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The triangle ABC is defined by
OA: 2
-3
AB: 3
4
AB*BC= 0 (the dot product)
AC is parallel to 0
1
What is BC and the vector OC?
Thanks a lot!
Oh, and one more question. The vectors i and j are unit vectors along the x- and y-axes respectively. THe vectors u = -i + 2j and v=3i+5j are given.
A vectors w has the same direction as u=-1+2j and has a magnitude of 26.
Find w in terms of i and j
THankss!
The triangle ABC is defined by
OA: 2
-3AB: 3
4AB*BC= 0 (the dot product)
AC is parallel to 0
1What is BC and the vector OC?
Thanks a lot!
It would help a lot if you would write out the entire problem. I assume that O is the origin of some coordinate system and the vector OA is <2, -3> and AB is <3, 4> . That means that OB= <2, -3>+ <3, 4>= <5, 1> . Specifically, in this coordianate system, A is the point (2, -3) and B is the point (5, 1). BC is perpendicular to AB (because their dot product is 0) and C is directly above A (because AC is parallel to <0, 1>). C must be the point (2, y) for some y. There are now a number of ways to determine the point C. For example, since ABC is a right triangle with right angle at B, we could use the Pythagorean theorem: the length of the hypotenuse, AC is y+ 3 and its square must equal the square of the distances from A to B and from B to C- that is math](y+3)^2= 25+ 9+ (y-1)^2[/math]. That should be easy to solve.
Even simpler is to use symmetry. Since AB is <3, 4> and BC is perpendicular to it (and C is above A), BC must be <-4, 3>. Then C is the point (5-4,1+3)= (1, 4) and OC is <1, 4>.
Oh, and one more question. The vectors i and j are unit vectors along the x- and y-axes respectively. THe vectors u = -i + 2j and v=3i+5j are given.
A vectors w has the same direction as u=-1+2j and has a magnitude of 26.
Find w in terms of i and jTHankss!
This is easier than the first problem. u has length
. Divide each component by to get a unit vector in that direction and then multiply by 26 to get a vector with magnitude 26 in that direction.