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.

Answer 1

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.