💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 71 Hard Difficulty

Find the distance from the point to the given plane.

$ (1, -2, 4) , 3x + 2y + 6z = 5 $


Distance $=\frac{18}{7}$


You must be signed in to discuss.

Video Transcript

we have a question and this we need to find the distance of the plane. The X. Let's do I Plus six. That equal to five From a .1 Common -2 comma four. Okay so we have a formula for this if point is excellent. Common violin Commons that one. And plane is A X plus B. Y plus C. Said plus day. Then distance will be equal to excellent plus B. Y. One plus sees that one Bless D. Mud by a square must be a squared plus B squared the route. No. In our question that distance will be called to Excellent. First of all that is tragedies. Yeah plus two I plus six dead minus five equals to zero. So now let's split in. Excellent. Why won't comment that one? As here in this equation in this formula So three into 1 bless two and 2 -2 plastics into four minus five, divide by and models of course A193 square to square Plus six square under the scheduled So mark this is 3 -4 Plus 24 -5, divide by nine. Platform plus 36. So this is 24 plus 3, -9. That is 18 martin, divide by 36-plus 4. 40 49 in the root so 18 is always positive. So more mode can be removed. Models can be removed 18 by square root of 49 is seven. So distances 18 x seven units. Thank you