Vector in component form

Vector in component form is the the coordinate form in which vector can be broken in x-component and y-component.
$\vec{V} = < v_{x},v_{y}>$ . this representative of vector v is in standard position.
A vector whose initial point is the origin so its coordinates are(0,0) and its terminal point has coordinates $(v_{1},v_{2}$ then the component of the vector v is $(v_{1},v_{2}$.The coordinates $(v_{1}$ and $v_{2})$ are the components of 'v'. If both the initial point and the terminal point both lie on the origin then the component of such vectors are <0,0> and it is called zero vector.

Vector in component form

Let the initial point P$(p_{1}, p_{2})$ and the terminal point Q($(q_{1}, q_{2})$ then the component for of the vector PQ is given by
$\vec{PQ} = \left \langle q_{1} -p_{1} , q_{2} - p_{2} \right \rangle $
The magnitude of vector PQ is given by

$\left \| PQ \right \| =\sqrt (q_{1} -p_{1})^2 + (q_{2} -p_{2})^2$

Example 1: Find the component form and magnitude of the vector v whose initial point (4,-7) and the terminal point (-1,5).
Solution : Let P(4,-7) = $(p_{1}, p_{2})$ and Q(-1,5) =$(q_{1}, q_{2})$
$\vec{PQ} = \left \langle q_{1} -p_{1} , q_{2} - p_{2} \right \rangle $
$v_{1} = q_{1} -p_{1} $= -1 -4 = -5
$v_{2} = q_{2} -p_{2} $= 5 -(-7)= 12
$\vec{PQ} = \left \langle v_{1}, v_{2} \right \rangle $
$\vec{PQ} = \left \langle -5 , 12 \right \rangle $
Magnitude of vector PQ is
$\left \| PQ \right \| =\sqrt (v_{1}^2 + v_{2}^2)$
$\left \| PQ \right \| =\sqrt ((-5)^2 + (12)^2)$
$\left \| PQ \right \| =\sqrt (25 + 144) $
$\vec{PQ} $= 13

Example :2 Find the component form and magnitude of the vector v whose initial point (-2,3) and the terminal point (-7,9).
Solution : Let P(-2,3) = $(p_{1}, p_{2})$ and Q(-7,9) =$(q_{1}, q_{2})$
$\vec{PQ} = \left \langle q_{1} -p_{1} , q_{2} - p_{2} \right \rangle $
$v_{1} = q_{1} -p_{1} $= -7 -(-2) = -5
$v_{2} = q_{2} -p_{2} $= 9 -3= 6
$\vec{PQ} = \left \langle v_{1}, v_{2} \right \rangle $
$\vec{PQ} = \left \langle -5 , 6 \right \rangle $
Magnitude of vector PQ is
$\left \| PQ \right \| =\sqrt (v_{1}^2 + v_{2}^2)$
$\left \| PQ \right \| =\sqrt ((-5)^2 + (6)^2)$
$\left \| PQ \right \| =\sqrt (25 + 36) $
$\vec{PQ} = \sqrt 61 $
$\vec{PQ}$= 7.81

11th grade math

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