Question

The acute angle between the line `vecr = (hati + 2hatj + hatk) + λ(hati + hatj + hatk)` and the plane `vecr xx (2hati - hatj + hatk)` is ______.

Options

• `cos^-1(sqrt(2)/3)`

• `sin^-1(sqrt(2)/3)`

• `tan^-1(sqrt(2)/3)`

• `sin^-1(sqrt(2)/sqrt(3))`

Answer 1

The acute angle between the line `vecr = (hati + 2hatj + hatk) + λ(hati + hatj + hatk)` and the plane `vecr xx (2hati - hatj + hatk)` is `underlinebb(sin^-1(sqrt(2)/3))`.

Explanation:

The angle between the line

`vecr = veca + λvecb` and the plane `vecr.hatn` = d is given by

sin θ = `(hatn.vecb)/(|hatn||vecb|)`

The equation of given line is

`vecr = (hati + 2hatj + hatk) + λ(hati + hatj + hatk)` and equation of plane is

`vecr.(2hati - hatj + hatk)` = 5

As, `vecb = hati + hatj + hatk` and `hatn = 2hati - hatj + hatk`

So, sin θ = `((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/(sqrt(6)sqrt(3))`

= `sqrt(2)/3`

Hence, θ = `sin^-1(sqrt(2)/3)`