SOLUTION: ABCD is a sqaure. The midpoints of BC and CD are M and N respectively. a) Express vector AM and vector An as linear combinations of vector AB and vector AD. b) Express vector AB

Algebra ->  College  -> Linear Algebra -> SOLUTION: ABCD is a sqaure. The midpoints of BC and CD are M and N respectively. a) Express vector AM and vector An as linear combinations of vector AB and vector AD. b) Express vector AB       Log On


   

Question 18656: ABCD is a sqaure. The midpoints of BC and CD are M and N respectively.
a) Express vector AM and vector An as linear combinations of vector AB and vector AD.
b) Express vector AB and vector AD as linear combinations of vector AM and vector AN.
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
ABCD is a sqaure. The midpoints of BC and CD are M and N respectively.
a) Express vector AM and vector An as linear combinations of vector AB and vector AD.
b) Express vector AB and vector AD as linear combinations of vector AM and vector AN.
ABCD being a square we have vector AB= vector DC
and vector AD=vector BC
but M is the mid point of BC .so vector BM=vector BC/2=vector AD/2
anfd N is the midpoint of CD.so vector NC=vector DC/2=vector AB/2
taking triangle ABM..we have vector AB+vector BM=vector AM
or...vector AM=vector AB+vector AD/2...............(1)
similarly we have from the polygon ABCN..
vector AB+vector BC+vector CN=vector AN
vector AN=vector AB+vector AD-vector NC=vector AB+vector AD-vector AB/2..or
vector AN=vector AB/2+vector AD..............(2)
from 2*(2)-(1) we get ..2*vector AN-vector AM=(2-1/2)(vector AD)=(3/2)vector AD
hence ...vector AD=(2/3)(2*vector AN-vector AM)..similarly by taking 2*(2)-(1),we can find AB in terms of AM and AN
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How do the graphs of y=X+3 (in
square root) compare the graph of y=x (insquare root) where x is greater
or equal to 0. And can you also graph y=f(x)+5
y=SQRT(x+3)...for every value of x we get 2 values for y ..one +ve and anpther -ve.further x cannot be less than -3 as it will lead to -ve number under square root which is imaginary.so the tabulation reads as follows.now you can plot the graph,i suppose.
x.....0.......1......2......3......-1......-2......-3
y.....3^0.5...2......5^0.5..6^0.5...2^0.5..1.......0
y....-3^0.5..-2.....-5^0.5.-6^0.5..-2^0.5.-1.......0
+graph%28+300%2C+300%2C+-3%2C+5%2C+-5%2C+5%2C+%2B%28x%2B3%29%5E0.5%2C-%28x%2B3%29%5E0.5%29+
y=sqrtx
x.....0.......1......2......3.......4
y.....0.......1......2^0.5..3^0.5...2
y.....0......-1.....-2^0.5.-3^0.5..-2
+graph%28+300%2C+300%2C+-3%2C+5%2C+-5%2C+5%2C+%2B%28x%29%5E0.5%2C-%28x%29%5E0.5%29+
graph y=f(x)+5...WHAT IS f(x)???????cant do wiyhout that