The angle between the line r.(i + j - 3k) + λ(2i + 2j + k) and the plane r.(6i - 3j + 2k) = 5 is ​

Question

The angle between the line \(\bar{r}\).(\(\hat{i}\) + \(\hat{j}\) - 3\(\hat{k}\)) + λ(2\(\hat{i}\) + 2\(\hat{j}\) + \(\hat{k}\)) and the plane \(\bar{r}\).(6\(\hat{i}\) - 3\(\hat{j}\) + 2\(\hat{k}\)) = 5 is

A. cos^-1\((\frac{8}{21})\)

B. cos^-1\((\frac{5}{21})\)

C. sin^-1\((\frac{5}{21})\) 

D. sin^-1\((\frac{8}{21})\)

Answer 1

Given: Equation of line is \(\bar{r}\).(\(\hat{i}\) + \(\hat{j}\) - 3\(\hat{k}\)) + λ(2\(\hat{i}\) + 2\(\hat{j}\) + \(\hat{k}\)) 

Equation of plane is \(\bar{r}\).(6\(\hat{i}\) - 3\(\hat{j}\) + 2\(\hat{k}\)) = 5

To find: angle between line and plane 

Formula Used: If θ is the angle between a line with direction ratio b₁:b₂:b₃ and a plane with direction ratio of normal n₁:n₂:n₃, then

Explanation: 

Here direction ratio of the line is 2 : 2 : 1 

Direction ratio of normal to the plane is 6 : -3 : 2 

Therefore,

Therefore, angle between the line and plane is sin^-1\(\frac{8}{21}\)