# Video: MATH-ALG+GEO-2018-S1-Q10

If the vector 𝐴 = (−2, 4, 6) and the vector 𝐵 = (0, 𝑘, 3) such that 𝑘 ∈ ℤ⁺ and the magnitude of the vector 𝐴𝐵 = 7. What is the value of 𝑘?

04:49

### Video Transcript

If the vector 𝐴 has components negative two, four, six and the vector 𝐵 has components zero, 𝑘, three such that 𝑘 is a positive integer and the magnitude of the vector 𝐴𝐵 is seven. What is the value of 𝑘?

The vector 𝐵 is given in terms of 𝑘 whose value we have to find. And so, the vector 𝐴𝐵 will be in terms of 𝑘 also. And if you find the magnitude of the vector 𝐴𝐵, that will be in terms of 𝑘 as well. And so, we’ll get an equation involving 𝑘 which hopefully we can solve for 𝑘. But our first step is to find the vector 𝐴𝐵. If we think of the vectors 𝐴 and 𝐵 as being the position vectors of points 𝐴 and 𝐵, then the vector 𝐴𝐵 that we’re looking for goes from the point 𝐴 to the point 𝐵.

Another way to get from 𝐴 to 𝐵 is to first go along the vector negative 𝐴 from 𝐴 to the origin 𝑂 and from there along the vector 𝐵 to the point 𝐵. So we see that the vector 𝐴𝐵 is equal to the vector negative 𝐴 plus the vector 𝐵. Or rearranging, it’s equal to the vector 𝐵 minus the vector 𝐴.

We know the components of the vector 𝐵. They are zero, 𝑘, three. And we know the components of 𝐴. They are negative two, four, six. We can find the components of the vector 𝐴𝐵 then by subtracting these components. The difference of the first components zero minus negative two is zero plus two which is two. The difference of the second components is 𝑘 minus four which is 𝑘 minus four. And difference of the third components is three minus six which is negative three. We found the components of the vector 𝐴𝐵 in terms of the unknown value 𝑘.

Our next step along the way to solving the equation the magnitude of 𝐴𝐵 equal seven is to find the magnitude of 𝐴𝐵 in terms of 𝑘. How do we find this magnitude? Well, the magnitude of a vector is the square root of some of the squares of the components of that vector. The components of the vector 𝐴𝐵 are two 𝑘 minus four and negative three. And so, its magnitude is the square root of two squared plus 𝑘 minus four squared plus negative three squared.

We can simplify inside the square root. Two squared is four. 𝑘 minus four squared is 𝑘 squared minus eight 𝑘 plus 16. That’s using the fact that 𝑥 minus 𝑎 all squared is 𝑥 squared minus two 𝑎𝑥 plus 𝑎 squared. And finally, negative three squared is nine. Now, we can collect the constant terms four plus 16 plus nine is 29. And so, we see that the magnitude of the vector 𝐴𝐵 is the square root of 𝑘 squared minus eight 𝑘 plus 29.

We are now ready to solve the equation the magnitude of 𝐴𝐵 equals seven. We substitute the expression we found for the magnitude of 𝐴𝐵. So we have the square root of 𝑘 squared minus eight 𝑘 plus 29 equals seven. We then square both sides. On the left, we just get rid of the square root sign. And on the right, we get seven squared which is 49. And if we then subtract 49 from both sides, we get 𝑘 squared minus eight 𝑘 minus 20 equals zero, which hopefully we know how to solve. We can do this by inspection. We’re looking for two numbers whose sum is negative eight and whose product is negative 20.

The values two and negative 10 work. Two plus negative 10 is negative eight and two times negative 10 is negative 20. And we see that there are two solutions then: 𝑘 equals negative two or 𝑘 equals 10. But we’re told in the question that 𝑘 is a positive integer and negative two is not a positive integer. 𝑘 equals negative two is not our answer then. Our answer is 𝑘 equals 10 as 10 is a positive integer.

To recap, we found this value of 𝑘 by solving the equation the magnitude of 𝐴𝐵 equals seven. This required first finding the components of 𝐴𝐵 in terms of 𝑘 and then its magnitude in terms of 𝑘. Substituting this expression for the magnitude of 𝐴𝐵 and squaring both sides gave a quadratic which we could solve. Our answer was the positive integer solution to this quadratic, 10.