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Given the vector v has an initial point at (7,2) and a terminal point at (2,6), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (7,2) (7,2) and a terminal point at (2,6) (2,6) , 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 (7,2) (7,2) and a terminal point at (2,6) (2,6) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Distance Formula: To find the magnitude of the 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 ∣∣v∣∣=(x2−x1)2+(y2−y1)2||v|| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
  2. Calculate Differences: First, we calculate the differences in the x and y coordinates: Δx=x2−x1\Delta x = x_2 - x_1 and Δy=y2−y1\Delta y = y_2 - y_1. For our vector v\mathbf{v} with initial point (7,2)(7,2) and terminal point (2,6)(2,6), we have Δx=2−7=−5\Delta x = 2 - 7 = -5 and Δy=6−2=4\Delta y = 6 - 2 = 4.
  3. Square Differences: Next, we square the differences: (Δx)2=(−5)2=25(\Delta x)^2 = (-5)^2 = 25 and (Δy)2=42=16(\Delta y)^2 = 4^2 = 16.
  4. Add Squares: Now, we add the squares of the differences: (Δx)2+(Δy)2=25+16=41(\Delta x)^2 + (\Delta y)^2 = 25 + 16 = 41.
  5. Find Magnitude: Finally, we take the square root of the sum to find the magnitude of the vector vv: ∣∣v∣∣=41||v|| = \sqrt{41}.

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