Question

The magnitude of a given vector with end points (4, -4, 0) and (-2, -2, 0) must be:
a)6
b)$5\sqrt{2}$
c)4
d)$2\sqrt{10}$

Answer 1

The magnitude of a vector is basically the length of the vector when represented in space. Usually vectors that are free vectors are represented in the three dimensional coordinate system with the help of coordinates. Hence the distance between the coordinates is basically the magnitude of the vector.
Formula used:
$\left| \overrightarrow{r} \right|=\sqrt{{{(x-{{x}_{\circ }})}^{2}}+{{(y-{{y}_{\circ }})}^{2}}+{{(z-{{z}_{\circ }})}^{2}}}$

Complete answer:
Let us say we have a vector in space where in the endpoints of a vector are given by the coordinates (x, y, z) and $({{x}_{\circ }},{{y}_{\circ }},{{z}_{\circ }})$ respectively. The direction of the vector is given by the unit vector whose magnitude is unity and points in the direction of the vector in space. Basically vectors magnitude is represented by the size in the coordinate system. Hence using the distance formula, the magnitude of the vector given the coordinates is given by,
$\left| \overrightarrow{r} \right|=\sqrt{{{(x-{{x}_{\circ }})}^{2}}+{{(y-{{y}_{\circ }})}^{2}}+{{(z-{{z}_{\circ }})}^{2}}}$
The coordinates of the vector in the question are given to be as (4, -4, 0) and (-2, -2, 0). Looking at the coordinates we can imply that the vector is in the X-Y plane. Using the distance formula the length of the above vector in space more specifically the magnitude of the vector is equal to,
$\begin{aligned} & \left| \overrightarrow{r} \right|=\sqrt{{{(x-{{x}_{\circ }})}^{2}}+{{(y-{{y}_{\circ }})}^{2}}+{{(z-{{z}_{\circ }})}^{2}}} \\\ & \Rightarrow \left| \overrightarrow{r} \right|=\sqrt{{{\left( 4-(-2) \right)}^{2}}+{{\left( -4-(-2) \right)}^{2}}+{{(0-0)}^{2}}} \\\ & \Rightarrow \left| \overrightarrow{r} \right|=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\\ & \Rightarrow \left| \overrightarrow{r} \right|=\sqrt{40} \\\ & \therefore \left| \overrightarrow{r} \right|=2\sqrt{10} \\\ \end{aligned}$

Therefore the correct answer of the above question is option d.

Note:
It is to be noted that a vector is given by the product of its magnitude times the direction i.e. the unit vector. The vector can have any units depending on the type of vector taken for the convenience. A vector in space can also be represented in polar form as well.