### Video Transcript

Consider the points π΄ five, negative one, negative eight and π΅ negative three, negative nine, negative six. Find the component of the vector π equals negative five π’ minus two π£ plus two π€ in the direction ππ rounded to the nearest hundredth.

Okay, in this exercise, we have this vector π and points π΄ and π΅ that all exist in three-dimensional space. If we say that this is point π΄ and this point π΅, then the vector ππ goes from point π΄ to point π΅ like this. We want to figure out how much this given vector π overlaps with the vector ππ. To start solving for this, letβs figure out the components of this vector ππ. This is a vector given by the coordinates of π΅ minus the coordinates of π΄. We substitute those given coordinates in. And when we perform this subtraction, we have negative three minus five, thatβs negative eight, negative nine minus negative one, that also is negative eight, and then negative six minus negative eight. That equals positive two. The components of vector ππ are negative eight, negative eight, two.

Knowing this, we can now write vector π in the same notation using brackets. Itβs negative five, negative two, positive two. Letβs now remember that if we want to solve in general for the component of a vector π in the direction of another vector π, that equals the dot product of the vectors divided by the magnitude of the second vector. When we calculate this, weβre computing whatβs called the scalar projection of π onto π. Applying this relationship to our scenario, what we want to calculate is π dot ππ over the magnitude of ππ. This will give us the component of vector π in the direction of ππ. Plugging in for the known values of these vectors gives us this.

Note that weβre using the fact that the magnitude of a vector equals the square root of the sum of the squares of its components. We can carry out this calculation first in our numerator by multiplying together corresponding components of these vectors and second in our denominator, squaring these components of the vector ππ. Carrying this out gives us in our numerator 40 plus 16 plus four and in our denominator the square root of 64 plus 64 plus four. This equals 60 divided by the square root of 132. This is our exact answer, but we want to report our final result rounded to the nearest hundredth. Calculating this fraction then, to the nearest hundredth, itβs equal to 5.22. This is the component of the vector π in the direction ππ.