Video Transcript
Given that π΄π΅πΆπ· is a square
having a side length of 53 centimeters, calculate the algebraic projection of ππ
in the direction of ππ.
Okay, to start out here, letβs say
that this is our square with corners marked out π΄, π΅, πΆ, and π·. Weβre told that the length of each
side of the square is 53 centimeters. And we want to calculate the
algebraic projection of this vector ππ in the direction of another vector
ππ.
Letβs begin by sketching in these
two vectors. ππ is a vector that goes from
point πΆ to point π΄. And then vector ππ starts at
point π΅ and ends at point πΆ. We want then to figure out the
projection of this vector onto this one. To help us do that, letβs recall
that in general if we project a vector π onto another vector π, this is called the
scalar or algebraic projection of π onto π, then thatβs equal to the dot product
of these vectors divided by the magnitude of the vector being projected onto.
Our next step, then, will be to
solve for the components of these two vectors defined on our square. To do this, letβs let point π΄,
this corner of our square, represent the origin of an π₯π¦-plane. The positive π₯-axis then moves
horizontally to the right from this point, and the positive π¦-axis points
vertically upward from it. With this framework, we can now
define the coordinates of our three points of interest π΄, π΅, and πΆ. Point π΄ is at the origin, so its
coordinates are zero, zero. Point π΅ lies at distance of one
side length of our square out along the π₯-axis. We know thatβs 53 centimeters. And itβs π¦-coordinate is zero. And lastly point πΆ has π₯- and
π¦-coordinates of 53.
Now vector ππ is equal to the
vector form of the coordinates of point π΄ minus those of point πΆ. Zero, zero minus 53, 53 gives us a
final result of negative 53, negative 53. These are the π₯- and π¦-components
of the vector ππ. Similarly for the vector ππ, this
is equal to the vector form of the difference between the coordinates of point πΆ
and point π΅. Point πΆ has coordinates 53,
53. And point π΅ has coordinates 53,
zero. So we get a vector with components
zero, 53. These are the π₯- and π¦-components
of ππ.
Weβre now ready to go about
calculating this projection of vector ππ in the direction of ππ. Our equation shows us this equals
the dot product of ππ and ππ divided by the magnitude of ππ. Clearing some space for this
calculation, in our numerator, weβll calculate this dot product. And in our denominator, we remember
that the magnitude of a vector is equal to the square root of the sum of the squares
of its components. Up top, multiplying our vectors out
component by component, we get negative 53 quantity squared. And downstairs, we have the square
root of 53 squared.
But then in our denominator, this
square and the square root cancel one another out. And then the remaining factor of 53
in denominator cancels with one factor in numerator so that after all the
cancelation all that remains is negative 53. And this is our answer. This is the algebraic projection of
ππ in the direction of ππ.