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Note on Magnitude and Direction of Vector

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Magnitude of a vector

Magnitude of a vector
Source:www.mathplanet.com
Magnitude of a vector

The magnitude of the vector is a positive number which represents the length of the vector or directed line segment. It is also known as modulus of the vector. If \(\overrightarrow {OP}\) is the position vector of P, then its magnitude is denoted by \(|\overrightarrow {OP}|\).If \(|\overrightarrow {AB}|\)is a vector, then magnitude is denoted by \(|\overrightarrow {AB}|\). If the vector is written in the single letter from \(\overrightarrow {a}\), then its magnitude is denoted by \(\overrightarrow {a}\).

If the initial point is at the origin and the terminal point at P (x,y).

\(\overrightarrow {OP}\) = \(\frac{x}{y}\) is given by \(|\overrightarrow {OP}|\) = \(\sqrt{(x)^2+(y)^2}\)

If A (x1,y1) and B(x2,y2) are two points,

We know, \(\overrightarrow {AB}\) =\(\begin{pmatrix} x_2-x_1\\y_2-y_1\end{pmatrix}\). Then \(|\overrightarrow {AB}|\) = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Thus\(|\overrightarrow {AB}|\) = \(\sqrt{(x-component)^2+ (y-component)^2}\)

If the initial point is at A(x1,y1) and terminal point at B(x2,y2).

Here, For \(\overrightarrow{AB}\)

x component of \(\overrightarrow{AB}\)= AE = x2 - x1

y component of \(\overrightarrow{AB}\) = BE = y2 - y1

Using pythagoras theorem,

(AB)2 = (AE)2 + (BE)2

or, AB2 = (x2 - x1)2 + (y2 - y1)2

or, AB = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

or, AB = \(\sqrt{(x\;component)^2 + (y\;component)^2}\)

\(\therefore\) \(|\overrightarrow {AB}|\) = AB =\(\sqrt{(x\; comp)^2 + (y\; comp)^2}\)

 

Direction of Vector

Direction of Vector
Source:mathinsight.org
Direction of Vector

Direction of a vector is the value of angle made by the vector with x-axis in positive direction. It is generally measured in degree which ranges from 0º to 360º. Direction of a vector is denoted by θ. In the figure, \(\overrightarrow {OP}\) makes an angle θwith x-axis in anticlockwise direction. Draw PM perpendicular to x-axis.

When the initial point is at the origin and terminal point at P(x,y).

 

Then, OM = x and MP =y.

In ΔPOM, tan θ = \(\frac{Perpendicular}{base}\)

or, tan θ =\(\frac{MP}{OM}\)

or, tan θ = \(\frac{y}{x}\)

∴ θ = tan-1 \((\frac{y}{x}\))

Which gives the direction of the vector.

Hence, direction of vector (θ) =tan-1 \((\frac{y-component}{x-component}\))

When the initial point is at A(x1,y1) and the terminal point is at B(x2,y2)

 

If A(x1,y1) and B(x2,y2) are two points, then direction of \(\overrightarrow {AB}\) is given by θ = tan-1 \(\begin{pmatrix}y_2-y_1\\x_2-x_1\\\end{pmatrix}\).

 

  • If a vector is parallel to x-axis from left to right, \(\theta\) = 0o
  • If a vector is parallel to x-axis from right to left, \(\theta\) = 180o
  • If a vector is parallel to y-axis down to up, then \(\theta\) = 90o
  • If a vector is parallel to y-axis up to down, then \(\theta\) = 270o
.

Very Short Questions

Here, \(\overrightarrow{OA}\) = {3,4}

\(\therefore\) x component = 3 and y component = 4

\(\therefore\) |\(\overrightarrow{OA}\)| = OA = \(\sqrt{(x comp)^2+(y comp)^2}\)

 = \(\sqrt{(3)^2 + (4)^2}\)

 = \(\sqrt{9+16}\) = \(\sqrt(25)\) = 5

\(\therefore\) \(\overrightarrow{OA}\) = 5 units

Here, P = (\(\sqrt{3}\),\(\sqrt{3}\))

\(\therefore\) x-component = \(\sqrt{3}\)

 y - component = \(\sqrt{3}\)

Let \(\theta\) be the angle made by \(\overrightarrow{OP}\) with the positive direction of x-axis.

tan\(\theta\) = \(\frac{y - component}{x - component}\) = \(\frac{\sqrt{3}}{\sqrt{3}}\) = 1

\(\therefore\) tan\(\theta\) = tan45

\(\theta\) = 45 [\(\therefore\) x-component and y-component are +ve the value must lie in 1st quadrant.]

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  • What is the value of column vector if (overrightarrow {AB}) = (egin {pmatrix} 3 \ 4 end {pmatrix}) , (overrightarrow {CD}) = (egin {pmatrix} -4 \ -3 end {pmatrix})  then express (overrightarrow {AB}) + (overrightarrow {CD})?

    (egin {pmatrix} 2 \ 2 end {pmatrix})
    (egin {pmatrix} -1 \ 1 end {pmatrix})
    (egin {pmatrix} 1 \ 1 end {pmatrix})
    (egin {pmatrix} -2 \ 2 end {pmatrix})
  •  What is the magnitude of direction if (overrightarrow {AB}) = (egin {pmatrix} 3 \ 4 end {pmatrix}) , (overrightarrow {CD}) = (egin {pmatrix} -4 \ -3 end {pmatrix})  then express (overrightarrow {AB}) + (overrightarrow {CD}) ?

    ( heta) = 110o
    ( heta) = 100o
    ( heta) = 135o
    ( heta) = 120o
  • What is the value of  column vector if (overrightarrow {PQ}) = (egin {pmatrix} -2 \ 7 end {pmatrix}) and (overrightarrow {RS}) = (egin {pmatrix} 3 \ -2 end {pmatrix})  , then express (overrightarrow {PQ}) + (overrightarrow {RS}) ?

    (egin {pmatrix} 2 \ 5 end {pmatrix})
    (egin {pmatrix} 2 \ 3 end {pmatrix})
    (egin {pmatrix} 1 \ 3 end {pmatrix})
    (egin {pmatrix} 1 \ 5 end {pmatrix})
  •  What is the magnitude if (overrightarrow{OA}) = {3,4} ?

    (overrightarrow{OA}) = 8 units
    (overrightarrow{OA}) =2 units
    (overrightarrow{OA}) = 5 units
    (overrightarrow{OA}) = 4 units
  • What is the direction of (overrightarrow{OP}) where P = ((sqrt{3}),(sqrt{3}))?

    ( heta) = 90°
    ( heta) = 30°
    ( heta) = 45°
    ( heta) = 60°
  • What is the column vector of (overrightarrow{AB}) if the vector (overrightarrow{AB}) displaces a point A(2,4) to B(5,7)?

    (egin {pmatrix} 2 \ 2 end {pmatrix})
    (egin {pmatrix} -2\ 2 end {pmatrix})
    (egin {pmatrix} -3 \ 3 end {pmatrix})
    (egin {pmatrix} 3 \ 3 end {pmatrix})
  • What is the direction of (overrightarrow{AB}) if the vector (overrightarrow{AB}) displaces a point A(2,4) to B(5,7)?

    ( heta) = 60°
    ( heta) = 90°
    ( heta) = 30°
    ( heta) = 45°
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parallel vector

what happen when two vector are parallel to each other


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