### Video Transcript

Find the different forms of the equation of the straight line that passes through the point with coordinates three, two, negative one and makes equal angles with the positive directions of the coordinate axes.

Weβre looking for the equation of the straight line. In fact, weβre looking for all different forms of this equation. And the easiest of the forms to start with is the vector form of the straight line. The vector form of a general straight line through a point π΄ with the direction vector π is π equals π΄ plus a parameter π‘ times the direction vector π.

Now weβre told in the question that our line passes through the point three, two, negative one. And so we can take this to be the point π΄ whose position vector appears in the equation. Substituting these components for π΄, we see that our vector equation has the form π equals three, two, negative one plus π‘ times the direction vector π. But we still need to find the direction vector π. How can we do this?

Well, thereβs one bit of information in the question that we havenβt used. Itβs that the line makes equal angles with the positive directions of the coordinate axes. If we call these angles ππ₯, ππ¦, and ππ§, then we can write that ππ₯ is equal to ππ¦ is equal to ππ§. And this is useful to know because we can get a direction vector π of our straight line by using the direction cosines, which is just the cosines of these angles.

Using the direction cosines, our direction vector is the vector with components cos ππ₯, cos ππ¦, cos ππ§. And a direction vector made in this way using the direction cosines of the line is a unit vector. This is equivalent to the fact that cos squared ππ₯ plus cos squared ππ¦ plus cos squared ππ§ is equal to one. This is true for any line. The angles that a lines makes with the positive directions of the π₯-, π¦-, and π§-axes β thatβs ππ₯, ππ¦, and ππ§ β will always satisfy this equation.

We know something special about our line, which is that these angles β ππ₯, ππ¦, and ππ§ β are all equal. And as ππ¦ and ππ§ are equal to ππ₯, we can replace them in the equation by ππ₯. So now we have cos squared ππ₯ plus cos squared ππ₯ plus cos squared ππ₯ equals one. So three cos squared ππ₯ equals one. And dividing by three, cos squared ππ₯ equals a third.

Now we can take square root on both sides. So cos squared ππ₯ just becomes cos of ππ₯. And on the right, a third becomes the square root of a third. Well, perhaps we should have a plus or minus sign here. But weβll see that we donβt need to.

Now that we have the value of cos ππ₯, we can substitute this value in the components of our direction vector π. And of course as ππ¦ and ππ§ are equal to ππ₯, cos of ππ¦ and cos of ππ§ are equal to cos of ππ₯. And so we can substitute into all the components of our direction vector. So our direction vector has components square root a third, square root a third, square root a third. And our line has vector equation π equals the vector with components three, two, negative one plus π‘ times the vector with components square root a third, square root a third, square root a third.

Going back to our question about whether we should have a plus or minus sign here at cos of ππ₯, we see that, had we chosen cos of ππ₯ to be negative the square root of a third, then our direction vector would just point in the opposite direction. This plus or minus sign it tells us that thereβs a choice we have to make as to which direction the direction vector π points along the line, does it point up the line or down the line.

Anyway, having found the components of the direction vector and hence the vector form of the equation of the straight line, we can clear some room and find the other forms of the straight line. One quick change we can make though is to write the square root of a third as one over the square root of three. One over the square root of three is just a more conventional way of writing this number. We get the parametric form of the equation of our straight line by taking the vector π, which is the position vector of an arbitrary point on our line, and writing it in components.

So calling these components π₯, π¦, and π§, we just replace the left-hand side with a vector with components π₯, π¦, and π§. Now on the right-hand side, we have a scalar multiple of the vector one over root three, one over root three, one over root three. And we can bring that scalar multiple of the parameter π‘ inside the vector, getting components one over root three π‘, one over root three π‘, one over root three π‘.

Weβve just multiplied each of the components by π‘. And now the sum of two vectors in component form is just the vector whose components are the sum of their components. Adding the π₯-components, we get three plus one over root three π‘. Adding the π¦-components, we get two plus one over root three π‘. And adding the π§-components, we get negative one plus one over root three π‘.

Now as this vector is equal to the vector on the left-hand side, its components must be equal to the components of the vector on the left-hand side. Comparing the π₯-components, we see that π₯ must be equal to three plus one over root three times π‘. Comparing the π¦-components, we see that π¦ must be equal to two plus one over root three π‘. And comparing the π§-components, we see that π§ must be equal to negative one plus one over root three π‘. These three equations form the parametric form of the equation of our straight line.

Letβs clear our working and find the third and final form of the equation of our straight line. The final form is the Cartesian form, which we get by eliminating π‘ from the equations in the parametric form. We can rearrange the equation π₯ equals three plus one over root three times π‘ so that its subject is π‘. We do this by subtracting three from both sides, dividing both sides by one over root three, and then swapping the sides. This gives π‘ equals π₯ minus three over one over root three.

We can do the same for the second equation π¦ equals two plus one over root three times π‘, where we find that π‘ is equal to π¦ minus two over one over root three. And doing this for the third equation, we get thatβs π‘ equals π§ plus one over one over root three. As these three expressions on the right-hand side are equal to π‘, theyβre also equal to each other. π₯ minus three over one over root three equals π¦ minus two over one over root three equals π§ plus one over one over root three. We have no need now to mention the parameter π‘. And so we get rid of this, leaving the Cartesian form of our equation.

If youβre wondering why we didnβt simply multiply by root three rather than dividing through by one over root three, the reason is to match the formula for the Cartesian form of the equation of the line through the point with coordinates π, π, π with direction vector having components π, π, π. Here the components of our direction vector are the denominators in our equation. You can check that applying this formula to our question does indeed give the Cartesian form we found.

But as we can derive the Cartesian form from the vector form and the parametric form, we can get away without memorizing this formula. So here we have all three different forms of the equation of our straight line which passes through the point three, two, negative one and makes equal angles with the positive directions of the coordinate axes.

One thing to note here is that we chose our direction vector to be the vector one over root three, one over root three, one over root three, which is the unit vector we get by considering the direction cosines. However, we could choose our direction vector to be any multiple of this vector. It doesnβt have to be a unit vector. So we could choose to multiply each component by root three to get the direction vector one, one, one. This is also a direction vector of our line. And you might argue that this direction vector is simpler as its components are all integers which have no common factor.

Using this direction vector, the vector form of our equation becomes π equals three, two, one plus π‘ times one, one, one. The parametric form becomes π₯ equals three plus π‘, π¦ equals two plus π‘, π§ equals negative one plus π‘. And the Cartesian form becomes π₯ minus three equals π¦ minus two equals π§ plus one. This is also, therefore, a correct answer.