Video Transcript
Given that the vectors six, 𝑘, one and 12, negative six, two are parallel, determine the value of 𝑘.
There are a couple of different ways we can go about this solution. Knowing that our two given vectors are parallel, that means for one thing that the ratio of their 𝑥-values equals the ratio of their 𝑦-values equals the ratio of their 𝑧-values. That means that these two equations are true, and we could use these to solve for 𝑘. Another way we could find 𝑘, though, is to recognize that since these vectors are parallel, that also means that their cross product is zero.
In general, if we have two three-dimensional vectors, we’ll call them 𝐮 and 𝐯, then the cross product of 𝐮 and 𝐯 is given by the determinant of this matrix. For our given vectors, that matrix would look like this. And as we mentioned, since the vectors are parallel, the determinant of this matrix must be zero. To show that these methods agree, let’s solve for 𝑘 both ways. In our first method, there are actually two different ways we could solve for 𝑘 using either one of these equations.
Say that we choose the one in pink that 𝑘 over negative six equals one over two. If we cross multiply, multiplying both sides by two and negative six, we find that two 𝑘 equals negative six or, in other words, that 𝑘 equals negative three. Moving on to our second method that the cross product of our two vectors must be zero and therefore the determinant of this matrix is zero, we can recognize that since the first row of this matrix contains the 𝐢, 𝐣, and 𝐤 hat unit vectors, the overall determinant will itself be a vector with 𝑥-, 𝑦-, and 𝑧-components.
Moreover, since the determinant overall equals zero — we know that’s true because our vectors are parallel — that means the components each individually must also be zero. If any of 𝑅 𝑥, 𝑅 𝑦, or 𝑅 𝑧 were not zero, then the magnitude of the vector overall wouldn’t be zero. To solve for 𝑘 then, using this method, we really only have to compute one of these three components. It just needs to be a component that contains 𝑘.
Thinking along these lines, we see that the 𝐢-component of our resultant vector will involve 𝑘. The magnitude of that 𝐢-component is given by the determinant of this two-by-two matrix. We can say that 𝑅 sub 𝑥 equals 𝑘 times two minus one times negative six or, in other words, two 𝑘 plus six. Recalling now that this component of our vector equals zero, we can note that two 𝑘 plus six equaling zero implies that two 𝑘 is equal to negative six or that 𝑘 equals negative three. We see then that both of these methods agree that the value of 𝑘 is negative three.
