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|dw:1360298829035:dw| find this vector
your x axis coordinate is incorrect
it should be (4,0)

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Other answers:

yes , thanks
That is a beautiful triangle you have there.
then use mid point formula your gonna get, (2,1) as coordinates of bisector n then your vector is , V= 2i +j
modphys doesn't want the mid point. He want the vector that's perpendicular.
well thats perpendicular bisector that means dividing equally that means mid point
What I would do, is write the line in the form \(y=-x/2+2\). The slope of the intersecting line will be \[-\frac{1}{-\frac{1}{2}}=2.\] So you have a line that looks like \(y=2x\) as the line that intersects. Set them equal to each other to get \(2x=-x/2+2\implies 5x/2=2\implies x=1\).
Plug that in again, \(x=1\implies y=2(1)=2\), so the point of intersection is \((1,2)\).
Which in turn means the vector is \[\vec{i}+2\vec{j}\]
I got same answer but here what my book did
1 Attachment
1 Attachment
If \(x=.8\), then the vector is \[.8\vec{i}+1.4\vec{j}\] Taking the dot product of that with the vector \[4\vec{i}-2\vec{j},\] we get \[.8\cdot4-2\cdot1.4=0.4\neq0\]so that solution is wrong.
|dw:1360299909016:dw| just check the magnitude of vector A : (4*2)/sqrt(4^2+2^2) = 8/sqrt(20) = 8/2sqrt(5) = 4/5 * sqrt(5) 0.8 and 1.6 has the same magnitude
  • phi
Just as in your book, we can say the direction vector B (from pt (0,2) to (4,0) is (4,-2) let vector A be perpendicular to B. A dot B = 0 (x,y) dot (4,-2) =0 we get 4x = 2y pick a nice number for x, like 1. then y =2 vector A = (1,2) is perpendicular to vector B make A unit length, by dividing by its length. |A|= sqrt(5) Au = (1,2)/sqrt(5) now to get the length to scale Au so it reaches the line between (0,2) and (4,0), we notice that the vector (4,0) projected onto Au has the correct length Au dot (4,0) = 4/sqrt(5) The vector we want is 4/sqrt(5) * (1,2)/sqrt(5) = 4/5 * (1,2) or (0.8, 1.6)

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