Question

The scalar product of the vector ` vec a= hat i+ hat j+ hat k` with a unit vector along the sum of the vectors ` vec b=2 hat i+4 hat j-5 hat k a n d vec c=lambda hat i+2 hat j+3 hat k` is equal to 1. Find the value of `lambda` and hence find the unit vector along ` vec b+ vec cdot`

Answer 1

Correct Answer - value of `lambda`.
Let `vec(a) = hat(i)+hat(j)+hat(k), vec(b) = 2hat(i) +4hat(j) -5hat(k)` and `vec(c) = lambdahat(i)+2hat(j) + 3hat(k)`.
Now, `vec(b) + vec(c) = 2hat(i) + 4hat(j) - 5hat(k) + lambdahat(i) + 2hat(j) + 3hat(k)`
`=(2+lambda)hat(i) + 6hat(j) -2hat(k)`
`implies |vec(b)+vec(c )| = sqrt((2+lambda)^(2) + (6)^(2) + (-2)^(2))`
`=sqrt(4+lambda^(2) + 4lambda+36 + 4)`
`=sqrt(lambda^(2) + 4lambda + 44)`
Unit vector along `(vec(b) + vec(c ))` is
`=(vec(b) + vec(c))/(|vec(b) + vec(c)|)=((2+lambda) hat(i) + 6hat(j) -2hat(k))/(sqrt(lambda^(2) + 4lambda+44))`
The scalar product of the above unit vector to the vector `(hat(i) + hat(j) + hat(k))` is 1.
`therefore (hat(i) + hat(j) + hat(k)). (vec(b) + vec(c))/(|vec(b) + vec(c)|) = 1`
`implies (hat(i) + hat(j) + hat(k)).((2+lambda)hat(i)+ 6hat(j) -2hat(k))/(sqrt(lambda^(2) + 4lambda+44))=1`
`implies (1(2+lambda)+1(6)+1(-2))/(sqrt(lambda^(2) + 4lambda +44))=1`
`implies (2+lambda+6 -2)/(sqrt(lambda^(2) +4lambda +4))=1`
`implies lambda + 6 = sqrt(lambda^(2) + 4lambda+44)`
`implies (lambda + 6)^(2) =lambda^(2) + 4lambda +44`
`implies lambda^(2) + 12lambda + 36 = lambda^(2) + 4lambda + 44`.
`implies 8lambda = 8 implies lambda =1`