Question

Write the vector v in the form aiplusbj​, given its magnitude left norm Bold v right norm and the angle alpha it makes with the positive​ x-axis.Left norm Bold v right normequals12​, alphaequals150

Answer 1

Answer:

$6\sqrt{3}i-6j$

Step-by-step explanation:

Given:

The magnitude, v=12

The angle $\alpha$ it makes with the positive​ x-axis =$150^0$

Now for a vector (x,y)

• $\alpha =tan^{-1}(\frac{y}{x} )$ ; and

• $v=\sqrt{x^2+y^2}$

Therefore:

$150 =tan^{-1}(\frac{y}{x} )\\\frac{y}{x} =tan 150\\\frac{y}{x}=-\frac{\sqrt{3} }{3}$

$y=-\frac{x \sqrt{3}}{3}$

Similarly:

$v=\sqrt{x^2+y^2}\\12=\sqrt{x^2+y^2}\\12^2=x^2+y^2\\144=x^2+y^2$

$144=x^2+(-\frac{x \sqrt{3}}{3} )^2\\144=x^2+\frac{3x^2}{9} \\144=\frac{4x^2}{3}\\4x^2=144 X 3=432\\x^2=108\\x=6\sqrt{3}$

Recall that:

$y=-\frac{x \sqrt{3}}{3}\\y=-\frac{6\sqrt{3} \sqrt{3}}{3}=-6$

Therefore: $(x,y)=(6\sqrt{3}, -6)$

The vector in the form ai+bj is therefore:

$6\sqrt{3}i-6j$