# Question Video: Finding the Measure of the Smaller Angle between a Straight Line and a Plane in Three Dimensions given Their Equations Mathematics

Find, to the nearest second, the measure of the smaller angle between the straight line (π₯ β 7)/7 = (π¦ β 7)/β5 = (π§ β 4)/1 and the plane 6π₯ β 8π¦ β 5π§ β17 = 0.

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### Video Transcript

Find, to the nearest second, the measure of the smaller angle between the straight line π₯ minus seven over seven equals π¦ minus seven over negative five equals π§ minus four over one and the plane six π₯ minus eight π¦ minus five π§ minus 17 equals zero.

Okay, so here we have a straight line and a plane. And we want to solve for the angle between them. Specifically, we want to calculate the measure of the smaller angle between these two objects. That is, if we imagine that our plane looked like this, looking at it edge on, and that our line passed through the plane like this, then we see that there are two different angles we could report as the angle between the plane and the line. Our problem statement tells us to report the smaller of these two values. So we know it wonβt be greater than 90 degrees.

Calling this smaller angle π, letβs recall the general mathematical relationship giving us the angle between a straight line and a plane. If π© is a vector that is parallel to a straight line and π§ is a vector normal to a plane, then the sine of the angle between this line and plane is given by this expression.

To use this equation then, weβll want to solve for a vector that is parallel to our line and one that is normal to our plane. Beginning by looking for a vector parallel to our line, if we look at the equation of the line given to us, we know that all three of these fractions are equal to one another because each one is equal to the same scale factor that we can call π‘. Knowing this lets us write out the equation of our line in whatβs called parametric form. We express our line in this form by writing one equation each for π₯, π¦, and π§.

If we look at the fraction involving π₯, this tells us that π₯ minus seven over seven is equal to π‘. And therefore, π₯ is equal to seven times π‘ plus seven. Similarly, the fraction π¦ minus seven over negative five is equal to π‘. So π¦ equals negative five π‘ plus seven. Then, lastly, since π§ minus four over one equals π‘, we can write that π§ is equal to π‘ plus four.

We can now take this parametric form of our lineβs equation and write it as a vector, where this vector π« represents the π₯-, π¦-, and π§-components of our line. Written this way, this second vector, the one by which we multiply our scale factor, is a vector that we know to be parallel to our line. We can say then that our π© vector, the one parallel to the given line, has components seven, negative five, and one.

Knowing this, letβs now look at the equation of our given plane. Itβs given to us in whatβs called general form, where the planeβs equation is written as ππ₯ plus ππ¦ plus ππ§ plus π equaling zero. We can see what each of those constants π, π, π, and π corresponds to in our given plane. And looking again at the general form of a planeβs equation, we can recall that a vector normal to such a plane will have components π, π, and π, the values that multiply π₯, π¦, and π§. We can say then that a vector that is normal to our given plane will have components six, negative eight, and negative five.

Weβve now got the two vectors we need to use this equation for the angle between our plane and line. When we substitute into this expression our π© and π§ vectors, we get this result. And our next step will be to calculate the dot product in our numerator by multiplying together the respective components of these vectors. And also in the denominator, we can square the various components of these two vectors. This gives us the magnitude of 42 plus 40 minus five divided by the square root of 49 plus 25 plus one times the square root of 36 plus 64 plus 25, which simplifies to 77 divided by the square root of 75 times 125. This expression, we recall, is equal to the sine of the angle we want to solve for.

To get that angle by itself then, weβll take the inverse sine of both sides. And when we enter this expression in our calculator and report it to the nearest second, we find a result of 52 degrees, 40 minutes, and 45 seconds. Note that this is less than 90 degrees. So itβs indeed the smaller of the two angles between our line and plane.

Our final answer then is that the smaller angle between this straight line and the given plane is 52 degrees, 40 minutes, and 45 seconds.