Question

How do you do this???

Answer 1

Answer:

<11,33>

Step-by-step explanation:

If it is perpendicular to <3,-1>  We can use  <0,0> as the other point to get the slope

m = (y2-y1)/(x2-x2)

The slope of the vector is

  (-1-0)/(3-0)

=  -1/3

Then we want a vector perpendicular, so we take the negative reciprocal of the slope

- (-3/1) =3

So the perpendicular slope is 3

Our new vector y/x = 3/1

    Using cross products

3x=y

The magnitude has to be 11 sqrt (10)

magnitude = sqrt( ((x)^2 + y^2) = 11sqrt(10)

magnitude = sqrt( ((x^2 + y^2) = 11sqrt(10)

Square both sides

x^2+y^2 = 11^2*10

x^2 + y^2 =1210

We have 2 equations and 2 unknowns

x^2 +y^2 =1210  and

3x=y

Replace y with 3x

x^2 + (3x)^2 =1210

x^2 +9x^2 =1210

10x^2 =1210

Divide both sides by 10

x^2 =121

Take the square root of each side

sqrt(x^2) = sqrt(121)

x=±11

But it problems states it has to be in the first quadrant so x=11

x =11

Now we can find y

y =3x

y =11*3

y =33

The vector is <11, 33>

Answer 2

$\text{Let}\ \vec{v}=\ \text{and}\ \vec{q}=,\ \text{then}\\\\\vec{q}\ \perp\ \vec{v}\iff \vec{q}\ \circ\ \veq{v}=ac+bd=0\\\\\text{The formula of magnitude}\ |\vec{v}|=\sqrt{a^2+b^2}.\\--------------------------------\\\text{We have}\ \vec{v}=.\ \text{Let}\ \vec{q}=.\\\\\vec{q}\ \perp\ \vec{v}\iff\ \circ\ =0\\\\3x-1y=0\\\\3x-y=0\qquad\text{add y to both sides}\\\\3x=y\to\boxed{y=3x}\\\\\text{The magnitude is}\ 11\sqrt{10}.\ \text{Therefore we have second equation:}$

$\sqrt{x^2+y^2}=11\sqrt{10}\qquad\text{square of both sides}\\\\x^2+y^2=11^2(\sqrt{10})^2\\\\x^2+y^2=121(10)\\\\x^2+y^2=1210\qquad\text{substitute}\ y=3x\\\\x^2+(3x)^2=1210\\\\x^2+3^2x^2=1210\\\\x^2+9x^2=1210\\\\10x^2=1210\qquad\text{divide both sides by 10}\\\\x^2=121\to x=\pm\sqrt{121}\to x=-11\ or\ x=11.\\\\\text{In first quadrant}\ x>0.\ \text{Therefore}\ x=11.\\\\\text{Put the value of x to}\ y=3x:\\\\y=3(11)=33\\\\Answer:\ \boxed{}$