The angle between line r = a +λ b and plane r ⋅ m =d is given by

Question

The angle between line r = a +λ b and plane r ⋅ m =d is given by (A)  cos -1|( b ⋅ n /| b | ⋅ n mid)| (B)  sin -1 mid ( b × n / b | n |) (C)  sin -1|( b ⋅ n /| b || n |)| (D)  cos -1|( b × n /| b | n |)|

Tardigrade Admin · Accepted Answer

If the equation of the line is r=a+λ b and the equation of the plane is r ⋅ n=d. Then, the angle θ between the line and the normal to the plane is cos θ=|(b ⋅ n/|b||n|)| and so the angle φ between the line and the plane is given by 90-θ, i.e., sin (90°-θ) = cos θ text i.e., sin φ =|(b ⋅ n/|b||n|)| text or φ= sin -1|(b ⋅ n/|b||n|)|

Q. The angle between line $r = a + λb$ and plane $r \cdot m = d$ is given by

$cos^{- 1} ∣ ∣ \frac{b \cdot n}{∣ b ∣ \cdot n ∣} ∣ ∣$

$sin^{- 1} ∣ \frac{b \times n}{b ∣ n ∣}$

$sin^{- 1} ∣ ∣ \frac{b \cdot n}{∣ b ∣∣ n ∣} ∣ ∣$

$cos^{- 1} ∣ ∣ \frac{b \times n}{∣ b ∥ n ∣} ∣ ∣$

Q. The angle between line r=a+λb and plane r⋅m=d is given by

cos−1∣∣​∣b∣⋅n∣b⋅n​∣∣​

sin−1∣b∣n∣b×n​

sin−1∣∣​∣b∣∣n∣b⋅n​∣∣​

cos−1∣∣​∣b∥n∣b×n​∣∣​

Q. The angle between line $r = a + λb$ and plane $r \cdot m = d$ is given by

$cos^{- 1} ∣ ∣ \frac{b \cdot n}{∣ b ∣ \cdot n ∣} ∣ ∣$

$sin^{- 1} ∣ \frac{b \times n}{b ∣ n ∣}$

$sin^{- 1} ∣ ∣ \frac{b \cdot n}{∣ b ∣∣ n ∣} ∣ ∣$

$cos^{- 1} ∣ ∣ \frac{b \times n}{∣ b ∥ n ∣} ∣ ∣$