Question

7.100 find the angle at which vector A= 100i+ 100j subtended along positive x-avis counter clock wise ?
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Answer 1

Answer:

Let O(0,0) be the origin and A(10,0),B(0,100)

So by distance formula we have,

Distance between two points =

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

OA

2

=(10−0)

2

+(0−0)

2

=100,OB

2

=(0−0)

2

+(100−0)

2

=10000

OA

2

+OB

2

=100+10000=10100---------(1)

AB

2

=(10−0)

2

+(0−100)

2

=10100----------(2)

From (1) and (2) we get

OA

2

+OB

2

=AB

2

In △OAB, OA

2

+OB

2

=AB

2

∴∠AOB=90°

Therefore, angle subtended the line segment AB at origin is 90°