FP2: Vectors

Question:

Find the equation of the line of intersection of the planes 1 and 2 with equations: r.(5i-1j-2k)=16, r.(16i-5j-4k)=53

working, so far:

change to cartesian:

5x-y-2z=16 (1)
16x-5y-4z=53 (2)

multiply (1) by 2, so the z column can be eliminated.

10x-2y-4z=32 (3)

(3)-(2)=-6x+3y=-21

x=\frac{y+7}{2}

substitute this result into (2)

z=\frac{3y+3}{4}

let,

y=\lambda

x=\frac{\lambda+7}{2}\implies\lambda=\frac{x-\frac{7}{2}}{2}

also,

z=\frac{3\lambda+3}{4}\implies\lambda=\frac{z-\frac{3}{4}}{\frac{4}{3}}

\therefore \frac{x-\frac{7}{2}}{2}=y=\frac{z-\frac{3}{4}}{\frac{4}{3}}=\lambda

or,

r=(\frac{7}{2}i+\frac{3}{4}k)+\lambda (2i+j+\frac{4}{3}k)

the answers don't agree with me, their final result is:

r=(3i-j)+\lambda(\frac{2}{3}i+\frac{4}{3}j+k)

Reply 1

Clarity Incognito

KeineHeldenMehr

z=\frac{3\lambda+3}{4}\implies\lambda=\frac{z-\frac{3}{4}}{\frac{4}{3}}

\therefore \frac{x-\frac{7}{2}}{2}=y=\frac{z-\frac{3}{4}}{\frac{4}{3}}=\lambda

or,

r=(\frac{7}{2}i+j+\frac{3}{4}k)+\lambda (2i+j+\frac{4}{3}k)

Should be

\frac{x-\frac{7}{2}}{\frac{1}{2}}=y=\frac{z-\frac{3}{4}}{\frac{3}{4}}=\lambda

Reply 2

KeineHeldenMehr

Clarity Incognito

Should be

\frac{x-\frac{7}{2}}{\frac{1}{2}}=y=\frac{z-\frac{3}{4}}{\frac{3}{4}}=\lambda

i agree, my mistake, i just realised there's another error on my final answer
should be 0j on the position vector.

so really, it should be:

r=(\frac{7}{2}i+\frac{3}{4}k)+\lambda(\frac{1}{2}i+j+\frac{3}{4}k)

still doesn't quite match the given answer...

Reply 3

Clarity Incognito

KeineHeldenMehr

or, r=(\frac{7}{2}i+j+\frac{3}{4}k)+\lambda (2i+j+\frac{4}{3}k)

the answers don't agree with me, their final result is:

r=(3i-j)+\lambda(\frac{2}{3}i+\frac{4}{3}j+k)

i have huge amounts of doubt when it comes to eliminating one of the variables to solve via simultaneous equation and have suspect it has some kind of meaning, i've trialled eliminating y terms, i get another strange result. i dont really understand what i'm doing, kind of brainlessly following a textbook example.

please help!

There shouldn't be a j in this line r=(\frac{7}{2}i+j+\frac{3}{4}k)+\lambda (2i+j+\frac{4}{3}k)

So your vector equation should turn out as

r=(\frac{7}{2}i+\frac{3}{4}k)+\lambda (\frac{1}{2}i+j+\frac{3}{4}k)

EDIT: you might be wondering why that doesn't look like the answer? Well, it describes exactly the same line as what the answer is. If you look, the direction vector is a multiple of the answer and they have just given it relative to an arbitrary point that's on the line.

Reply 4

Clarity Incognito

KeineHeldenMehr

i agree, my mistake, i just realised there's another error on my final answer
should be 0j on the position vector.

See above, I caught on to that at about the same time too :p:

Reply 5

KeineHeldenMehr

Clarity Incognito

There shouldn't be a j in this line r=(\frac{7}{2}i+j+\frac{3}{4}k)+\lambda (2i+j+\frac{4}{3}k)

So your vector equation should turn out as

r=(\frac{7}{2}i+\frac{3}{4}k)+\lambda (\frac{1}{2}i+\frac{4}{3}j+k)

EDIT: you might be wondering why that doesn't look like the answer? Well, it describes exactly the same line as what the answer is. If you look, the direction vector is a multiple of the answer and they have just given it relative to an arbitrary point that's on the line.

for your direction vector, you havent happened to misplace the fraction onto j instead of k.

i thought the final result should be:

r=(\frac{7}{2}i+\frac{3}{4}k)+\lambda (\frac{1}{2}i+j+\frac{3}{4}k)

since y=lambda, y/1. i dunno.

Reply 6

Clarity Incognito

KeineHeldenMehr

for your direction vector, you havent happened to misplace the fraction onto j instead of k.

i thought the final result should be:

r=(\frac{7}{2}i+\frac{3}{4}k)+\lambda (\frac{1}{2}i+j+\frac{3}{4}k)

since y=lambda, y/1. i dunno.

Yep sorry I had just copy and pasted your vector equation and tried to make amendments. I can confirm that the vector equation should be what you have written above.

Multiply your direction vector by