Solveeit Logo

Question

Question: If a vector has an X- component of \( - 25\) units and a Y- component of \(40\) units, then the magn...

If a vector has an X- component of 25 - 25 units and a Y- component of 4040 units, then the magnitude and direction of this vector are:
A) 589  units;sin1589  with  x - axis5\sqrt {89} \;{\text{units;si}}{{\text{n}}^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }}\;{\text{with}}\;{\text{x - axis}}
B) 589  units;cos1589  with  x - axis5\sqrt {89} \;{\text{units;co}}{{\text{s}}^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }}\;{\text{with}}\;{\text{x - axis}}
C) 45  units;cos159  with  x - axis45\;{\text{units;co}}{{\text{s}}^{ - 1}}\dfrac{{ - 5}}{9}\;{\text{with}}\;{\text{x - axis}}
D) 45  units;sin159  with  x - axis45\;{\text{units;si}}{{\text{n}}^{ - 1}}\dfrac{{ - 5}}{9}\;{\text{with}}\;{\text{x - axis}}

Explanation

Solution

The magnitude of a vector is the square root of the sum of components of X- direction, Y- direction, and Z- direction. And the quadrant where the vector belongs has to be found. Thus the angle of the vector made with the X-axis can be found by using the tangent function.

Complete step by step answer:
Given the X-component of the vector is 25 - 25 units and the Y- component of the vector is 4040 units.
The expression for the magnitude of the vector having X- component and Y- component is given as,
X2+Y2\sqrt {{X^2} + {Y^2}}
Thus the magnitude of the vector by substituting the values in the above expression gives
(25)2+402=2225 =25×89 =589  units  \sqrt {{{\left( { - 25} \right)}^2} + {{40}^2}} = \sqrt {2225} \\\ = \sqrt {25 \times 89} \\\ = 5\sqrt {89} \;{\text{units}} \\\
Thus the magnitude of the vector is 589  units5\sqrt {89} \;{\text{units}}.
The angle the vector makes with the horizontal line can be defined as the direction of the vector.
And, the angle the vector makes with the X-axis is given as,
The sign of the X- component is negative and the sign of Y- component is positive. It implies that the vector is in the second quadrant. Thus the angle it makes with the X-axis is
θ=tan1(yx) =tan1(4025) =tan1(85) =cos1589  \theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) \\\ = {\tan ^{ - 1}}\left( {\dfrac{{40}}{{ - 25}}} \right) \\\ = {\tan ^{ - 1}}\left( {\dfrac{8}{{ - 5}}} \right) \\\ = {\cos ^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }} \\\

Therefore the angle it made with the X- axis is cos1589{\cos ^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }}. Hence, the answer is option B.

Note:
The X- component of the vector will be the unit that makes the vector to left or right. And, the Y- component of the vector will be the unit that makes the vector up or down. And the direction depends on both the components.