Question
(1 point) Find the distance the point P(-1, -5, 1), is to the plane through the three points Q(-3, 22, 4) , R(, 6,-7) , and S(O, 3, 0).
(1 point) Find the distance the point P(-1, -5, 1), is to the plane through the three points Q(-3, 22, 4) , R(, 6,-7) , and S(O, 3, 0).
Answers
Find the distance between the point and the plane. $$\begin{aligned} &(1,3,-1)\\ &3 x-4 y+5 z=6 \end{aligned}$$
Okay, So we asked upon the distance from the points P one comma two, comma three neglect you and the equation. It was an X minus three. But to why was one is equal to r minus. Worthing, you don't thing. So we can actually write equations to this. Exposed to wind my fourth e minus one itself. Okay, you see that Are normal. Back through for our plane is in physical, too. 1234 s. So we know that the points Q, which is one comma zero company Roland lives on our plane. So our component form of the vector from cute to be nickel to one when it's one college, two minutes. Zero come in three minutes. Did you get that secret to Joe? Come on, Number two, number three. No, it's fine. This thief on the plane to appoint me So we have d is equal to projection of u P on to end. But it's the same man the camp of Q P on to end magnitude, u p times and overtime. I do, Ben, That gives me You're 16 over Scarlett's 21
Hello. So here the distance from the point to one negative one. Um From the plane here x minus two Y plus two Z plus five equals zero. That is going to be given by the absolute value of two minus two um Times one plus two times negative one plus five. All divided by the square root of one squared plus two Squared plus two squared an absolute value. So this here is going to be equal with the top. This just becomes equals to 3/3. So this is what we evaluate this. This is equal to 3/3 which is equal to one. So here the distance is just equal to one unit.
For this problem, we want to find the distance from the point to the plane. So we know the point and we know the plane. So now we just need to do the calculations. We know that the distance is going to be equal to and peanut dot p over magnitude of an so based on what we have here we have three, There's gonna be three times 1 Because three comes from the plane and one comes from the point. It has to be plus two times negative, two plus six times four And then we have -5 because that's the constant term, that's all divided by the magnitude of n. Where an is 3- six, the normal vector. So the magnitude van is going to be the square root of three squared plus two squared plus six squared. That ends up giving us 18 over route 49, Which is 18/7. Final answer.
For this problem. We know the point one negative. 24 we know three X plus two I plus six z equals five. And we know our distance is going to be equal to the amplitude or the magnitude of end dot peanut dot p all over the magnitude of n. So when we do our calculations, um, the question gives us everything we need to plug in our formula. Um, so we're going to use an our normal vector, but we have to rise that and is going to equal 3 to 6 based on the equation of the plane that were given. So our distance is going to be equal to 18. Over the magnitude of end is Route 49. So that's the same thing as 18/7. And that is going to be our final answer for the distance between that line or that point and the plane