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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`.
Solution
`(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.
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