The points A and B have position vectors, relative to the origin O, given by

Question

−−→OA = i+2j+3k and −−→OB = 2i+j+3k.
The line l has vector equation
r = (1−2t)i+ (5+t)j+ (2−t)k.
(i) Show that l does not intersect the line passing through A and B. [4]
(ii) The point P lies on l and is such that angle PAB is equal to 60◦
. Given that the position vector
of P is (1 − 2t)i + (5 + t)j + (2 − t)k, show that 3t
2
+ 7t + 2 = 0. Hence find the only possible
position vector of P.

I got everything else except that Idk how to find the possible vector of P.
t=-2 and -3

so how do you know which to use?

Ans 5i + 3j +4k, means they used t=-2

Please help!
Thank you!

Answer 1

For easier typing I will use <...> to represent a vector and (.. ..)to represent a point.

I can write the line l in parametric form
x = 1-2t
y = 5+t
z = 2-t

let P(x,y,z) be the point we are looking for on line l
or P is (1-2t, 5+t, 2-t)
vector AP = <1-2t-1, 5+t-2, 2-t-3> or
vector AP = <-2t , 3+t , -1-t >
similarly
vector AB = <1,-1,0>
AB dot AP= |AB||AP|cos60°
-2t -3-t + 0 = √2 √(4t^2 + (3+t)^2 + (-1-t)^2) (1/2)
-3-3t = √2 √(4t^2 + 9+6t+t^2 + 1+2t+t^2) (1/2)
-6-6t = √2 √(6t^2 + 8t + 10) ******
square both sides
36 + 72t + 36t^2 = 12t^2 + 16t + 20
24t^2 +56t + 16 = 0
divide by 8
3t^2 + 7t + 2 = 0 -----> as requested

(t+2)(3t + 1) = 0
t = -2 or t = -1/3

check: in ******
if t = -2
LS = -6+12 = 6
RS = √2 √(24 -16+10)
= √2√18
= √36 = 6 = LS

If t + -1/3
LS = -6 + 2 = -4
But by definition of √, √(of anything) is the Positive result,
so LS ≠ RS

only t = -1 works, which gives us vector OP = <5,3,4>
and P is the point (5,3,4)

Answer 2

Reiny amazing answer