C4 Vectors

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They simplified (-4,2,2) into the simplest ratio of (-2,1,1). (-4,2,2) = 2(-2,1,1). 2 is a scaler parameter that is included as multiple of u.

So, lets say r=(1,5,2) + t(-4,2,2) = (1,5,2) + u(-2,1,1)

u=2t

How would I go about doing the last question there? I know it starts off as r=(3,2,-5).. But then what? The unknown there really puts me off and confuses me :frown:

(edited 10 years ago)

Reply 21

BabyMaths

Original post by krisshP

How would I go about doing the last question there? I know it starts off as r=(3,2,-5).. But then what?

Look at the equation of the line that it has to be parallel to. This is r=i+3j+lambda*(......).

The bit in brackets gives you the direction.

Reply 22

zaback21

Original post by krisshP

How would I go about doing the last question there? I know it starts off as r=(3,2,-5).. But then what? The unknown there really puts me off and confuses me :frown:

Rewrite the equation as:

r= i+ 3j + lambda ( 2i -4j + 5k)

Reply 23

krisshP

Original post by BabyMaths

Look at the equation of the line that it has to be parallel to. This is r=i+3j+lambda*(......).

The bit in brackets gives you the direction.

Original post by zaback21

Rewrite the equation as:

r= i+ 3j + lambda ( 2i -4j + 5k)

That made me feel dumb :colondollar:

Now I got r=(3,2,-5) +lambda(2,-4,5) :smile:

Thanks

Reply 24

zaback21

Original post by krisshP

That made me feel dumb :colondollar:

Now I got r=(3,2,-5) +lambda(2,-4,5) :smile:

Thanks

Asking never hurts. Now you will never make this mistake again. We all did when we first started learning. You just get better with practice.

Reply 25

krisshP

Original post by zaback21

Asking never hurts. Now you will never make this mistake again. We all did when we first started learning. You just get better with practice.

Well said, it does feel a bit mind boggling with the vector forms of i, j and k stuff and lambda coming all at you and I couldn't see a way forward I was clueless. I guess I'll get much better with a lot of practice.

One must fail to learn I believe :rolleyes:

Reply 26

zaback21

Original post by krisshP

Haha take the Further Maths and you will see how interesting it gets in 3D Vector Space ( a must if you plan to study science, maths or engineering).

Reply 27

krisshP

Original post by zaback21

Haha take the Further Maths and you will see how interesting it gets in 3D Vector Space ( a must if you plan to study science, maths or engineering).

Is that in A2 or at AS?

Reply 28

zaback21

Original post by krisshP

Is that in A2 or at AS?

Its in FP1-FP3. Check the content. Most likely A2. But you can take them in AS too if you want I guess. Well I did. I took two Maths at A Level.

Reply 29

krisshP

Original post by zaback21

find the angle in degrees between the vectors a=-8i+j+4k and b=6i+3j+6k

I got 105°. What did you get? My textbook says it is 75°, why? :confused:

Thanks

Reply 30

BabyMaths

Original post by krisshP

find the angle in degrees between the vectors a=-8i+j+4k and b=6i+3j+6k

I got 105°. What did you get? My textbook says it is 75°, why? :confused:

Thanks

\arccos \left(-\frac{7}{27}\right)=105.026..

degrees.

Did the question ask for the acute angle?

Reply 31

krisshP

Original post by BabyMaths

\arccos \left(-\frac{7}{27}\right)=105.026..

degrees.

Did the question ask for the acute angle?

Nope, a bit annoying :frown:

Anyway thanks

Reply 32

krisshP

Original post by BabyMaths

I feel so confused on this and I check it like 3 times :frown:

AB.AC=12z+26

|AB|=13

|AC|=√(z^2 + 40)

Cos60=1/2 from special triangles

(12z+26)/[13√(z^2 + 40)] = 1/2

2(12z+26)=13√(z^2 +40)

24z+52=√(169z^2 +6760)

576z^2 + 2704=169z^2 +6760

407z^2 - 4956=0

Using the quadratic formula gives me z=3.2 or z=-3.2 both to 2SF

Where did I go wrong? Textbook says z=1.33 or z=-7.47 :confused:

WHY?

Thank you VERY much :smile:

Reply 33

zaback21

Original post by krisshP

I feel so confused on this and I check it like 3 times :frown:

WHY?

Thank you VERY much :smile:

You made a mistake in the bold line. When you square (24z+52)^2, use (a+b)^2 formula=a^2+2ab+b^2

So, it should be 576z^2 +2496z+ 2704=169z^2 +6760

(edited 10 years ago)

Reply 34

krisshP

Original post by zaback21

You made a mistake in the bold line. When you square (24z+52)^2, use (a+b)^2 formula=a^2+2ab+b^2

So, it should be 576z^2 +2496z+ 2704=169z^2 +6760

Okay, thanks. I feel really dumb now :colondollar:

Reply 35

krisshP

Original post by BabyMaths

Part a was fine for me and I got r=(4,-11,4)+lambda(1,4,1)

In part b I hit a brick wall. The dot product of MA and MO must be 0 as both vectors are perpendicular. So

x(x-4)+y(11+y)+z(z-4)=0

I only have 1 equation with 3 unknowns. It would be okay if I had to equations with 3 unknowns that an be solved simultaneously, but this isn't the case. :confused:

help me out please

Thanks :smile:

Reply 36

BabyMaths

Original post by krisshP