Video Transcript
Given that 𝐴𝐵𝐶𝐷 is a square
having a side length of 53 centimeters, calculate the algebraic projection of 𝐂𝐀
in the direction of 𝐁𝐂.
Okay, to start out here, let’s say
that this is our square with corners marked out 𝐴, 𝐵, 𝐶, and 𝐷. We’re told that the length of each
side of the square is 53 centimeters. And we want to calculate the
algebraic projection of this vector 𝐂𝐀 in the direction of another vector
𝐁𝐂.
Let’s begin by sketching in these
two vectors. 𝐂𝐀 is a vector that goes from
point 𝐶 to point 𝐴. And then vector 𝐁𝐂 starts at
point 𝐵 and ends at point 𝐶. We want then to figure out the
projection of this vector onto this one. To help us do that, let’s recall
that in general if we project a vector 𝐀 onto another vector 𝐁, this is called the
scalar or algebraic projection of 𝐀 onto 𝐁, then that’s 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, let’s let point 𝐴,
this corner of our square, represent the origin of an 𝑥𝑦-plane. The positive 𝑥-axis then moves
horizontally to the right from this point, and the positive 𝑦-axis points
vertically upward from it. With this framework, we can now
define the coordinates of our three points of interest 𝐴, 𝐵, and 𝐶. Point 𝐴 is at the origin, so its
coordinates are zero, zero. Point 𝐵 lies at distance of one
side length of our square out along the 𝑥-axis. We know that’s 53 centimeters. And it’s 𝑦-coordinate is zero. And lastly point 𝐶 has 𝑥- and
𝑦-coordinates of 53.
Now vector 𝐂𝐀 is equal to the
vector form of the coordinates of point 𝐴 minus those of point 𝐶. Zero, zero minus 53, 53 gives us a
final result of negative 53, negative 53. These are the 𝑥- and 𝑦-components
of the vector 𝐂𝐀. Similarly for the vector 𝐁𝐂, this
is equal to the vector form of the difference between the coordinates of point 𝐶
and point 𝐵. Point 𝐶 has coordinates 53,
53. And point 𝐵 has coordinates 53,
zero. So we get a vector with components
zero, 53. These are the 𝑥- and 𝑦-components
of 𝐁𝐂.
We’re now ready to go about
calculating this projection of vector 𝐂𝐀 in the direction of 𝐁𝐂. Our equation shows us this equals
the dot product of 𝐂𝐀 and 𝐁𝐂 divided by the magnitude of 𝐁𝐂. Clearing some space for this
calculation, in our numerator, we’ll 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
𝐂𝐀 in the direction of 𝐁𝐂.
