11:25 If you decide to project OP onto PQ Suppose the line is parallel to the vector a (OA) or vector q (OQ) Suppose further this line has the equation ax + by + c = 0 ---- (*) Suppose P = (x_1, y_1) Suppose our 'origin' is (x_0, y_0) which lies on the line, so ax_0 + by_0 + c = 0 or equivalently = + lambda OP = (x_1 - x_0 , y_1 - y_0) the directional vector parallel to vectors a and q can be represented as (b, -a) the normal vector (perpendicular to this directional vector of the line we're interested in) can be written as (a, b) Confirm this by taking the dot product of (b,-a) and (a,b) This means the vector PQ is parallel to (a, b) The unit vector of (a,b) is 1/sqrt(a^2 + b^2) (unit normal vector) Projecting OP onto PQ will give you QP, or equivalently, projecting OP onto the unit normal vector 1/sqrt(a^2 + b^2) still gives you QP proj_(unit normal vector)^(OP) = (OP.(unit normal vector)) = QP (dot product of OP and unit normal vector) We want the distance, |PQ| or |QP| This means |proj_(unit normal vector)^(OP)| = |QP| = |(OP.(unit normal vector))| = |(OP.(unit normal vector))||| (we can leave our answer like this I guess = |(OP.(unit normal vector))|(1) = |(OP.(unit normal vector))| (absolute value of the dot product of OP and 1/sqrt(a^2 + b^2) ) ^we could leave the formula like that, as is it a legitimate formula, but expanding it further we get perpendicular distance from point to line = |(OP.(unit normal vector))| = |.1/sqrt(a^2 + b^2) | =|(1/sqrt(a^2+b^2))(.)| = |(1/sqrt(a^2+b^2))||(.)| = 1/sqrt(a^2 + b^2)|(.)| since 1/sqrt(a^2 + b^2) is non-negative where a,b are real. = |(.)|/sqrt(a^2+b^2) = |a(x_1 - x_0) + b(y_1 - y_0)|/sqrt(a^2+b^2) = |ax_1 - ax_0 + by_1 - by_0|/sqrt(a^2 + b^2) = |ax_1 - ax_0 + by_1 - by_0 + c - c|/sqrt(a^2 + b^2) = |ax_1 + by_1 + c - ax_0 - by_0 - c|/sqrt(a^2 + b^2) = |ax_1 + by_1 + c - (ax_0 + by_0 + c)|/sqrt(a^2 + b^2) = |ax_1 + by_1 + c - 0|/sqrt(a^2 + b^2) from (*) d = |ax_1 + by_1 + c|/sqrt(a^2 + b^2) Showing equivalence to |ax_1 + by_1 + c|/sqrt(a^2 + b^2) by projecting OP onto PQ seems easier than projecting OP onto a (but is still possible). This formula |(OP.(unit normal vector))| can be extended to 3 dimensions (perpendicular distance from a point to a plane) eqn of plane Ax + By + Cz + D = 0 normal vector to plane is unit normal vector is /sqrt(A^2 + B^2 + C^2) vector OP would be where (X_0, Y_0, Z_0) is a point that lies on the plane Ax + By + Cz + D = 0 and (X_1, Y_1, Z_1) is just some arbitrary point the equivalent formula leads to |AX_1 + BY_1 + CZ_1 + D|/sqrt(A^2 + B^2 + C^2)
11:25
If you decide to project OP onto PQ
Suppose the line is parallel to the vector a (OA) or vector q (OQ)
Suppose further this line has the equation ax + by + c = 0 ---- (*)
Suppose P = (x_1, y_1)
Suppose our 'origin' is (x_0, y_0) which lies on the line, so ax_0 + by_0 + c = 0 or equivalently = + lambda
OP = (x_1 - x_0 , y_1 - y_0)
the directional vector parallel to vectors a and q can be represented as (b, -a)
the normal vector (perpendicular to this directional vector of the line we're interested in) can be written as (a, b)
Confirm this by taking the dot product of (b,-a) and (a,b)
This means the vector PQ is parallel to (a, b)
The unit vector of (a,b) is 1/sqrt(a^2 + b^2) (unit normal vector)
Projecting OP onto PQ will give you QP, or equivalently, projecting OP onto the unit normal vector 1/sqrt(a^2 + b^2) still gives you QP
proj_(unit normal vector)^(OP) = (OP.(unit normal vector)) = QP (dot product of OP and unit normal vector)
We want the distance, |PQ| or |QP|
This means |proj_(unit normal vector)^(OP)| = |QP| = |(OP.(unit normal vector))|
= |(OP.(unit normal vector))||| (we can leave our answer like this I guess
= |(OP.(unit normal vector))|(1)
= |(OP.(unit normal vector))| (absolute value of the dot product of OP and 1/sqrt(a^2 + b^2) )
^we could leave the formula like that, as is it a legitimate formula, but expanding it further we get
perpendicular distance from point to line = |(OP.(unit normal vector))|
= |.1/sqrt(a^2 + b^2) |
=|(1/sqrt(a^2+b^2))(.)|
= |(1/sqrt(a^2+b^2))||(.)|
= 1/sqrt(a^2 + b^2)|(.)| since 1/sqrt(a^2 + b^2) is non-negative where a,b are real.
= |(.)|/sqrt(a^2+b^2)
= |a(x_1 - x_0) + b(y_1 - y_0)|/sqrt(a^2+b^2)
= |ax_1 - ax_0 + by_1 - by_0|/sqrt(a^2 + b^2)
= |ax_1 - ax_0 + by_1 - by_0 + c - c|/sqrt(a^2 + b^2)
= |ax_1 + by_1 + c - ax_0 - by_0 - c|/sqrt(a^2 + b^2)
= |ax_1 + by_1 + c - (ax_0 + by_0 + c)|/sqrt(a^2 + b^2)
= |ax_1 + by_1 + c - 0|/sqrt(a^2 + b^2) from (*)
d = |ax_1 + by_1 + c|/sqrt(a^2 + b^2)
Showing equivalence to |ax_1 + by_1 + c|/sqrt(a^2 + b^2) by projecting OP onto PQ seems easier than projecting OP onto a (but is still possible).
This formula |(OP.(unit normal vector))| can be extended to 3 dimensions (perpendicular distance from a point to a plane)
eqn of plane Ax + By + Cz + D = 0
normal vector to plane is
unit normal vector is /sqrt(A^2 + B^2 + C^2)
vector OP would be
where (X_0, Y_0, Z_0) is a point that lies on the plane Ax + By + Cz + D = 0
and (X_1, Y_1, Z_1) is just some arbitrary point
the equivalent formula leads to |AX_1 + BY_1 + CZ_1 + D|/sqrt(A^2 + B^2 + C^2)
if this got released a week before my exam instead of now 😔
It help me a lot. Thanks for your job and your videos! I like to learn some math in free time. Greetings from Uruguay
If we can extend vector ã along its direction, then why can't we move it and place it right upon point P to make the dist. 0
Ok, so where is a good place to start for someone who wants to understand all of this but have a pretty basic understanding of math?
I would recommend looking up introduction to vectors.
@@carultch Will do, thank you. :)
Hey love your vidz!
Such a good guy but doesn't reply to comments