Find the cosine of the angle between the planes $ x + y + z = 0 $ and $ x + 2y + 3z = 1 $
Hello. So the question is straight and from vectors and geometry of space fine. And we have to find a cousin of the angle between the planes. So there are two planes. The funniest X plus Y plus that is equal to zero and the other is X plus two Y plus trees. It is equal to them. And the direction because I know vector corresponding peddler to this plane is I plus J. Plus key is equal. So this is the vector corresponding to let this is victor a corresponding to first plane. And retro be corresponding to second plane is I plus to the G Plus three. Okay so a sign of feta between these two is vector a vector B divided by model vector A and model that will be okay. So it'll be will be I plus G plus k. Don't I place two J last two. Okay divided by magnitude of I plus J plus case square with one plus one plus one. So that will be squared. Okay. Okay so and the magnitude of vector B is one plus four plus. Nice. So like my father solving it we get the value of course of three to is equal to my daughter is one. Jojo is one. Otherwise I dont jay z I don't gaze you so we get one plus two. Last three Divided by Square with three and Square 14. So that will be six or 1 squared what to do. Okay which is the so which is the required value of the sine of the angle between them. So angle will be caused in world six. So we'll spare 42. So its historic wired solution of the equation. Hope this grazers.