In the given figure, prove that: vec.AB + vec.BC + vex.CD + vec.DE = vec.AE. | Vector Geometry | SciPiPupil

Question: In the given figure, prove that: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = \vec{AE}$

Solution:

To prove: $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} = \vec{AE}$

Construction: Join AC and join CE.

Using ∆ law of vector addition in ∆ ABC,

$\vec{AC} = \vec{AB} + \vec{BC}$ - (i)

Using ∆ law of vector addition in ∆ CDE,

$\vec{CE} = \vec{CD} + \vec{DE}$ - (ii)

Using ∆ law of vector addition in ∆ ACE,

$\vec{AE} = \vec{AC} + \vec{CE}$ - (iii)

Taking RHS

$= \vec{AE}$

[ Put $\vec{AE} = \vec{AC} + \vec{CE}$ from equation (iii) ]

$= \vec{AC} + \vec{CE}$

[ Put $\vec{AC} = \vec{AB} + \vec{BC}$ from equation (i) ]

$= \vec{AB} + \vec{BC} + \vec{CE}$

[ Put $\vec{CE} = \vec{CD} + \vec{DE}$ from equation (ii) ]

$= \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE}$

= LHS

#proved