Relative to an origin O, the position vectors of points A, B and C are given by #vec(OA)= 0uli + 2ulj + (-3)ulk # #vec(OB)= 2uli + 5ulj + (-2)ulk # #vec(OC)= 3uli + pulj + qulk # (i) In the case where ABC is a straight line, find the values of *p* and *q?

Relative to an origin O, the position vectors of points A, B and C are given by
#vec(OA)= 0uli + 2ulj + (-3)ulk #
#vec(OB)= 2uli + 5ulj + (-2)ulk #
#vec(OC)= 3uli + pulj + qulk #

(i) In the case where ABC is a straight line, find the values of p and q.
(ii) In the case where angle BAC is #90^@#, express q in terms of p .
(iii) In the case where #p=3# and the lengths of AB and AC are equal, find possible values of q ?

2 Answers
Douglas K.
Mar 13, 2018

Part (i)

The vector form of the line that includes points A,B,C is:

#(x,y,z) = (0,2,-3)+ t{(2-0)hati+(5-2)hatj+(-2--3)hatk}#

Simplify the vector:

Line #L = (x,y,z) = (0,2,-3)+ t{2hati+3hatj+hatk}#

The parametric equations are:

#x = 2t#
#y = 3t+2#
#z = t-3#

#vec(OC)= 3hati + p hat j + q hatk# tells us that we must set #x = 3#, solve for t, and then use the value of t to compute the values of p and q.

Set #x = 3#:

#3 = 2t#

Solve for t:

#t = 3/2#

Use the value of t to find the value of p and q:

#p = y = 3(3/2)+2#

#p = 13/2#

#q = z = 3/2-3#

#q = -3/2#

Part (ii)

#vec(AB)= (2-0)hati+(5-2)hatj+(-2--3)hatk#

Simplify:

#vec(AB)=2hati+3hatj+hatk#

#vec(AC) = (3-0)hati+(p-2)hatj+(q+3)hatk#

Simplify:

#vec(AC) = 3hati+(p-2)hatj+(q+3)hatk#

For #angle BAC = 90^@#, #AB * AC = 0#:

#AB*AC = 2(3)+ 3(p-2)+q+3=0#

#q = -3p-3#

Part (iii)

The length of #vec(AB)# is

#|vec(AB)| = sqrt(2^2+3^2+1^2)#

#|vec(AB)| = sqrt(17)#

In #vec(AC)#, set #p=3#:

#vec(AC) = 3hati+(3-2)hatj+(q+3)hatk#

#vec(AC) = 3hati+hatj+(q+3)hatk#

Set the magnitudes equal:

#sqrt17= sqrt(3^2+1^2+(q+3)^2)#

Square both sides:

#17 = 3^2+1^2+(q+3)^2#

#(q+3)^2= 7#

#q+3 = +-sqrt7#

#q = -3-sqrt7# and #q = -3+sqrt7#

Monzur R.
Mar 13, 2018

#p=13/2,q=-3/2#

Explanation:

We have

#bb(OA)=bba=((0),(2),(-3))#

#bb(OB)=bb(b)=((2),(5),(-2))#

#bb(OC)=bbc=((3),(p),(q))#

corresponding to the points #A,B,C#.

If #ABC# is on a straight line, then we can right that straight line in vector form as

#l:bba+lambda(bb(b)-bba)=((0),(2),(-3))+lambda(((2),(5),(-2))-((0),(2),(-3)))#

#=((0),(2),(-3))+lambda((2),(3),(1))#

which #bbc# will satisfy. In particular,

#((3),(p),(q))=((0),(2),(-3))+lambda((2),(3),(1))#

or

#3=2lambdarArrlambda=3/2#

#p=2+3lambda=2+3(3/2)=13/2#

#q=-3+lambda=-3+3/2=-3/2#

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