Find the magnitude and direction of a vector F = 23i + 9J.


Find the magnitude and direction of a vector F = 23i + 9J.

Magnitude and Direction of a Vector:

The magnitude of a vector can be calculated by taking the square root of the summation of the square of both X and Y components of a vector, and direction can find out by taking the inverse of the tangent of the fraction of Y and X component.

Answer and Explanation:

Given data

  • The given vector is: {eq}{\vec v} = 23\hat i + 9\hat j{/eq}.
  • X-component of the vector is: {eq}x = 23{/eq}.
  • Y-component of the vector is: {eq}y = 9{/eq}.

Calculate the magnitude of the vector,

{eq}\begin{align*} \left| {\vec v} \right| &= \sqrt {{x^2} + {y^2}} \\ &= \sqrt {{{\left( {23} \right)}^2} + {{\left( 9 \right)}^2}} \\ &= \sqrt {529 + 81} \\ &= \sqrt {610} \\ &= 24.7 \end{align*}{/eq}

Calculate the direction of the vector,

{eq}\begin{align*} \tan \theta &= \dfrac{y}{x}\\ \theta &= {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\\ &= {\tan ^{ - 1}}\left( {\dfrac{9}{{23}}} \right)\\ &= 21.37^\circ \end{align*}{/eq}

The direction of the vector will be at 1st quadrant because, both the components have positive value.

Thus, the magnitude of the vector is {eq}24.7{/eq} and direction is {eq}21.37^\circ{/eq} from positive X-axis in the 1st quadrant.

Learn more about this topic:

Vectors: Definition, Types & Examples

from Common Entrance Test (CET): Study Guide & Syllabus

Chapter 57 / Lesson 3

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