Question

The scalar product of the vector `hati + hatj + hatk` with a unit vector along the sum of vectors `2hati + 4hatj - 5hatk` and  `lambdahati + 2hatj +  3hatk` is equal to one. Find the value of `lambda`.

Answer 1

`(2hati + 4hatj - 5hatk) + (lambdahati + 2hatj + 3hatk)` 

= `(2 + lambda)hati + 6hatj - 2hatk`

Therefore, unit vector along `(2hati + 4hatj - 5hatk) + (lambdahati + 2hatj + 3hatk)` is given as:

`((2 + lambda)hati + 6hatj - 2hatk)/sqrt((2 + lambda)^2 + 6^2 + (-2)^2) `

`= ((2 + lambda)hati + 6hatj - 2hatk)/sqrt(4 + 4lambda + lambda^2 + 36 + 4) `

`= ((2 + lambda)hati + 6hatj - 2hatk)/sqrt(lambda^2 + 4lambda + 44)`

Scalar product of `(hati + hatj + hatk)` with this unit vector is 1.

⇒ `(hati + hatj + hatk) . ((2 + lambda)hati + 6hatj - 2hatk)/sqrt(lambda^2 + 4lambda + 44) = 1`

⇒ `((2 + lambda) + 6 - 2)/sqrt(lambda^2 + 4lambda + 44) = 1`

⇒ `sqrt(lambda^2 + 4lambda + 44) = lambda + 6`

⇒ `lambda^2 + 4lambda + 44 = (lambda + 6)^2`

⇒ `lambda^2 + 4lambda + 44 = lambda^2 + 12lambda + 36`

⇒ 8λ = 8

⇒ λ = 1

Hence, the value of λ is 1.