- #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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