Question

Which could be the initial point of vector $\text{u}$ if it has a magnitude of $17$ and the terminal point $\left(3,13\right)$$?$

• A. $(-5,-2)$
• B. $(-5,2)$
• C. $(5,-2)$
• D. $(5,2)$

Answer 1

The correct option is A

$(-5,-2)$

Explanation for correct option:

Option(A):

To find the initial point for the vector $\text{u}$.

Consider the given magnitude of the vector $\text{u}$ is $17$ and the terminal point $\left(3,13\right)$.

Let $\left(x,y\right)$ be the initial point.

The Cartesian form of the vector $\text{u}$ is:

$\text{u}=\left(3-x\right)\text{i}+\left(13-y\right)\text{j}\left(\text{Terminal point-Initial point}\right)$

The magnitude of the vector $\text{u}$ is:

$\begin{array}{rcl}17& =& \sqrt{{\left(3-x\right)}^2+{\left(13-y\right)}^2}\left(\text{Magnitude of a vector formula}\right)\\ & \Rightarrow & {17}^2={\left(\sqrt{{\left(3-x\right)}^2+{\left(13-y\right)}^2}\right)}^2\left(\text{Take square both sides}\right)\\ \therefore 289& =& {\left(3-x\right)}^2+{\left(13-y\right)}^2...\left(1\right)\end{array}$

Substitute $x=-5,y=-2$ in the equation $\left(1\right)$:

$\begin{array}{rcl}& \Rightarrow & 289={\left(3-\left(-5\right)\right)}^2+{\left(13-\left(-2\right)\right)}^2\\ & \Rightarrow & 289={\left(3+5\right)}^2+{\left(13+2\right)}^2\left(\text{Simplify}\right)\\ & \Rightarrow & 289={\left(8\right)}^2+{\left(15\right)}^2\\ & \Rightarrow & 289=64+225\\ \therefore 289& =& 289\end{array}$

Since the point $\left(-5,-2\right)$ are satisfying the equation (1), so the point $\left(-5,-2\right)$ is the initial point of the vector $\text{u}$.

Therefore, option $\left(\text{A}\right)$ is correct.

Explanation for incorrect options:

Option $\left(\text{B}\right):\left(-5,2\right)$

Substitute $x=-5,y=2$ in the equation $\left(1\right)$:

$\begin{array}{rcl}& \Rightarrow & 289={\left(3-\left(-5\right)\right)}^2+{\left(13-2\right)}^2\\ & \Rightarrow & 289={\left(3+5\right)}^2+{\left(13-2\right)}^2\left(\text{Simplify}\right)\\ & \Rightarrow & 289={\left(8\right)}^2+{\left(11\right)}^2\\ & \Rightarrow & 289=64+121\\ \therefore 289& \ne & 185\end{array}$

Since the point $\left(-5,2\right)$ is not satisfying the equation (1), so the point $\left(-5,2\right)$ is not the initial point of the vector $\text{u}$.

Therefore, option $\left(\text{B}\right)$ is incorrect.

Option $\left(\text{C}\right):\left(5,-2\right)$

Substitute $x=5,y=-2$ in the equation $\left(1\right)$:

$\begin{array}{rcl}& \Rightarrow & 289={\left(3-5\right)}^2+{\left(13-\left(-2\right)\right)}^2\\ & \Rightarrow & 289={\left(-2\right)}^2+{\left(13+2\right)}^2\left(\text{Simplify}\right)\\ & \Rightarrow & 289=4+{\left(15\right)}^2\\ & \Rightarrow & 289=4+225\\ \therefore 289& \ne & 229\end{array}$

Since the point $\left(5,-2\right)$ is not satisfying the equation (1), so the point $\left(5,-2\right)$ is not the initial point of the vector $\text{u}$.

Therefore, option $\left(\text{C}\right)$ is incorrect.

Option $\left(\text{D}\right):\left(5,2\right)$

Substitute $x=5,y=2$ in the equation $\left(1\right)$:

$\begin{array}{rcl}& \Rightarrow & 289={\left(3-5\right)}^2+{\left(13-2\right)}^2\\ & \Rightarrow & 289={\left(-2\right)}^2+{\left(11\right)}^2\left(\text{Simplify}\right)\\ & \Rightarrow & 289=4+121\\ \therefore 289& \ne & 125\end{array}$

Since the point $\left(5,2\right)$ is not satisfying the equation (1), so the point $\left(5,2\right)$ is not the initial point of the vector $\text{u}$,

Therefore, option $\left(\text{D}\right)$ is incorrect.

Hence, option $\left(\text{A}\right)$ is correct answer.