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Given the vector v has an initial point at (-3,4) and a terminal point at (-6,4), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−3,4) (-3,4) and a terminal point at (−6,4) (-6,4) , 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 (−3,4) (-3,4) and a terminal point at (−6,4) (-6,4) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Calculate Magnitude of Vector: The magnitude of a vector vv, denoted as ∣∣v∣∣||v||, is calculated using the distance formula between its initial and terminal points. The formula is ∣∣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 Given Points: Substitute the given points into the distance formula. The initial point is (−3,4)(-3,4) and the terminal point is (−6,4)(-6,4). So, x1=−3x_1 = -3, y1=4y_1 = 4, x2=−6x_2 = -6, and y2=4y_2 = 4.
  3. Calculate Differences: Calculate the differences: x2−x1=−6−(−3)=−6+3=−3x_2 - x_1 = -6 - (-3) = -6 + 3 = -3, and y2−y1=4−4=0y_2 - y_1 = 4 - 4 = 0.
  4. Square the Differences: Square the differences: (−3)2=9(-3)^2 = 9 and (0)2=0(0)^2 = 0.
  5. Add Squared Differences: Add the squared differences: 9+0=99 + 0 = 9.
  6. Find Magnitude of Vector: Take the square root of the sum to find the magnitude of vector vv: 9=3\sqrt{9} = 3.

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