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
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

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