Video: Finding the Measure of the Angle between a Straight Line and a Plane

Find, to the nearest second, the angle between the straight line π‘₯ = 3𝑑 βˆ’ 1, 𝑦 = βˆ’2𝑑 + 4, 𝑧 = 5 and the plane 3π‘₯ βˆ’ 4𝑦 + 𝑧 = 2.

02:40

Video Transcript

Find, to the nearest second, the angle between the straight line π‘₯ equals three 𝑑 minus one, 𝑦 equals negative two 𝑑 plus four, 𝑧 equals five and the plane three π‘₯ minus four 𝑦 plus 𝑧 equals two.

Okay, so here we’re given the equation of a straight line. And it’s given to us in what’s called parametric form, written as three separate equations: one for π‘₯, one for 𝑦, and one for 𝑧. We’re also given the equation of a plane. And this is nearly in the form that’s called general form.

If we were to subtract two from both sides, we would get this result, which is indeed known as the general form of the equation of this plane. So we have our plane in general form and our line written in parametric form. Having all this, what we want to do first is solve for a vector that is parallel to our line and second for a vector that is normal or perpendicular to our plane.

If we call a vector in general that is parallel to a line 𝐩 and a vector that is normal to a plane 𝐧, then the sine of the angle between that line and that plane is given by this expression. This is why we want to solve for a vector that’s parallel to our given line and a vector that’s normal to our given plane.

Coming back to the equation of our line in parametric form, there’s a way that we can combine all three of these equations into one. We could say that the π‘₯-, 𝑦-, and 𝑧-values of this line are represented by a vector 𝐫. Furthermore, the point negative one, four, five lies on the line, and the vector three, negative two, zero is parallel to it. This means that we’ve solved for a vector 𝐩 that’s parallel to our line. It has components three, negative two, zero.

Next, to find the vector 𝐧 that is normal to our plane, we can recall that since our plane is in general form, the values by which we multiply π‘₯, 𝑦, and 𝑧 are the components of a vector normal to this plane. In other words, one vector normal to the plane has components three, negative four, one.

Now that we have our parallel and normal vectors, we can substitute them into this expression to ultimately solve for πœƒ. That gives us this expression, where in our numerator we have a dot product to calculate and in our denominator two vector magnitudes to compute. As we compute and simplify this fraction, we get to a point of having 17 divided by the square root of 13 times 26. This, we recall, is the sin of πœƒ. So to solve for πœƒ itself, we take the inverse sine of both sides.

Entering this expression into our calculator and rounding to the nearest second, our result is 67 degrees, 37 minutes, and 12 seconds. This is the angle between our line and plane to the nearest second.

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