Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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?

Sagot :

Kailes

[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

We would calculate the magnitude by applying pythagorean theorem:

[tex]\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}[/tex]

[tex]\longrightarrow \sf{Magnitude = 33.12}[/tex]

[tex]\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}[/tex]

The angle between two vectors is given by the formula:

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

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

[tex]\small\sf{\longrightarrow (b1, b2) = (1, 0) }[/tex]

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

[tex]\small\longrightarrow \sf{ \dfrac{14}{33.12} }[/tex]

[tex]\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}[/tex]

[tex]\small\longrightarrow \sf{\emptyset= 115^\circ}[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

[tex] \small\bm{The \: angle \: between \: the \: vector \: }[/tex]

[tex]\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}[/tex]