### Video Transcript

If vector π is five, seven, two and vector π is six, two, one, determine the magnitude of π plus π rounded to one decimal place.

When we read this notation with vertical bars, we called it the magnitude of the sum of the vectors. So letβs remind ourselves what we mean when we talk about the magnitude of a three-dimensional vector. Letβs call that vector π₯π’ plus π¦π£ plus π§π€. The magnitude of this vector is the square root of π₯ squared plus π¦ squared plus π§ squared. So it follows that to find the magnitude of π plus π, weβll first need to find the vector described by π plus π. The sum of these two vectors is five, seven, two plus six, two, one. And to add vectors, we just add their components. Five plus six is 11, seven plus two is nine, and two plus one is three.

So letβs rewrite the general vector π in the form we have; itβs π₯, π¦, π§. Then, the magnitude of π plus π is the square root of the sum of the squares of the components. Thatβs root 11 squared plus nine squared plus three squared. So we need to evaluate the square root of 121 plus 81 plus nine. And thatβs root 211. Thatβs 14.5258 and so on. We need to evaluate this correct to one decimal place. Since two is less than five, we round down to 14.5. So the magnitude of π plus π rounded to one decimal place is 14.5.