
If a vector has an X- component of $ - 25$ units and a Y- component of $40$ units, then the magnitude and direction of this vector are:
A) $5\sqrt {89} \;{\text{units;si}}{{\text{n}}^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }}\;{\text{with}}\;{\text{x - axis}}$
B) $5\sqrt {89} \;{\text{units;co}}{{\text{s}}^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }}\;{\text{with}}\;{\text{x - axis}}$
C) $45\;{\text{units;co}}{{\text{s}}^{ - 1}}\dfrac{{ - 5}}{9}\;{\text{with}}\;{\text{x - axis}}$
D) $45\;{\text{units;si}}{{\text{n}}^{ - 1}}\dfrac{{ - 5}}{9}\;{\text{with}}\;{\text{x - axis}}$
Answer
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Hint: 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$ units and the Y- component of the vector is $40$ units.
The expression for the magnitude of the vector having X- component and Y- component is given as,
$\sqrt {{X^2} + {Y^2}} $
Thus the magnitude of the vector by substituting the values in the above expression gives
$
\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 $5\sqrt {89} \;{\text{units}}$.
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
$
\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 ${\cos ^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }}$. 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.
Complete step by step answer:
Given the X-component of the vector is $ - 25$ units and the Y- component of the vector is $40$ units.
The expression for the magnitude of the vector having X- component and Y- component is given as,
$\sqrt {{X^2} + {Y^2}} $
Thus the magnitude of the vector by substituting the values in the above expression gives
$
\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 $5\sqrt {89} \;{\text{units}}$.
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
$
\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 ${\cos ^{ - 1}}\dfrac{{ - 5}}{{\sqrt {89} }}$. 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.
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