### Video Transcript

Suppose that the lines for vector 𝑟 equals five, negative three, four plus 𝑡 times negative three, negative one, 𝑔 and 𝑥 minus five over ℎ equals 𝑦 minus four over negative four equals 𝑧 minus two over four are parallel, what are 𝑔 and ℎ?

Our first line is given to us in vector form and our second line is given in symmetric equation form. We want to take this vector form equation and rewrite it in symmetric equation form. Vector form is given as 𝑥 nought, 𝑦 nought, 𝑧 nought plus 𝑡 times 𝑎, 𝑏, 𝑐. And under those conditions, the symmetric equation form is then 𝑥 minus 𝑥 nought over 𝑎 equals 𝑦 minus 𝑦 nought over 𝑏 equals 𝑧 minus 𝑧 nought over 𝑐.

We have the vector form of the vector 𝑟. And so, now we need to plug in what we know. Our first term would be 𝑥 minus five over negative three and the next term 𝑦 minus negative three over negative one. We can rewrite the numerator as 𝑦 plus three. Then, finally, we have our 𝑧 term which will be 𝑧 minus four over our variable 𝑔. In order for these lines to be parallel, there denominators must be proportional to one another.

If we consider the denominators of our 𝑥 term, on the left, we have negative three. And on the right, we have the variable ℎ. The denominators of our 𝑥 terms, on the right is four and on the left is the variable 𝑔. We can determine a proportion from these values with variables. So, we consider the 𝑦 term. On the left, we have a denominator of negative one. And on the right, we have a denominator of negative four.

How do we go from negative one to negative four? We multiply by four. And that means, to find a value of our ℎ variable, we would multiply negative three times four. Negative three times four equals negative 12. ℎ equals negative 12. We also know that the 𝑔 variable times four equals four. If 𝑔 times four equals four, then 𝑔 must be equal to one. One times four equals four. Now that we know ℎ and 𝑔, we can take this information and write our symmetric equation in vector form.

Our parallel vector 𝑡 would be five, four, two, remember, we take those from the numerator, whatever is being subtracted from the variable, plus 𝑡 times 𝑎, 𝑏, and 𝑐 where 𝑎, 𝑏, and 𝑐 are the denominators. And we have negative 12, negative four, and four. We can take out a factor of four. And we’ll have negative three, negative one, one, which are the same 𝑎, 𝑏, and 𝑐 terms from vector 𝑟. And it proves that these two vectors are indeed parallel under the conditions ℎ equals negative 12 and 𝑔 equals one.