Angle of intersection between a line and a plane

Question

I have a line $L$ given by $x = 2 -t$, $y = 1 + t$, $z = 1 + 2t$, which intersects a plane $2x + y - z = 1$ at the point $(1,2,3)$. I have to find the angle which the line makes with the plane. I know that, to do this, I should use the following formula: $cos\theta = \frac{\vec{u}\cdot\vec{v}} {||{\vec{u}}||\cdot||{\vec{v}}||}$. However, do I first need to find an equation for the plane using the derivative of $L$ and the point? What would be my $\vec{u}$ and what would be my $\vec{v}$ if this were the case?

Answer 1

The line can be written as

$$\pmatrix{x\\y\\z}=\pmatrix{2\\1\\1}+t\pmatrix{-1\\1\\2}\;,$$

and the plane can be written as

$$\pmatrix{2\\1\\-1}\cdot\pmatrix{x\\y\\z}=1\;.$$

The angle between the direction vector $\pmatrix{-1\\1\\2}$ of the line and the normal vector $\pmatrix{2\\1\\-1}$ of the plane is complementary to the angle between the line and the plane.

Answer 2

Confusing question. Maybe deliberately. The point of intersection on the plane is irrelevant, and the point on the line is irrelevant. All that matters is the direction vector of the line and the normal vector of the plane.

Answer 3

The equation of the line is, $$\frac{x-2}{-1}=\frac{y-1}{1}=\frac{z-1}{2}=t$$

the direction ratios of the line are $(-1,1,2)$

and the direction ratios of the normal vector of the plane are $(2,1,-1)$

the angle between these $2$ vectors gives the angle between the planes.