Video: EG19M2-ALGANDGEO-Q18

EG19M2-ALGANDGEO-Q18

08:10

Video Transcript

Prove that the points 𝐴: one, three, five; 𝐡: four, four, zero; and 𝐢: negative one, two, four are not collinear. Then, find the different forms of the equation of the plane that passes through these points.

We have three points, and we want to show that they are not collinear. Three points are collinear if they all lie on the same line. And they are not collinear if they dont all lie on the same line, and instead form a triangle. To show that our three points are not collinear, we could find the equation of the line through the points 𝐴 and 𝐡, and so that the point 𝐢 doesnt satisfy this equation and hence must lie off the line. But later on in the question, we’re asked about the equation of the plane that passes through the three points. And this makes us think about vectors. So what well do instead is well show that the vectors 𝐴𝐡 and 𝐴𝐢 dont point along the same line. And as two vectors only point along the same line if one is a multiple of the other, well show that 𝐴𝐡 is not a multiple of 𝐴𝐢.

To do this, we find the components of the vectors 𝐴𝐡 and 𝐴𝐢. The vector 𝐴𝐡 is the difference of the position vector of 𝐡 and the position vector of 𝐴. And the components of the position vector of 𝐡 are just the coordinates of the point 𝐡 which we’re told are four, four, zero. Similarly, we see that the components of 𝐴 are one, three, five. And to subtract these vectors, we just subtract their components, getting the vector three, one, negative five. We find that the components of the vector 𝐴𝐢 are negative two, negative one, negative one in the same way.

Now that we have the components of 𝐴𝐡 and 𝐴𝐢, we can see that 𝐴𝐡 is not a multiple of 𝐴𝐢, as the quotients of corresponding components are not all equal. So we have shown, as required, that the points 𝐴, 𝐡, and 𝐢 are not collinear. We can move on to the second part of this question, which is to find the different forms of the equation of the plane that passes through these points.

To find the equation of the plane, we first need to find a normal to the plane. And this normal vector is perpendicular to the plane. And so, its perpendicular to the vectors 𝐴𝐡 and 𝐴𝐢 which both lie in the plane. And so, we can take it to be their cross product. We can write this cross product as a three by three determinant whose first row is composed of 𝑖, 𝑗, and π‘˜, the unit vectors in the π‘₯-, 𝑦-, and 𝑧-directions, respectively. The second row is formed of the components of 𝐴𝐡. And the third row is formed of the components of 𝐴𝐢. We expand the determinant along the first row, evaluate each two-by-two determinant, and get the vector negative six, 13, negative one. Lets clear some working.

This is a normal to our plane. If you got any multiple of this, then you are also right. And as this normal vector, which is the cross product of the vectors 𝐴𝐡 and 𝐴𝐢, is not the zero vector, we have another proof of the fact that 𝐴𝐡 is not a multiple of 𝐴𝐢. And hence that 𝐴, 𝐡, and 𝐢 are not collinear. If they were all collinear and 𝐴𝐡 was a multiple of 𝐴𝐢, then this cross product would be the zero vector. But we found this normal vector to find the different forms of the equation of the plane through points 𝐴, 𝐡, and 𝐢. The equation of the plane passing through the point π‘₯ one, 𝑦 one, 𝑧 one with normal vector 𝑛 equals 𝐴, 𝐡, 𝐢 is, in vector form, 𝑛 dot π‘Ÿ equals 𝑛 dot π‘₯ one, 𝑦 one, 𝑧 one. In standard form: π‘Ž times π‘₯ minus π‘₯ one plus 𝑏 times 𝑦 minus 𝑦 one plus 𝑐 times 𝑧 minus 𝑧 one equals zero. And in general form, π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero.

Lets start with the vector form of the equation. We have the normal vector 𝑛. Its the vector negative six, 13, negative one. And so, the left-hand side of the vector form is negative six, 13, negative one dot π‘Ÿ, where π‘Ÿ is the position vector of an arbitrary point on the plane. On the right-hand side, we have 𝑛 dot the vector π‘₯ one, 𝑦 one, 𝑧 one which is the position vector of a point on our plane. Any point on our plane will do. Which point on our plane should we choose? We have three of them, the points 𝐴, 𝐡, and 𝐢. Lets choose the point 𝐴 with coordinates one, three, five. And so, on the right-hand side of the vector form, we have the dot product of the normal vector 𝑛, negative six, 13, negative one with the position vector of 𝐴, one, three, five. And we can compute this dot product on the right-hand side. Its negative six times one, which is negative six, plus 13 times three, which is 39, plus negative one times five, which is negative five. And this sum is 28.

You can check that have we chosen point 𝐡 or point 𝐢 instead of point 𝐴, we still wouldve got a dot product of 28 on the right-hand side. The vector form of our equation is negative six, 13, negative one dot π‘Ÿ equals 28. We could multiply both sides of this equation by some number. And wed still have the vector form of the equation of this plane. And you might have found a multiple of this equation that weve just found if you used a different method. We find the standard form of the equation of our plane by substituting the components of our normal vector for 𝐴, 𝐡, and 𝐢 and the coordinates of the point in the plane that we choose for π‘₯ one, 𝑦 one, and 𝑧 one. The components of our normal are negative six, 13, negative one. And so, the standard form of our equation is negative six times π‘₯ minus π‘₯ one plus 13 times 𝑦 minus 𝑦 one minus one times 𝑧 minus 𝑧 one, where π‘₯ one, 𝑦 one, and 𝑧 one are the coordinates of a point on our plane.

Again, we have a choice for which point in our plane to pick. If we choose the point 𝐴 again, then π‘₯ one is one, 𝑦 one is three, and 𝑧 one is five. Changing minus one times minus five to just minus 𝑧 minus five, we get the standard form as negative six times π‘₯ minus one plus 13 times 𝑦 minus three minus 𝑧 minus five equals zero.

Finally, we need to get the general form of the equation of our plane. We Know what 𝐴, 𝐡, and 𝐢 are. They’re the components of our normal vector. But whats this 𝑑? Well, if we expand the brackets on the left-hand side of the standard form, 𝑑 will be our constant term. And the π‘₯, 𝑦, and 𝑧 terms will naturally fall out. Another way to find the general form is to start with the vector form, writing the vector π‘Ÿ in component form as π‘₯, 𝑦, 𝑧. We can then write the dot product on the left-hand side in terms of π‘₯, 𝑦, and 𝑧. Its negative six π‘₯ plus 13𝑦 minus 𝑧. And if we also subtract 28 from both sides, we get zero on the right-hand side. And comparing this to our formula for the general form, we see that 𝑑 is negative 28.

These are the three different forms of the equation of our plane. Lets clear away some working to allow us to write them together. These three forms of the equation along with the proof that the points 𝐴, 𝐡, and 𝐢 are not collinear form the answer to this question. The answers given here arent the only possible correct answers. There are other valid proofs for 𝐴, 𝐡, and 𝐢 not being collinear. And any multiple of the vector, standard, and general form given here would be valid. And we could replace the coordinates for 𝐴 that we used in our standard form of the equation with the coordinates of 𝐡 or 𝐢 or indeed any other point on the plane.

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