# Question Video: Finding Unknown Coefficients in the Equation of Two Straight Lines in Three Dimensions given That They Are Parallel Mathematics

If the straight line (π₯ β 10)/6 = (π¦ + 6)/8 = (π§ + 2)/π is parallel to (π₯ β 1)/β12 = (π¦ + 3)/π = (π§ + 1)/14, find π + π.

04:30

### Video Transcript

If the straight line π₯ minus 10 all over six is equal to π¦ plus six all over eight is equal to π§ plus two all over π is parallel to π₯ minus one all over negative 12 is equal to π¦ plus three all over π is equal to π§ plus one all over 14, find π plus π.

In this question, weβre given the equation of two lines and weβre told that these two lines are parallel. We need to use these to determine the value of π plus π. To do this, we start by seeing that our lines are given in symmetric form. And we recall the symmetric form of a straight line is the form π₯ minus π₯ zero all over π is equal to π¦ minus π¦ zero all over π is equal to π§ minus π§ zero all divided by π. And this is a very useful form to write our straight line in because we can read off some information about our line.

First, we see if we substitute π₯ is equal to π₯ zero, π¦ is equal to π¦ zero, and π§ is equal to π§ zero into this equation, then we get zero is equal to zero is equal to zero. Therefore, any straight line given in this form will pass through the point π₯ zero, π¦ zero, π§ zero. However, there is another useful piece of information we can get from the symmetric form of this line. The direction vector of this straight line is going to be the vector π, π, π. In this question, we need to use the fact that the two lines are parallel to determine the value of π plus π. And to do this, weβre going to need to recall exactly what we mean when we say two straight lines are parallel.

We recall that we say that two straight lines are parallel if their direction vectors are nonzero scalar multiples of each other. So to determine whether these two straight lines are parallel or not, weβre going to need to determine whether their direction vectors are scalar multiples of each other. And because both of these lines are given in symmetric form, we can just read off their direction vectors.

For our first straight line, the direction vector is going to be the vector six, eight, π. And we can do exactly the same for our second straight line. Its direction vector is going to be the vector negative 12, π, 14. Finally, because we know these two straight lines are parallel to each other, we know that these two vectors have to be nonzero scalar multiples of each other. In other words, the vector six, eight, π must be equal to π multiplied by the vector negative 12, π, 14, where π is some nonzero scalar because we know that these lines are parallel.

So letβs solve these two vectors being equal to each other. We need to evaluate the scalar multiplication. Remember, we need to do this by multiplying every component in our vector by π. Multiplying through by π, we get the vector six, eight, π is equal to the vector negative 12π, ππ, 14π. And remember, for two vectors to be equal to each other, they must have the same dimension and their components must be equal. We can use this to find the values of π, π, and π.

Letβs start by finding the value of π by setting the first component of these two vectors to be equal. Doing this, we get that six should be equal to negative 12π. And of course, we can solve for π by dividing through by negative 12. We get that π is equal to negative six divided by 12, which we can simplify to give us that π is equal to negative one-half. Now that we know the value of π, we can substitute this into our vector.

Clearing some space and substituting in π is equal to negative one-half into our vector equation, we get that the vector six, eight, π is equal to the vector negative 12 times negative one-half, negative π over two, 14 times negative one-half. And of course, we can simplify this. In the first component on the vector on the right-hand side, negative 12 times negative one-half is equal to six. And in the third component of this vector, 14 multiplied by negative one-half is equal to negative seven. So weβve simplified the vector on the right-hand side to give us the vector six, negative π over two, negative seven.

Now, since these two vectors are equal, all of their components must be equal. We can now equate the second components of these two vectors. This gives us that eight is equal to negative π divided by two. And of course, we can solve for our value of π by multiplying this equation through by negative two. We get that π is equal to negative two times eight, which is, of course, equal to negative 16. Of course, we can also equate the third components of these two vectors. And doing this, we see that our value of π has to be equal to negative seven.

So weβve found the value of π and the value of π. And weβre asked in this question to find the value of π plus π. So we add these values together. π plus π is negative seven plus negative 16, which we can calculate is negative 23. Therefore, we were able to show if the straight line π₯ minus 10 all over six is equal to π¦ plus six all over eight is equal to π§ plus two all over π is parallel to the straight line π₯ minus one all over negative 12 is equal to π¦ plus three all over π is equal to π§ plus one all over 14, then π plus π must be equal to negative 23.