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Given the vector v has an initial point at (-1,0) and a terminal point at (2,-3), find the exact value of ||v||. Answer:

Given the vector v \mathbf{v} has an initial point at (−1,0) (-1,0) and a terminal point at (2,−3) (2,-3) , 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 (−1,0) (-1,0) and a terminal point at (2,−3) (2,-3) , find the exact value of ∥v∥ \|\mathbf{v}\| .\newlineAnswer:
  1. Calculate Components of Vector: To find the magnitude of vector v, we need to calculate the difference in the x-coordinates and the y-coordinates of the terminal and initial points to get the components of the vector.\newlineThe x-component of vector v is the difference between the x-coordinate of the terminal point and the x-coordinate of the initial point: xcomponent=2−(−1)=3 x_{component} = 2 - (-1) = 3 .\newlineThe y-component of vector v is the difference between the y-coordinate of the terminal point and the y-coordinate of the initial point: ycomponent=−3−0=−3 y_{component} = -3 - 0 = -3 .
  2. Use Pythagorean Theorem: Now that we have the components of vector v, we can calculate its magnitude using the Pythagorean theorem. The magnitude of vector v, denoted as ∣∣v∣∣ ||v|| , is the square root of the sum of the squares of its components.\newlineSo, ∣∣v∣∣=xcomponent2+ycomponent2 ||v|| = \sqrt{x_{component}^2 + y_{component}^2} .\newlineSubstitute the values we found for the components: ∣∣v∣∣=32+(−3)2 ||v|| = \sqrt{3^2 + (-3)^2} .
  3. Square Components and Add: Perform the squaring of the components and add them together: ∣∣v∣∣=9+9 ||v|| = \sqrt{9 + 9} .
  4. Simplify Square Root: Simplify the expression under the square root: ∣∣v∣∣=18 ||v|| = \sqrt{18} .
  5. Factor Out Perfect Square: We can simplify 18 \sqrt{18} by factoring out the square of a whole number. The number 1818 can be written as 9×2 9 \times 2 , and since 99 is a perfect square, we can take its square root out of the radical.\newlineSo, ∣∣v∣∣=9×2=9×2=32 ||v|| = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2} .

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