Question 88187
To calculate the norm of the vector *[Tex \LARGE \v{v}=4i-2j] use the following formula:


*[Tex \LARGE \left|\left|\v{a}\right|\right|=sqrt{\v{a}\bullet\v{a}}] where *[Tex \Large {\v{a}\bullet\v{a}}] is the dot product of the given vector with itself


Remember the dot product of the vector *[Tex \LARGE \v{a}] with itself is:


*[Tex \LARGE \v{a}\bullet\v{a}= (a_{0})*(a_{0})+(a_{1})*(a_{1})]



*[Tex \LARGE sqrt{\v{a}\bullet\v{a}}=sqrt{(4)*(4)+(-2)*(-2)}] Calculate the dot product of the radicand


*[Tex \LARGE sqrt{16+4}] Multiply


*[Tex \LARGE sqrt{20}] Add


*[Tex \LARGE 2sqrt{5}] Simplify if possible


So *[Tex \LARGE \left|4i-2j\right|=2sqrt{5}]




Check:

Lets use Pythagorean's Theorem to check our work


Notice if we draw the vector *[Tex \LARGE \v{v}=4i-2j] we get

{{{drawing(500, 500, -5+2, 5+2, -5+2, 5+2,
graph(500, 500, -5+2, 5+2, -5+2, 5+2, 0),
green(line(0,0,4,0)),
green(line(4,0,4,-2)),
arrow(0,0,4,-2)
)}}} Plot of the vector (black line) with the vector components (green)

We can see that the vector has x and y components, which form the legs of the triangle. We can also see that the legs are 4 units and 2 units


Since we have a triangle with legs of 4 , 2 and a hypotenuse of x(our unknown side), we can use Pythagoreans theorem to find the unknown side.

Pythagoreans theorem
{{{a^2+b^2=c^2}}}


{{{4^2+2^2=c^2}}}  Plug in a=4 and b=2 and lets solve for c

{{{1 6 + 4 =  c  ^ 2}}} Square each individual term




{{{2 0 =  c  ^ 2}}} Combine like terms



{{{s q r t ( 2 0 ) = s q r t (  c  ^ 2 )}}} Take the square root of both sides


{{{2*sqrt(5)=c}}} simplify


So the length of the hypotenuse is {{{2*sqrt(5)}}}. This verifies our answer