Given that point A, B, C, E are (0,0,6), (-1,-7,11), (-8,-6,6) and (15,-20,6). BD is horizontal and parallel to AE. Find a vector equation of the line BD. Given that the length of BD is 15 metres, find the coordinates of D. How to do?

1 Answer
Steve M
Mar 18, 2018

The vector equation of the line #BD# is :

# bb(ul r) = ( (-1),(-7),(11) ) + lamda ( (3),(-4),(0) )#

#D# has coordinates #(8,-19,11)#, or #(-10,5,11) #

Explanation:

We have the following coordinates:

# A =(0,0,6) #
# B = (-1,-7,11) #
# C = (-8,-6,6) #
# D = (a,b,c) # (unknown)
# E = (15,-20,6) #

Using vector notation, we have:

# bb vec(BD) = bb vec(OD) - bb vec(OB) #
# \ \ \ \ \ \ = ( (a),(b),(c) ) - ( (-1),(-7),(11) ) #
# \ \ \ \ \ \ = ( (a+1),(b+7),(c-11) ) #

Similarly:

# bb vec(AE) = bb vec(OE) - bb vec(OA) #
# \ \ \ \ \ \ = ( (15),(-20),(6) ) - ( (0),(0),(6) ) #
# \ \ \ \ \ \ = ( (15),(-20),(0) ) #

We are given that # bb vec(BD) # is horizontal and therefore the #z# component is zero, giving us:

# c-11 = 0 => c = 11 #

We also know that #BD# is parallel to #AE#, and so #EE m in RR# where #m# is a constant, st

# bb vec(BD) = m \ bb vec(AE) #

From this we can establish that:

# ( (a+1),(b+7),(c-11) ) = m ( (15),(-20),(0) ) => { (a+1=15m), (b+7=-20m) :}#

# :. (a+1)/15 = (b+7)/(-20) #

# :. (a+1)/3 = (b+7)/(-4) #

# :. -4a-4 = 3b+21 #

# :. 3b = -4a-25 #

# :. b = -(4a+25)/3 #

Hence, we have:

# bb vec(BD) = ( (a+1),(7-(4a+25)/3),(0) ) = ( (a+1),((-4a-4)/3),(0) ) #
# \ \ \ \ \ \ \ = ( (a+1),(-4/3(a+1)),(0) )#
# \ \ \ \ \ \ \ = (a+1)/3( (3),(-4),(0) )#

And so using #B# as the reference point (we could equally use #D#), we can form the vector equation of the line #BD#:

# bb(ul r) = bb vec(OB) + lamda bb vec(BD)#
# \ \ \ = ( (-1),(-7),(11) ) + lamda ( (3),(-4),(0) )#

Finally, we know that #BD=15 => ||bb vec(BD)||=15 #, and so we have:

# || (a+1)/3( (3),(-4),(0) ) || = 15 #

# :. |(a+1)/3| || ( (3),(-4),(0) ) || = 15 #

# :. |(a+1)|sqrt(9+16+0)= 45 #

# :. |a+1|sqrt(25)= 45 #

# :. 5|a+1|= 45 #

# :. |a+1|= 9 #

# :. a+1= +-9 => a = -10, 8 #

With #a=8 => b = -(32+25)/3 = -19 #

Giving us the coordinate of #D#:

# D = (a,b,c) = (8,-19,11) #

And , with #a=-10 => b = -(-40+25)/3 = 5 #

Giving us the coordinate of #D#:

# D = (a,b,c) = (-10,5,11) #

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