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Mathematics 22 Online
OpenStudy (callisto):

Vector question, please see the attachment

OpenStudy (callisto):

Why are AB and MN perpendicular to each other?

OpenStudy (radar):

Just reading that gave me a headache! Hopefully you will get some assistance, but it is above my pay grade.

OpenStudy (amistre64):

can you paste the text?

OpenStudy (callisto):

@amistre64 lol Thank you!

OpenStudy (amistre64):

for reference: i = <1,0,0> j = <0,1,0> k = <0,0,1> i+j+k = <1,1,1>

OpenStudy (callisto):

Wait :S Let OA = i , OB = j and OC = i + j+ k (see Figure 2). Let M and N be points on the straight lines AB and OC respectively such that AM : MB = (a :1 − a) and ON : NC = (b:1 − b) , where 0 < a <1 and 0 < b < 1 . Suppose that MN is perpendicular to both AB and OC .

OpenStudy (callisto):

Okay, nothing. Thanks again

OpenStudy (amistre64):

|dw:1335445432735:dw|

OpenStudy (callisto):

The second question is that why MN is the shortest distance between line OC and line AB?

OpenStudy (anonymous):

Ok. AB = OB - OA = j - i = - i + j MN = (a+b-1)i + (b-a)j+ b k Dot product of AB and MN vector = 0 (if Ab is perpendicular to MN) AB . MN = 0 (-i + j ) . ( (a+b-1)i + (b-a)j + b k)=0 = (-1)(a+b-1)( i . i) + (1)(b-a) (j.j) + 0=0 we know that i.i = j.j =1 -a-b+1+b-a=0 -2a+1=0 a=1/2

OpenStudy (anonymous):

my proof is based on assumption that you have proved part a of the question

OpenStudy (amistre64):

Let OA = i OB = j OC = i + j+ k (see Figure 2). AM : MB = (a :1 − a) ON : NC = (b:1 − b) |dw:1335445555446:dw| , where 0 < a <1 and 0 < b < 1 . Suppose that MN is perpendicular to both AB and OC .

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