Question

Find the angle of intersection of the plane −1x+2y−4z=5 with the plane −1x−3y+2z=1.

Answer 1

i found 2 normal vectors

n1 = <-1, 2, -4>
n2 = <-1, -3, 2>

i used the formula
$$cos(\theta) = \frac{\vec{n_{1}}\cdot \vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|}$$

Answer 2

so i have theta = arccos(-13/(sqrt(21)sqrt(14)))

Answer 3

which is incorrect, what am i doing wrong?

Answer 4

(arccos(-13/(sqrt(21)sqrt(14))))(180/pi) is also incorrect

Answer 5

I had someone else tell me my answer is correct but maybe we both missed something

Answer 6

why do you think it is wrong ?

Answer 7

because the thing i type it into that validates my answer says it is, but its possible there is an error in the answer key maybe, @ganeshie8

Answer 8

what value are you entering ?

Answer 9

im entering the exact value arccos(-13/(sqrt(21)sqrt(14))) which gives me 2.43130661686978, ive also tried rounding and tried using degrees instead of radians

Answer 10

try entering 40.7

Answer 11

40.7 degrees

Answer 12

lol thats correct, nice!!, why is that correct?

Answer 13

it wants the acute angle between the planes

Answer 14

let me ask you a question maybe

Answer 15

hmm interesting

Answer 16

whats the angle between those lines ?
is it \(\theta\) or \(\alpha\) ?

Answer 17

well its either imo, but i would probably initially say theta

Answer 18

it coule be either of them
depends on what the grader wants :P

Answer 19

that makes sense so i just need to subtract my answer from 180

Answer 20

usually we stick to acute angles when the question doesn't explicitly asks for anything

Answer 21

ah ok

Answer 22

Yep \(\theta +\alpha = 180\)

Answer 23

perfect

Answer 24

because they make a straight line

Answer 25

ty :)