Question Video: Finding the Measure of the Angle between a Straight Line and the Normal to a Plane Mathematics

Find, to the nearest second, the measure of the angle between the straight line 𝐫 = <9𝐒 + 8𝐣 βˆ’ 5𝐀> + 𝑑<𝐒 + 𝐣 βˆ’ 3𝐀> and the normal to the plane 𝐫 β‹… <βˆ’3𝐒 + 2𝐣 + 5𝐀> = βˆ’7.


Video Transcript

Find, to the nearest second, the measure of the angle between the straight line 𝐫 equals nine 𝐒 plus eight 𝐣 minus five 𝐀 plus 𝑑 times 𝐒 plus 𝐣 minus three 𝐀 and the normal to the plane 𝐫 dot negative three 𝐒 plus two 𝐣 plus five 𝐀 equals negative seven.

Okay, here we have the equation for a straight line and the equation for a plane. And we need to be careful about how we read our question because we’re being asked to solve for the measure of the angle, not between the line and the plane, but between the line and the normal to the plane. In other words, if this is a side-on view of our plane and this orange bar is our line, then it’s not this angle here that we’re looking to solve for. But rather if 𝐧 is a vector normal to our plane, we want to solve for this angle between that vector and our line.

We’ll call that angle πœƒ. And to help us begin solving for it, we can recall that the angle between a plane and a line is given by this expression. As we’ve seen though, if we use this expression in particular, we would calculate not what we’ve marked as πœƒ, but rather this angle here. The angle we actually want to calculate between the normal to the plane and the line is equal to this angle here subtracted from 90 degrees. That means when we go to calculate the angle between a plane normal and a line, we’ll shift our trigonometric function 90 degrees from sine to cosine.

So, now, the πœƒ in this expression and the πœƒ we want to solve for match up. And we can calculate this angle by solving for a vector that, first, is parallel to our line, that’s what we’ve called 𝐩, and one that is normal to our plane. We’ve called that 𝐧. We can find these vectors by looking at the given equations for our straight line and for our plane. Using slightly different notation, we can write the equation of our straight line like this. This equation tells us that our line passes through the point nine, eight, negative five and that it’s parallel to the vector one, one, negative three. We can say then that those are the components of our vector 𝐩 parallel to our line.

Next, let’s consider the equation of our plane, which is given in what’s called vector form. Written this way, this vector here with components negative three, two, five is known to be normal to the plane surface. So, we can write 𝐧 is equal to negative three, two, five. And now, we have both of the vectors we need to substitute into our equation for the cos of πœƒ.

Plugging in 𝐩 and 𝐧 gives us this expression, and we can start evaluating the right-hand side by computing this dot product as well as squaring out the various components of these two vectors. That gives us this fraction, which further simplifies so we can write it as 16 over the square root of 11 times 38. This is equal to the cos of πœƒ. So, to solve for πœƒ itself, we take the inverse cos of both sides. Calculating this expression and rounding to the nearest second, we get a result of 38 degrees, 30 minutes, and seven seconds. This is the measure of the angle between the straight line and the normal to our given plane.

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