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Question Video: Finding the General Equation of a Plane Passing through a Given Point and Perpendicular to Two Given Planes Mathematics

Find the general equation of the plane which passes through the point (2, 8, 1) and is perpendicular to the two planes βˆ’6π‘₯ βˆ’ 4𝑦 + 6𝑧 = βˆ’5 and 5π‘₯ + 3𝑦 βˆ’ 6𝑧 = 3.

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Video Transcript

Find the general equation of the plane which passes through the point two, eight, one and is perpendicular to the two planes negative six π‘₯ negative four 𝑦 plus six 𝑧 equals negative five and five π‘₯ plus three 𝑦 minus six 𝑧 equals three.

Okay, so in this exercise, we’re told about three planes, two of which we have the equations for. The third plane whose general equation we want to solve for is perpendicular to these two and it also passes through this given point. Now, in general, we can write the equation for a plane if we have two bits of information about it. First, if we know a vector that is normal to the surface of the plane and, second, if we know a point on the plane. In this scenario, we see that, indeed, we are given a point which lies on this unknown plane. But we don’t yet know the components of a vector normal to it.

However, notice this. We do know that this plane, whatever its equation, is perpendicular to these two planes given by these equations. This implies that the normal vector 𝐧 we want to solve for is also perpendicular to the normal vectors of these two planes, meaning if we can solve for the normal vectors of these planes and then find a vector perpendicular to them both, we’ll have an expression for 𝐧. We can begin doing this by recalling that when the equation of a plane is given in what’s called standard form, like we see it here, then the components of vectors normal to those planes are given by the factors by which we multiply the variables π‘₯, 𝑦, and 𝑧.

For example, if we call a vector normal to our first plane 𝐧 one, then this may have components negative six, negative four, six. And if we call a vector normal to our second plane 𝐧 two, this will have components five, three, negative six. And now that we know 𝐧 one and 𝐧 two, recall that 𝐧 is perpendicular to both of these vectors. So here’s what we can do. We can take the cross product of 𝐧 one and 𝐧 two. And if we recall that the cross product of two vectors β€” we’ve called them 𝐀 and 𝐁, which are three-dimensional β€” is equal to the determinant of this three-by-three matrix, then we can apply this relationship to our scenario, where the second and third rows of our matrix are populated by the respective components of our vectors 𝐧 one and 𝐧 two.

We see that the 𝐒-component of 𝐧 one is negative six, the 𝐣-component is negative four, and the 𝐀-component is positive six. Likewise, the 𝐒-component of 𝐧 two is five, its 𝐣-component is three, and its 𝐀-component negative six. We’re now ready to calculate this cross product. And we’ll do it component by component, starting with 𝐒. The 𝐒-component of the vector resulting from crossing 𝐧 one and __ two is equal to the determinant of this two-by-two matrix. That’s a negative four times negative six or 24 minus six times three or 18. We then move on to the 𝐣-component, which is negative the determinant of this two-by-two matrix. Negative six times negative six is positive 36. And we subtract from that six times five or 30.

And lastly then, we compute the 𝐀-component equal to the determinant of this two-by-two matrix. Negative six times three is negative 18 minus negative four times five or negative 20. Adding these three components together, we get the cross product of 𝐧 one and 𝐧 two. That’s equal to six 𝐒 minus six 𝐣 plus two 𝐀 or, written in vector form, six, negative six, two. Now that we know this cross product, let’s recall that we were calculating it because we knew that 𝐧 is perpendicular to 𝐧 one and 𝐧 two. And since this vector resulting from crossing 𝐧 one and 𝐧 two is perpendicular to them both by the rule of the cross product, we can say that it’s equal to the normal vector 𝐧.

Now that we know this and we also know a point on our plane 𝑃, we can make quick progress in finding the general equation of the plane. We can write the equation of our plane as the normal vector 𝐧 dotted with a vector to a general point on the plane with coordinates π‘₯, 𝑦, and 𝑧 being equal to that same normal vector dotted with the vector to the known point on the plane with coordinates two, eight, one. And since we know the vector form of 𝐧, we could substitute that into this equation. And having done that, if we then calculate both of these dot products, we find that six π‘₯ minus six 𝑦 plus two 𝑧 equals 12 minus 48 plus two. This is equal to negative 34.

So if we add 34 to both sides of our equation, we can write that six π‘₯ minus six 𝑦 plus two 𝑧 plus 34 equals zero. And noticing that all the terms on the left are divisible by two, we find that three π‘₯ minus three 𝑦 plus 𝑧 plus 17 equals zero. And this is the general equation of the plane which passes through the point two, eight, one and is perpendicular to the two planes given.

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