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# another P3 vector qu :( watch

1. (for reference rev ex2, p3 green heinman book....qu.33)
Referred to a fixed origin O, the point A and B have posn vectors (5i-j-k) and (i-5j+7k) respectively.
Find the posn vector of the point D, where D is not A, on AB such that OD=OA

Okay so if i know D is on AB, it must satisfy 5i-j-k+t(-4i-4j+8k).
But , if OD=OA, then 5i-j-k+t(-4i-4j+8k)=(5i-j-k), but that just gives me D=A[cuz t=o after coefficients are equated], and qu says this is not the case, so where am i going wrong?
(for reference rev ex2, p3 green heinman book....qu.33)
Referred to a fixed origin O, the point A and B have posn vectors (5i-j-k) and (i-5j+7k) respectively.
Find the posn vector of the point D, where D is not A, on AB such that OD=OA

Okay so if i know D is on AB, it must satisfy 5i-j-k+t(-4i-4j+8k).
But , if OD=OA, then 5i-j-k+t(-4i-4j+8k)=(5i-j-k), but that just gives me D=A[cuz t=o after coefficients are equated], and qu says this is not the case, so where am i going wrong?
When it says OA = OD it means (I think) that they have the same modulus!
So the triangle OAD is an isoceles triangle, with (lemgth of) OA = (length of) OD , if that helps
3. Yeh, just checked it - compare the modulus of OA, as given, and as from the formula you worked out.
4. (Original post by Fermat)
When it says OA = OD it means (I think) that they have the same modulus!
So the triangle OAD is an isoceles triangle, with (lemgth of) OA = (length of) OD , if that helps
ohhh lol that's why the book wrote it as |OA|=|OD|!!! D'oh!
5. ok soo if i write sqrt(27)=5i-j-k+t(-4i-4j+8k), ah wait xi+yj+zk=5i-j-k+t(-4i-4j+8k).
i: x=5-4t
j:y=-1-4t
k:z=-1+8t,

so (5-4t)i+(-1-4t)j+(-1+8t)k=sqrt(27)

Am i on the right track or am i just going round in circles here?i suspect the latter
ok soo if i write sqrt(27)=5i-j-k+t(-4i-4j+8k), ah wait xi+yj+zk=5i-j-k+t(-4i-4j+8k).
i: x=5-4t
j:y=-1-4t
k:z=-1+8t,

so (5-4t)i+(-1-4t)j+(-1+8t)k=sqrt(27)

Am i on the right track or am i just going round in circles here?i suspect the latter
Almost!
It should be modulus(lhs) = sqrt(27)
then you get a quadratic in t.
7. (Original post by Fermat)
Almost!
It should be modulus(lhs) = sqrt(27)
then you get a quadratic in t.
oooo 3i-3j+3k!!
the book answer! yay! thanks fermat!
oooo 3i-3j+3k!!
the book answer! yay! thanks fermat!
fermat strikes again.
9. (Original post by Fermat)
Almost!
It should be modulus(lhs) = sqrt(27)
then you get a quadratic in t.
what's lhs?
10. (Original post by wibble...)
what's lhs?
Left Hand Side?

~Darkness~
11. (Original post by Darkness)
Left Hand Side?

~Darkness~
yep

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