Question Video: Finding the Measure of Angles between Two Planes Mathematics • Third Year of Secondary School

Video Transcript

If the angle between the two planes 𝑎𝑥 minus four 𝑦 plus five 𝑧 equals negative five and 𝐫 dot seven, negative one, zero equals four is 60 degrees, find the value of the positive constant 𝑎.

Okay, so here we’re given the equations of two planes. Here’s one equation, and here is the second. For the first plane, we can say that a vector normal to its surface has components 𝑎, negative four, and positive five. Let’s call this vector 𝐧 one. And again, it’s normal or perpendicular to the first plane’s surface.

Our second plane is given to us in a form called vector form. Written this way, on the left-hand side, we have a dot product of a vector to an arbitrary point in the plane with a vector that is normal to the plane. We can say then that a vector normal to our second plane, we’ll call it 𝐧 two, has components seven, negative one, and zero.

Let’s recall now that the general equation for the cosine of the angle between two planes with normal vectors 𝐧 one and 𝐧 two is given by this expression. In this example, we’re actually told the value of what we’ve called 𝜃. It’s 60 degrees. What we don’t yet know and what we want to solve for is this value 𝑎 multiplying 𝑥 in the equation of our first plane. So if we substitute everything we know into this equation for the cosine of the angle between our planes, we can now begin simplifying the right- and the left-hand sides with the goal of solving for 𝑎, which we’re told is a positive constant value.

First, carrying out the dot product in our numerator, we get a result of seven 𝑎 plus four plus zero. And squaring the components of our normal vectors in our denominator, we have the square root of 𝑎 squared plus 16 plus 25 times the square root of 49 plus one plus zero. This equals the magnitude of seven 𝑎 plus four divided by the square root of 𝑎 squared plus 41 times the square root of 50.

Now, at this point, since we’re told that 𝑎 is a positive value, we know that seven times 𝑎 must be positive. And then when we add four to it, that will be positive two. This means we can remove the absolute value bars from our numerator since that number will be positive anyway. And now, looking at the left-hand side of this expression, let’s recall that the cos of 60 degrees is equal to exactly one-half.

Bringing it all together, we know that one-half is equal to this right-hand side. And again, it’s the value of 𝑎 that we want to solve for. To begin doing that, let’s multiply both sides of our equation by the entire denominator on the right. We then have that the square root of 50 over two times the square root of 𝑎 squared plus 41 equals seven 𝑎 plus four. And if we now square both sides of the equation, we have on our left side 50 over four times the quantity 𝑎 squared plus 41 is equal on the right to 49𝑎 squared plus 56𝑎 — that’s two times four times seven 𝑎 — plus 16.

On the left-hand side, if we multiply through by this factor of 50 over four and then subtract both 50𝑎 squared over four and 50 times 41 over four from both sides of this equation, then that makes the value on the left zero. And we have an equation that is quadratic in 𝑎. 49 minus 50 over four simplifies to 73 over two, while 16 minus 50 times 41 over four simplifies to negative 993 divided by two.

Our equation overall then is that zero equals seventy-three halves times 𝑎 squared minus 56𝑎 minus 993 over two. In terms then of an equation that lets us solve for the roots of a quadratic equation, where we solve for the roots of 𝑥 using these constant values 𝑎, 𝑏, and 𝑐, in our particular quadratic equation, we would say that 73 over two is 𝑎, negative 56 is 𝑏, and negative 993 over two is 𝑐.

We can use these labels so long as we’re careful not to be confused that this value for 𝑎 is not the same as the 𝑎 that we’re trying to solve for. That 𝑎 corresponds to 𝑥 in this equation for the roots of a quadratic equation. Knowing that, the 𝑎 that we do want to solve for is given by this equation. There are two solutions here because of this plus and minus sign in our numerator. When we calculate all this out, as expected, we get two results. When we use the minus sign in the numerator, we get negative 331 over 73. When we use the plus sign, our result is simply three.

At this point, let’s recall that our problem statement tells us that 𝑎 is a positive constant. Since it’s positive, that rules out our negative 331 over 73 result. As our final answer then, we say that 𝑎 is equal to positive three.