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Question
If `veca = 2hati + 2hatj + 3hatk, vecb = -veci + 2hatj + hatk and vecc = 3hati + hatj` are such that `veca + lambdavecb` is perpendicular to `vecc`, then find the value of λ.
Solution
The given vectors are `veca = 2hati + 2hatj + 3hatk, vecb = -hati + 2hatj + hatk`, and `vecc = 3hati + hatj`.
Now, `veca + λvecb = (2hati + 2hatj + 3hatk) + λ(-hati + 2hatj + hatk)`
= `(2- λ)hati + (2 + 2λ)hatj + (3 + λ)hatk`
If `(veca + λvecb)` is perpendicular to `vecc`, then `(veca + λvecb).vecc` = 0
⇒ `[(2 - λ)hati + (2 + 2λ)hatj + (3 + λ)hatk]*(3hati + hatj)` = 0
⇒ (2 – λ).3 + (2 + 2λ).1 + (3 + λ).0 = 0
⇒ 6 – 3λ + 2 + 2λ = 0
⇒ –λ + 8 = 0
⇒ λ = 8
Hence, the required value of λ is 8.
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