Answer:
Explanation:
Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula
[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]| \vec{A} |[/tex] its the magnitude and θ.
So, for our vectors, we will have:
[tex]\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )[/tex]
[tex]\vec{D}= ( 2.121 m , -2.121 m )[/tex]
and
[tex]\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )[/tex]
[tex]\vec{E}= ( 2.71 m , 3.59 m )[/tex]
Now, we can take the sum of the vectors
[tex]\vec{R} = \vec{D} + \vec{E}[/tex]
[tex]\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )[/tex]
[tex]\vec{R} = ( 2.121 \ m + 2.71 \ m , -2.121 \ m + 3.59 \ m ) [/tex]
[tex]\vec{R} = ( 4.831 \ m , 1.469 \ m ) [/tex]
This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem
[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]
[tex]|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}[/tex]
[tex]|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}[/tex]
[tex]|\vec{R}| = \sqrt{25.496 m^2}[/tex]
[tex]|\vec{R}| = 5.049 m[/tex]
To find the direction, we can use
[tex]\theta = arctan(\frac{R_y}{R_x})[/tex]
[tex]\theta = arctan(\frac{1.469 \ m}{4.831 \ m})[/tex]
[tex]\theta = arctan(0.304)[/tex]
[tex]\theta = 16\°54'33''[/tex]
As we are in the first quadrant, this is relative to the x axis.
