# How to find the acute angle between the tangent to the curve r(t) = ti + t2j - 3tk and the line normal to the plane 2x - y + z = -2 at each point of intersection?

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We need to find two vectors before we start to consider their angle of intersection. We need to find:

1) The normal vector to the plane

2) The vector representing the tangent to the curve

**Step 1**: **Find the normal vector**

Based on the definition of a plane that we see in vector calculus, we can determine the vector pretty easily from the equation you gave,

`2x-y+z = -2`

When you have a plane in the form Ax+By+Cz = D, your normal vector will be the following (see link below):

`vec(N) = <A,B,C>`

You may be wondering, "What about D?" Well, D doesn't matter so much for finding vectors, because it just gives you where exactly the plane is at instead of its orientation. Because we're trying to find angles, we only care about the orientation of the plane (and, by extension, it's normal vector).

So, just to wrap up this part, the normal vector we find based on the equation of the plane:

`vec(N) = <2, -1, 1>`

**Step 2: Finding the tangent to the curve r(t)**

Finding the tangent to the curve r(t) is much like finding the tangent to a curve in single-variable calculus. To find the rate of change of the curve in each dimension, we just take the derivative of the parameterized curve r(t) with respect to t!

`(dvec(r)(t))/dt = vec(i) + 2vec(j) - 3vec(k) = <1, 2, -3>`

That's pretty convenient. It turns out the tangent vector is constant! This will help significantly with how we're going to determine the angle of intersection.

**Step 3: Find the angle of intersection**

Now we have vectors defining the direction of the two lines that will intersect:

`vec(N) = <2, -1, 1> = 2vec(i) - vec(j) + vec(k)`

`vec(r)'(t) = <1, 2, -3> = vec(i) + 2vec(j) - 3vec(k)`

If you recall the properties of the dot (or scalar) product (if not, see link below), the following formula holds for two vectors:

`vec(a)*vec(b) = |vec(a)||vec(b)|costheta`

where `|vec(a)|` is the magnitude of vector a and `costheta` is the cosine of their angle of intersection.

We have our two vectors, but in order to use this equation, we'll need their magnitudes, calculated here:

`|vec(N)| = sqrt(2^2 + (-1)^2 + 1^2) = sqrt(6)`

`|vec(r)'(t)| = sqrt(1^2 + 2^2 + (-3)^2) = sqrt(14)`

Now, we have the information we need to find `costheta`

Using the dot product equation above:

`vec(N)*vec(r)'(t) = |vec(N)||vec(r)'(t)|costheta`

`<2,-1,1>*<1,2,-3> = sqrt(6)sqrt(14)costheta`

Now, solve for the dot product and simplify the square roots on the right:

`-3 = 2sqrt(21)costheta`

Now, just divide both sides by 2sqrt(21):

`-3/(2sqrt(21)) = costheta`

`-1/2 sqrt(3/7) = costheta`

This doesn't line up well with our trig identities, so it's time to go to the calculator and take the arccos (`cos^-1`) of both sides to find theta:

`cos^-1(-1/2sqrt(3/7)) = cos^-1(costheta)`

`theta` = 1.904 radians = 108.9 degrees

We're almost done. The problem asks specifically for the acute angle and we have an obtuse one.

When two lines intersect, we always get one obtuse angle and one acute angle (unless they are perpendicular). Thinking back to high school geometry, we know that both angles add to 180 degrees (or pi radians), so we'll calculate the angle they want by subracting the angle we found from 180 degress:

`theta_(acute) = 180 - theta_(obtuse)`

`theta_(acute) = 180 - 108.9 = 71.1`

There we go, the acute angle at which the tangent to the curve and the normal to the plane intersect is **71.1 degress**.

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