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

−−→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!

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)

Reiny amazing answer

Well, the reason they used t = -2 is because substituting t = -2 into the vector equation of l gives the position vector (1 - 2(-2))i + (5 + (-2))j + (2 - (-2))k which simplifies to 5i + 3j + 4k, which is the position vector of P.

As for why they used t = -2 instead of t = -3, I'm afraid I don't have an answer for that. Maybe t = -2 gave a more meaningful or logical result in the given context. Or maybe it was just a random choice made by the person who solved the problem. After all, life is full of choices, some of which may seem arbitrary, like deciding between t = -2 and t = -3. Just like choosing between a red nose and a blue nose. It's all a clown's game!

To find the possible position vector of P, we need to determine the value of t that satisfies the equation 3t^2 + 7t + 2 = 0.

We can factorize the quadratic equation to solve for t:
(3t + 1)(t + 2) = 0

Setting each factor equal to zero, we have two possibilities:
3t + 1 = 0 or t + 2 = 0

Solving the first equation:
3t + 1 = 0
3t = -1
t = -1/3

Solving the second equation:
t + 2 = 0
t = -2

So we have t = -1/3 and t = -2 as possible values for t.

Now, we substitute each value of t into the equation of the position vector of P to find the corresponding position vectors.

For t = -1/3:
P = (1 - 2(-1/3))i + (5 + (-1/3))j + (2 - (-1/3))k
P = (1 + 2/3)i + (15/3 - 1/3)j + (6/3 + 1/3)k
P = (3/3 + 2/3)i + (14/3)j + (7/3)k
P = 5/3i + 14/3j + 7/3k

For t = -2:
P = (1 - 2(-2))i + (5 + (-2))j + (2 - (-2))k
P = (1 + 4)i + (3)j + (4)k
P = 5i + 3j + 4k

Therefore, we have two possible position vectors of P: (5/3)i + (14/3)j + (7/3)k and 5i + 3j + 4k.

To find the possible position vector of point P, we need to solve the equation 3t^2 + 7t + 2 = 0. This is a quadratic equation that can be factored:

3t^2 + 7t + 2 = 0

(3t + 1)(t + 2) = 0

Now, we set each factor equal to zero and solve for t:

3t + 1 = 0 => t = -1/3

t + 2 = 0 => t = -2

So, we have two possible values for t: t = -1/3 and t = -2.

To determine which value of t to use, we can substitute each value into the position vector of P given in the question and check if it satisfies the given condition.

Let's substitute t = -1/3 into the position vector:
P = (1 - 2(-1/3))i + (5 + (-1/3))j + (2 - (-1/3))k
= (1 + 2/3)i + (5 - 1/3)j + (2 + 1/3)k
= (3/3 + 2/3)i + (15/3 - 1/3)j + (6/3 + 1/3)k
= (5/3)i + (14/3)j + (7/3)k

Now, let's substitute t = -2 into the position vector:
P = (1 - 2(-2))i + (5 + (-2))j + (2 - (-2))k
= (1 + 4)i + (5 - 2)j + (2 + 2)k
= (5)i + (3)j + (4)k
= 5i + 3j + 4k

So, we find that when t = -1/3, the position vector of P is (5/3)i + (14/3)j + (7/3)k, and when t = -2, the position vector of P is 5i + 3j + 4k.

Therefore, both values of t satisfy the condition, but since the question asks for a possible position vector of P, we can use either of them. In this case, they used t = -2 and the corresponding position vector of P is 5i + 3j + 4k.