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Question
If `veca and vecb` are two vectors such that `|veca+vecb|=|veca|,` then prove that vector `2veca+vecb` is perpendicular to vector `vecb`
Solution
Given |`veca+vecb|=|veca|`∣
`|veca+vecb|^2=|veca|^2`
`|veca|^2+2veca.vecb+|vecb|^2=|veca|^2`
`2veca.vecb+|vecb|^2=0 ................(1)`
`Now (2veca+vecb)(vecb)=2vecavecb+vecbvecb=2vecavecb+|vecb|^2=0 using(1)`
We know that, if the dot product of two vectors is zero, then either of the two vectors is zero or the vectors are perpendicular to each other.
Thus,
Given `|veca+vecb|=|veca|`
`|veca+vecb|^2=|veca|^2`
`|veca|^2+2veca.vecb+|vecb|^2=|veca|^2`
`2veca.vecb+|vecb|^2=0 ................(1)`
`Now (2veca+vecb)(vecb)=2vecavecb+vecbvecb=2vecavecb+|vecb|^2=0 using(1)`
We know that, if the dot product of two vectors is zero, then either of the two vectors is zero or the vectors are perpendicular to each other.
Thus, `2veca+vecb` is perpendicular to `vecb`
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