# Question Video: Determining the Components of a Vector Mathematics • 12th Grade

Determine, to the nearest hundredth, the component of vector π along ππ, given that π = β©β7, 2, 10βͺ and the coordinates of π and π are (1, β4, β8) and (3, 2, 0), respectively.

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### Video Transcript

Determine, to the nearest hundredth, the component of vector π along ππ given that π equals negative seven, two, 10 and the coordinates of π and π are one, negative four, negative eight and three, two, zero, respectively.

Okay, in this exercise, we have a three-dimensional vector π and points in three-dimensional space π΄ and π΅. Letβs say that point π΄ is located here and point π΅ is here. We want to solve for the component of this given vector π along a vector ππ. This vector ππ will go from point π΄ to point π΅ looking like this. And to calculate the component of vector π along ππ, weβll want to know the components of vector ππ.

To solve for those, we can subtract the coordinates of point π΄ from the coordinates of point π΅. In other words, we could write that vector ππ equals π minus π in vector form. Substituting in the coordinates of π΅ and π΄, we find subtracting those of π΄ from those of π΅ gives us a vector with components of three minus one or two, two minus negative four or six, and zero minus negative eight or eight. So then we now have our vector ππ. And as weβve seen, we want to solve for the component of vector π that lies along ππ.

We can begin to do this by recalling that the scalar projection of one vector onto another is equal to the dot product of those two vectors divided by the magnitude of the vector being projected onto. In our example, as we calculate the component of vector π along ππ, weβre computing the scalar projection of π onto ππ. Therefore, we can say that the quantity we want to solve for is given by π dot ππ over the magnitude of vector ππ.

Remembering that the magnitude of a vector is equal to the square root of the sum of the squares of the components of that vector, we see that what we want to calculate is this dot product over this square root. Carrying out this dot product, we start by multiplying the respective components of these two vectors together. And then working in our denominator, we know that two squared is four, six squared is 36, and eight squared is 64. So our fraction simplifies to negative 14 plus 12 plus 80 divided by the square root of four plus 36 plus 64. This equals 78 over the square root of 104.

And we could leave this as our answer, except that weβre told to determine this overlap to the nearest hundredth. If we enter this fraction on our calculator then, to the nearest hundredth, it equals 7.65. Thatβs the component of vector π along vector ππ.