# A level vectors HELP!!!

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I have been trying to answer these questions for a while but I’m struggling with them. Can someone please help??

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(Original post by

I have been trying to answer these questions for a while but I’m struggling with them. Can someone please help??

**G_ogh**)I have been trying to answer these questions for a while but I’m struggling with them. Can someone please help??

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#3

For:

10a) use the vector equation of a line

where

10b) You now two equations that have an equal

10c) Use the formula cos

where

11) intersections

Step 1: Make both equations equal to each other

Step 2: Simplify

Step 3: Create 3 linear simultaneous equations such that you solve two, and the third equation you test to see if the defined scalars equate LHS = RHS.

Step 4: Substitute your value of

Hope this helps massively

10a) use the vector equation of a line

**r**=**a**+ μ**b**where

**a**is a fixed point,**b**is the position vector (AB =**b**-**a**),**r**is the general point in the line and μ is a scalar (note that I use μ instead of lambda as it is indicated in the next question.10b) You now two equations that have an equal

**r**such that you should make an attempt to see if they*do*intersect via simultaneous equations. If LHS does not = RHS when you have defined scalars then they do not.10c) Use the formula cos

*θ*= (**b**.**d**)/mod(**b**)*mod(**d**)where

**b**and**d**are the*direction vectors*of the two lines, not that mod is modular signs indicating it cannot be negative.11) intersections

Step 1: Make both equations equal to each other

Step 2: Simplify

Step 3: Create 3 linear simultaneous equations such that you solve two, and the third equation you test to see if the defined scalars equate LHS = RHS.

Step 4: Substitute your value of

*s*or*t*to state a position vector of their point of intersection.Hope this helps massively

Last edited by delporto8; 2 years ago

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#5

what about :

create linear equations in X,Y & Z and solve simultaneously for s and t.

You'll only need 2 of these three equations to find s & t

and for intersection, these values need to be consistent in the 3rd eqn.

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#6

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#7

(Original post by

No, the vector equation of a line / dot product etc. have been moved to further maths so I'm guessing the OP is a further maths student.

**Notnek**)No, the vector equation of a line / dot product etc. have been moved to further maths so I'm guessing the OP is a further maths student.

thanks Notters

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(Original post by

For:

10a) use the vector equation of a line

where

10b) You now two equations that have an equal

10c) Use the formula cos

where

11) intersections

Step 1: Make both equations equal to each other

Step 2: Simplify

Step 3: Create 3 linear simultaneous equations such that you solve two, and the third equation you test to see if the defined scalars equate LHS = RHS.

Step 4: Substitute your value of

Hope this helps massively

**delporto8**)For:

10a) use the vector equation of a line

**r**=**a**+ μ**b**where

**a**is a fixed point,**b**is the position vector (AB =**b**-**a**),**r**is the general point in the line and μ is a scalar (note that I use μ instead of lambda as it is indicated in the next question.10b) You now two equations that have an equal

**r**such that you should make an attempt to see if they*do*intersect via simultaneous equations. If LHS does not = RHS when you have defined scalars then they do not.10c) Use the formula cos

*θ*= (**b**.**d**)/mod(**b**)*mod(**d**)where

**b**and**d**are the*direction vectors*of the two lines, not that mod is modular signs indicating it cannot be negative.11) intersections

Step 1: Make both equations equal to each other

Step 2: Simplify

Step 3: Create 3 linear simultaneous equations such that you solve two, and the third equation you test to see if the defined scalars equate LHS = RHS.

Step 4: Substitute your value of

*s*or*t*to state a position vector of their point of intersection.Hope this helps massively

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#9

(Original post by

Thank you but is it possible if you could tell me the actual answers because vectors are by far my weakest topic 😭

**G_ogh**)Thank you but is it possible if you could tell me the actual answers because vectors are by far my weakest topic 😭

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#10

**G_ogh**)

Thank you but is it possible if you could tell me the actual answers because vectors are by far my weakest topic 😭

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(Original post by

*sharp intake of breath*

**begbie68**)*sharp intake of breath*

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#12

for part b) because we have 2 dimensional vectors the only way for the lines not to intersect is if they are parallel....

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#13

**G_ogh**)

Thank you but is it possible if you could tell me the actual answers because vectors are by far my weakest topic 😭

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#14

any 2 lines in 3D will have at some point 2 of their coords being the same

think about looking up at the sky and seeing aeroplane vapour trails which appear to intersect, but one is way higher than the other.

so, from the vector line eqns, always we can find a pair of coords which will be the same, for some value of s & t (in this case)

so we should be able to equate any 2 pairs of eqns for ordinates

eg

2 lines are :

l1:

then think about the 3 eqns for each ordinate

X: 3 + p = 1 + 2q

Y: -2 + p = 1 - q

Z: 1 + p = -2 + q

Choose any 2 of these eqns to solve for p,q

Then, if the 2 lines actually intersect, these values for p,q will be consistent in the 3rd eqn which you've not yet used.

Then, if they do intersect, the pt of intersection will be found when you subst the value(s) of p,q back into their respective line eqns.

In this example I'm using i'd personally use the Y,Z eqns to eliminate q quite easily to give -1 + 2p = -1 , then p = 0, q = 3

sub these back into X-eqn gives 3 + (0) = 1 + 2(-3) which is not true, so the 2 lines do not meet.

You can always use geogebra to check your answers & to see what's happening.

good luck.

think about looking up at the sky and seeing aeroplane vapour trails which appear to intersect, but one is way higher than the other.

so, from the vector line eqns, always we can find a pair of coords which will be the same, for some value of s & t (in this case)

so we should be able to equate any 2 pairs of eqns for ordinates

eg

2 lines are :

l1:

**r**= (3**i**- 2**j**+**k**) + p(**i**+**j**+**k**) ; l2 :**r**= (**i**+**j**-2**k**) + q(2**i**-**j**+**k**)then think about the 3 eqns for each ordinate

X: 3 + p = 1 + 2q

Y: -2 + p = 1 - q

Z: 1 + p = -2 + q

Choose any 2 of these eqns to solve for p,q

Then, if the 2 lines actually intersect, these values for p,q will be consistent in the 3rd eqn which you've not yet used.

Then, if they do intersect, the pt of intersection will be found when you subst the value(s) of p,q back into their respective line eqns.

In this example I'm using i'd personally use the Y,Z eqns to eliminate q quite easily to give -1 + 2p = -1 , then p = 0, q = 3

sub these back into X-eqn gives 3 + (0) = 1 + 2(-3) which is not true, so the 2 lines do not meet.

You can always use geogebra to check your answers & to see what's happening.

good luck.

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#15

(Original post by

I totally get what you mean by this, but really, how do we make equations equal to each other?

what about :

create linear equations in X,Y & Z and solve simultaneously for s and t.

You'll only need 2 of these three equations to find s & t

and for intersection, these values need to be consistent in the 3rd eqn.

**begbie68**)I totally get what you mean by this, but really, how do we make equations equal to each other?

what about :

create linear equations in X,Y & Z and solve simultaneously for s and t.

You'll only need 2 of these three equations to find s & t

and for intersection, these values need to be consistent in the 3rd eqn.

*only two equations*get two values for s and t. Then the third equation you use to test to see if LHS = RHS. If they do, then they intersect, if not then they are skew.

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