Question

A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude of the vector? What is the angle between the vector and the positive x-axis?

Answer 1

`\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}`

We would calculate the magnitude by applying pythagorean theorem:

`\longrightarrow \sf{Magnitude= √((-14)^2 ) + 30^2}`

`\longrightarrow \sf{Magnitude = 33.12}`

`\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}`

The angle between two vectors is given by the formula:

`\sf{\longrightarrow \small \cos \emptyset = ((a1b1 + a2b2))/( √((a1)^2 + (a2)^2√(b1)^2 + (b2)^2) ) }`

In two dimensional, the x axis of vector form is:

`\small\sf{\longrightarrow (b1, b2) = (1, 0) }`

`\sf{\longrightarrow \small \cos \: \emptyset = ((14 * 1 + 30 x 0))/(( √((-14)^2 + (30)^2)(√(1)^2 + (0)^2)) ) }`

`\small\longrightarrow \sf{ (14)/(33.12) }`

`\small\longrightarrow \sf{\emptyset \: = arcCos (( - 14)/(33.12) )}`

`\small\longrightarrow \sf{\emptyset= 115^\circ}`

`\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}`

`\small\bm{The \: angle \: between \: the \: vector \: }`

`\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}`