Answer:
The magnitude of vector is:
[tex]\left|\begin{pmatrix}5&-6\end{pmatrix}\right|=\sqrt{61}[/tex]
v = <5, -6> means the vector has x-coordinate x = 5 and y-coordinate y = -6, so the vector v = <5, -6> is heading towards SE.
Thus, option ( j ) is correct.
i.e. [tex]\sqrt{61},\:SE[/tex]
Step-by-step explanation:
Given the vector
v = <5, -6>
Determining the magnitude of the vector
To find a magnitude of a vector v = (a, b) we use the formula
[tex]||v||\:=\:\sqrt{a^2+b^2}[/tex]
Magnitude of the vector is basically termed as the length of the vector, which is denoted by
[tex]|\left(5,\:-6\right)|=\sqrt{\left(5\right)^2+\left(-6\right)^2}[/tex]
[tex]=\sqrt{5^2+6^2}[/tex]
[tex]=\sqrt{25+36}[/tex]
[tex]=\sqrt{61}[/tex]
Thus, the magnitude of vector is:
[tex]\left|\begin{pmatrix}5&-6\end{pmatrix}\right|=\sqrt{61}[/tex]
As the vector v = <5, -6> lies in 4th quadrant.
v = <5, -6> means the vector has x-coordinate x = 5 and y-coordinate y = -6, so the vector v = <5, -6> is heading towards SE.
Thus, option ( j ) is correct.
i.e. [tex]\sqrt{61},\:SE[/tex]
