Answer:
The magnitude of vector d is 16 and the angle with the x-axis is 270°
Explanation:
Operations With Vectors
Given two vectors in rectangular components:
[tex]\vec a=(ax,ay)\ ,\ \vec b=(bx,by)[/tex]
The sum of the vectors is:
[tex]\vec a+\vec b=(ax+bx,ay+by)[/tex]
The difference between the vectors is:
[tex]\vec a-\vec b=(ax-bx,ay-by)[/tex]
The magnitude of [tex]\vec a[/tex] is:
[tex]|\vec a|=\sqrt{ax^2+ay^2}[/tex]
The angle [tex]\vec a[/tex] makes with the horizontal positive direction is:
[tex]\displaystyle \tan\theta=\frac{ay}{ax}\\[/tex]
The question provides the vectors:
[tex]\vec a=(2,6)[/tex]
[tex]\vec b=(2,22)[/tex]
[tex]\vec d=\vec a-\vec b[/tex]
Calculate:
[tex]\vec d=(2,6)-(2,22)=(0,-16)[/tex]
The magnitude of [tex]\vec d[/tex] is:
[tex]|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16[/tex]
The angle is calculated by:
[tex]\displaystyle \tan\theta=\frac{-16}{0}[/tex]
The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.
The magnitude of vector d is 16 and the angle with the x-axis is 270°
