2)||v|| = \sqrt{((-5 - (-3))^2 + (-5 - 0)^2)}
  3. Simplify Expression: Simplify the expression inside the square root:\newline||v|| = \sqrt{((\(-5\) + \(3\))^\(2\) + (\(-5\))^\(2\))}\(\newline||v|| = \sqrt{(22^22 + (−5-5)^22)}\newline||v|| = \sqrt{(44 + 2525)}
  4. Find Magnitude: Add the squares to find the magnitude:\newline∣∣v∣∣=29||v|| = \sqrt{29}\newlineSince 2929 is a prime number, it cannot be simplified further, so the exact value of the magnitude of vector vv is 29\sqrt{29}.

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