# Thread: Angle between line and plane

1. ## Angle between line and plane

1. l:r = 2i – 3j + 5k + t ( i – j – k ), point P(0,-1,7) lies on the line l. The angle between OP and the line l.

2. Point A(1,4,-3), B(5,6,-1) and C(-3,1,2). The plane π contains points A, B and C. Find the angle between the line OC and π.

2. ## Re: Angle between line and plane

Originally Posted by alexander9408
1. l:r = 2i – 3j + 5k + t ( i – j – k ), point P(0,-1,7) lies on the line l. The angle between OP and the line l.
Here is the first one: $\displaystyle \arccos \left( {\frac{{\overrightarrow {OP} \cdot \left\langle {1, - 1, - 1} \right\rangle }}{{\left\| {\overrightarrow {OP} } \right\|\left\| {\left\langle {1, - 1, - 1} \right\rangle } \right\|}}} \right)$

3. ## Re: Angle between line and plane

Originally Posted by alexander9408
1. l:r = 2i – 3j + 5k + t ( i – j – k ), point P(0,-1,7) lies on the line l. The angle between OP and the line l.
i- j- k is, of course, a vector in the direction of line l. The vector -j+ 7k is in the direction of the line OP. And, of course, $\displaystyle \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)|$ where $\displaystyle \theta$ is the angle between the two vectors.

2. Point A(1,4,-3), B(5,6,-1) and C(-3,1,2). The plane π contains points A, B and C. Find the angle between the line OC and π.
Again, a vector in the direction of OC is -3i+ j+ 2k. The vector AC is (-3-1)i+ (1-4)j+(2-(-3))k= -4i- 3j+ 5k and the vector AB is the (5- 1)i+ (6- 4)j+ (-1-(-3))k= 4i+ 2j+ 3k. The normal vector to the plane is the cross product of those two. After you find the angle between OC and the normal to the plane, the angle between OC and the plane is its complement.