- #1
ibysaiyan
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Homework Statement
The set of questions is:
The points A, B, C have position vectors
a = 5i+4j+2k
b = 4i−5j+3k
c = 2i−j−2k
(d) Find the vector equation for the line
passing through A and C
(e) Find the nearest distance from the
point B to this line.
Homework Equations
r=A+[itex]\lambda B[/itex]
The Attempt at a Solution
My working is as following:First I need to find out vector AC [direction vector] which is : -3i -5j -4k, so when I substitute in -3i -5j -4k into vector B and any of the given points, this should give me the equation of the line. Which's : r = 5i+4j+2k [itex]\lambda B[/itex] (-3i -5j-4k)
Part e )
Now I think there are two ways of finding out the shortest distance between a given point and a line.
The question's :
(e) Find the nearest distance from the
point B to this line.
My working:
Let H be the point which's perpendicular to the line and is at a shorter distance.
Since H is on the line , then using the equation of the line I get OH = ( 5-3 [itex]\lambda [/itex] , -5[itex]\lambda [/itex] +4, -4[itex]\lambda [/itex] +2k)
Therefore, Vector BH = H + (-B) = ( 1-3[itex]\lambda [/itex] , -5[itex]\lambda [/itex] +9 ,-4[itex]\lambda [/itex] -1)
The condition which arises is that BH is perpendicular to direction vector of the equation i.e : -3i -5j-4k
In other words dot product = 0 ,
which gives me the value of lamba = 22/25
In the end the distance of the length OH which i have found out is :about 2.7 to s.f)
Is my working right ? also I wanted to ask that.. on my summary booklet, there's a much easier way to find this length : The formula is :
d = | (p-a) x B^ | (equation 1)
Ironically, I know how to derive the above equation (1)
BUT i don't understand it :(
Any help , feedback will be appreciated! :)
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