Which of the following is the smaller angle between the straight line 𝐫 equals negative seven 𝐢 minus 𝐣 minus nine 𝐤 plus 𝑡 times two 𝐢 plus 𝐣 minus 𝐤 and the plane 𝐫 dot nine 𝐢 minus nine 𝐣 plus two 𝐤 equals 13? (A) The inverse sin of seven times the square root of 249 divided by 498. (B) The inverse cos of seven times the square root of 249 divided by 498. (C) The inverse sin of seven times the square root of 43 over 86. (D) The inverse cos of seven times the square root of 43 over 86.
Okay, in this exercise, we’re given the equation of a line and the equation of a plane. Let’s imagine that this is our line and this is our plane. We want to solve for the smaller angle between these two objects. We imagine this line and plane intersecting in three-dimensional space. And we see that that intersection forms two angles, this one here as well as this one. Our problem statement tells us to solve for the smaller of these two angles.
To do this, we’ll need to know a vector that is normal or perpendicular to our plane as well as a vector that is parallel to our line. Beginning with our plane, let’s say that a vector normal to its surface is called 𝐧. If we look at the form in which the equation of our plane is given, we see it’s presented to us in what’s called vector form. Written this way, a vector to a general point in the plane is dotted with a vector normal to the plane. From this then, we can read off the components of our vector 𝐧, nine, negative nine, two.
Now let’s look to solve for a vector that is parallel to our line. We’ll call this vector 𝐩. If we look at the form in which the equation of our line is given, we can find a vector that’s parallel to this line by looking for the scale factor; in this case it’s called 𝑡. The vector that multiplies the scale factor is parallel to our line. We can say then that the components of 𝐩 are two, one, and negative one. Now that we have our vector normal to our plane and a vector parallel to our line, we can recall the general equation that the sin of the angle 𝛼 between a plane and a line is equal to the magnitude of the dot product of a vector parallel to the line and one normal to the plane all divided by the product of the magnitude of each of these vectors.
To solve for 𝛼 then, we can go about calculating this numerator and denominator and then bringing them together. The magnitude of 𝐩 dot 𝐧 using our 𝐩 and 𝐧 vectors equals the magnitude of two, one, negative one dotted with nine, negative nine, two. This equals the magnitude of 18 minus nine minus two or seven. Knowing this result, now let’s calculate the denominator of our fraction. This equals the square root of the sum of the squares of two, one, and negative one times the square root of the sum of the squares of nine, negative nine, and two. That equals this expression or the square root of six times the square root of 166, which equals the square root of 996.
We can now say that the sin of the angle 𝛼 equals seven over the square root of 996. And if we rationalize our denominator, this equals seven times the square root of 996 over 996. And then we note that 996 is equal to four times 249. Written this way, we can bring the four outside of the square root, where it becomes a two. It turns out, though, that 996 is divisible by two. So if we divide numerator and denominator of this fraction by two, we find it equals seven times the square root of 249 divided by 498. So if the sin of 𝛼 is equal to this expression, then 𝛼 is the inverse sin of seven times the square root of 249 over 498.
Now, if we evaluate this on our calculator, we find a result of about 13 degrees. This confirms to us that, indeed, we’ve solved for the smaller of the two angles of intersection between our line and the plane. Our result matches up with answer option (A). The smaller angle between this line and plane is the inverse sin of seven times the square root of 249 divided by 498.