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:

$\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}$

$\longrightarrow \sf{Magnitude = 33.12}$

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

The angle between two vectors is given by the formula:

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

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

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

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

$\small\longrightarrow \sf{ \dfrac{14}{33.12} }$

$\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}$

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

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

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

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