Find the angle between the line `(x-1)/(3)=(y+1)/(2)=(z+2)/(4)` and the plane `2x + y - 3z + 4 = 0` .

Explanation

The angle `theta` between the line
`(x-x_(1))/(a_(1))=(y-y_(1))/(b_(1))=(z-z_(1))/(c_(1))` and the plane `ax + by + cz +d = 0` is given by
`sin theta=|(aa_(1)+"bb"_(1)+"cc"_(1))/(sqrt(a^(2)+b^(2)+c^(2))*sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2)))|`
Here, `a_(1) = 3, b_(1) =2, c_(1) = 4 and a = 2, b = 1, c = -3`
`therefore aa_(1)+"bb"_(1) + "cc"_(1)=2(3)+1(2)+(-3)(4)`
` = 6 + 2 -12 = -4`
`sqrt(a^(2)+b^(2)+c^(2))=sqrt(2^(2)+1^(2)+(-3)^(2))=sqrt(4+1+9)=sqrt(14)` and
`sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2))=sqrt(3^(2)+2^(2)+4^(2))=sqrt(9+4+16)=sqrt(29)`
`therefore` from (1), we have
`sin theta=|(-4)/(sqrt(14)sqrt(29))|=|(-4)/(sqrt(406))|=(4)/(sqrt(406))`
`therefore theta = sin^(-1)j ((4)/(sqrt(406)))`.