The acute angle between the line `r=(hati+2hatj+hatk)+lambda(hati+hatj+hatk)` and the plane `r.(2hati-hatj+hatk)=5`

Options

A. `cos^(-1)((sqrt2)/(3))`
B. `sin^(-1)((sqrt2)/(3))`
C. `tan^(-1)((sqrt2)/(3))`
D. `sin^(-1)((sqrt2)/(3))`

Answer

B

Explanation

We know that, the angle between the line `r=a+lambdab` and the planer r.n=d is
`sin theta =(n.b)/(|n||b|)`
Given equation of line is
`r=(hati+2hatj+hatk)+lambda(hati+hatj+hatk)` and plane is `r.(2hati-hatj+hatk)=5`
Here, `b=hati+hatj+hatk and n=2hati-hatj+hatk`
`therefore sin theta=((2hati-hatj+hatk).(hati+hatj+hatk))/(|2hati-hatj+hatk|hati+hatj+hatk))`
`=(2-1+1)/(sqrt(2^2+(-1)^(2)+(1)^(2)sqrt(1^(2)+1^(2)+1^(2))`
`(2)/(sqrt(4+1+1)sqrt(1+1+1))=(2)/(sqrt6 sqrt3))=(2)/(3sqrt2))`
`Rightarrow sin theta=(sqrt2)/(3)`
`Rightarrow theta=sin^(-1) ((sqrt2)/(3))`