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Given the vector
v has an initial point at
(5,1) and a terminal point at
(4,-4), find the exact value of
||v||.
Answer:

Given the vector $\mathbf{v}$ has an initial point at $(5,1)$ and a terminal point at $(4,-4)$ , find the exact value of $\|\mathbf{v}\|$ . $\newline$ Answer:

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Transformations of absolute value functions: translations and reflections

Full solution

Q. Given the vector

\mathbf{v}

has an initial point at

(5,1)

and a terminal point at

(4,-4)

, find the exact value of

\|\mathbf{v}\|

.

\newline

Answer:

Calculate Differences: To find the magnitude of vector $v$ , we need to calculate the difference in the $x$ -coordinates and the difference in the $y$ -coordinates between the terminal point and the initial point. The magnitude of vector $v$ , denoted as $||v||$ , is the square root of the sum of the squares of these differences. $\newline$ Let's calculate the differences: $\newline$ $\Delta x = x_{\text{terminal}} - x_{\text{initial}} = 4 - 5 = -1$ $\newline$ $\Delta y = y_{\text{terminal}} - y_{\text{initial}} = -4 - 1 = -5$
Use Pythagorean Theorem: Now, we will use the Pythagorean theorem to find the magnitude of vector $v$ . The magnitude $||v||$ is given by the formula: $\newline$ $||v|| = \sqrt{(\Delta x^2 + \Delta y^2)}$ $\newline$ Substitute the values of $\Delta x$ and $\Delta y$ into the formula: $\newline$ $||v|| = \sqrt{((-1)^2 + (-5)^2)}$
Perform Squaring and Summation: Perform the squaring of $\Delta x$ and $\Delta y$ and sum them up: $||v|| = \sqrt{1 + 25}$
Add Squared Values: Now, add the squared values to find the value under the square root: $||v|| = \sqrt{26}$
Final Magnitude Calculation: Since $26$ is not a perfect square, we leave the square root as it is. The exact value of $||v||$ is $\sqrt{26}$ .

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