Find the Equation of Plane & Distance from Point

[VooDoo]

Hi Guys,

I am stuck on a maths question for my first year of Engineering at University. Your help would be greatly appreciated!

I don't know where to start

1. Given that a = (3, 1, 2) and b = (1,−2,−4) are the position vectors of the points P and Q
respectively, find
(a) the equation of the plane passing through Q and perpendicular to PQ,
(b) the distance rom the point (−1, 1, 1) to the plane obtained in (a).

Thanks guys

 

[HallsofIvy]

You need to know: The equation of a plane containing $(x_0,y_0,z_0)$, perpendicular to the vector Ai+ Bj+ Ck, is $A(x-x_0)+ B(y- y_0)+ C(z- z_0)= 0$. (For any point (x,y,z) in that plane, $(x-x_0)i+ (y- y_0)j+ (z- z_0)k$ is in the plane and so is perpendicular to $Ai+ Bj+ Ck$. The dot product of the two is 0.)

In this problem, the vector PQ is (3-1)i+ (1-(-2))j+ (2-(-4))k= 2i+ 3j+ 6k

As far as (b) is concerned, I'll bet there is a formula for the distance between a point and a plane in this same section of your textbook. To do it without that formula, remember that the shortest distance between a point and a plane is along a line perpendicular to that plane.
i) Write the parametric equations, for x, y, z in terms of a paratmeter t, for a line through (-1, 1, 1) in the same direction as the vector 2i+ 3j+ 6k.
ii) Plug those equations into the equation of the plane, from (a), to get a single equation in the single variable t and solve for t.

iii) Put that value of t into the parametric equations to get (x,y,z) coordinates of the point on the line and plane.

iii) Calculate the distance between that point and (-1, 1, 1).

 

[tacman]

Hi there,

Sure, I'd be happy to help you with this question! Let's break it down step by step.

(a) To find the equation of a plane passing through two points, we need to find the normal vector of the plane. This can be done by taking the cross product of the two position vectors, PQ x QP. In this case, it would be (-2, -3, -3).

Next, we can choose any point on the plane to use in our equation. Since the plane passes through Q, we can use the coordinates of Q as our point. This gives us the equation of the plane as (-2, -3, -3) * (x-1, y+2, z+4) = 0.

Simplifying this equation gives us -2x - 3y - 3z + 17 = 0.

(b) To find the distance from a point to a plane, we can use the formula d = |ax + by + cz + d| / √(a^2 + b^2 + c^2), where (a, b, c) is the normal vector of the plane and (x, y, z) is the coordinates of the given point.

In this case, we can plug in the values from the equation we found in part (a) and the coordinates of the given point to get d = |-2(-1) - 3(1) - 3(1) + 17| / √((-2)^2 + (-3)^2 + (-3)^2), which simplifies to 7/√22.

I hope this helps! Let me know if you have any further questions or need clarification on any of the steps. Best of luck with your studies! :)