### Video Transcript

Find the different forms of the equation of the plane that passes through the point two, negative one, zero and has normal vector π’ equals four π plus 10π minus seven π.

There are three forms of equation of a plane: the vector form, the standard form, and the general form. Given a general plane which passes through the point π΄ with coordinates π₯ one, π¦ one, π§ one and which has a normal vector of π with components π, π, and π, the equation of that plane in vector form is π dot π equals π dot π΄. If youβre going to remember just one of the forms of the equation of a plane, make it this one. Itβs easy to get the other two forms of the equation, if we start from this form. We can write π in terms of its components π, π, and π. So π, π, π dot π equals π, π, π dot π΄. And we can choose the components of the vector π, which is the position vector of an arbitrary point on our plane, to be π₯, π¦, and π§, as is convention.

So on the left-hand side, we get π, π, π dot π₯, π¦, π§. We can do the same on the right-hand side using the components of the position vector π΄ which are, of course, just the coordinates of the point π΄, π₯ one, π¦ one, π§ one. And having done this, we can compute both dot products. On the left, we get π times π₯ plus π times π¦ plus π times π§. And on the right, we get π times π₯ one plus π times π¦ one plus π times π§ one. We can then subtract all the terms on the right-hand side from both sides, getting zero on the right-hand side. And we can then combine the two terms involving π, the two terms involving π, and the two terms involving π.

We find that π times π₯ minus π₯ one plus π times π¦ minus π¦ one plus π times π§ minus π§ one is zero. And this is the standard form of the equation of the plane. So letβs put it here. Finally, thereβs a general form of the equation, ππ₯ plus ππ¦ plus ππ§ plus π equals zero. Now, we know what π, π, and π are. They are the components of the normal vector π. But what is this π? This is new. The answer is that this is a constant you get when you expand out the standard form of the equation. We could write what π is explicitly in terms of π, π, π, π₯ one, π¦ one, and π§ one. But itβs easier to just compute π, once we have a concrete example.

So letβs turn to the problem in the question. The point the plane passes through is two, negative one, zero. And the normal vector is π’ equals four π plus 10π minus seven π. We can write this normal vector in component form as four ,10, negative seven. The π₯, π¦, and π§ components of this vector are just the coefficients of the terms involving the unit vectors π, π, and π, respectively.

Now that we have the components of the normal vector to the plane and the coordinates of a point on that plane, we can substitute them in to our vector form of the equation. We get four, 10, and negative seven dot π equals four, 10, negative seven dot two, negative one, zero. And this dot product on the right-hand side, we can compute. This is four times two plus 10 times negative one plus negative seven times zero, which is eight minus 10 plus zero, which is negative two. So the right-hand side is just negative two. And this is the vector form of our equation.

Now we turn to finding the standard form of the equation of our plane. Recall that π, π, and π are the components of our normal vector. And we found those components to be four, 10, and negative seven. And π₯ one, π¦ one, and π§ one were the coordinates at the point the plane passes through, which were two, negative one, zero. Substituting these values then, we get four times π₯ minus two plus 10 times π¦ minus negative one plus negative seven times π§ minus zero equals zero. But we can simplify. Subtracting negative one is the same as adding one. π§ minus zero is just π§. And adding negative seven times π§ is the same as subtracting seven π§. With the simplification, the standard form becomes four times π₯ minus two plus 10 times π¦ plus one minus seven π§ equals zero.

Finally, we turn to the general form of the equation of the plane which we get just by expanding the standard form. We copy down the standard form and expand each of the brackets. We get four π₯ minus eight plus 10π¦ plus 10 minus seven π§ equals zero. And hereβs where the constant π comes in. Itβs what we get by combining the constants. Negative eight plus 10 is two. And so, the general form of our equation is four π₯ plus 10π¦ minus seven π§ plus two equals zero. Notice that we also couldβve got this by taking our vector form and adding two to both sides.

Now, we found all three forms of the equation of our plane. To conclude, we just collect together the three forms that we found. If you know which numbers go where in one form of the equation of a plane, then itβs not too hard to find the other two forms of the equation of that plane. As a result, it makes sense to remember the vector form of a plane, which is not only the shortest form. Itβs also one that can be intuitively understood.