Last number from the sample test!
A plane P passes by the coordinates:
A(1,0,-7); B(-2,-1,0); C(0,0,-3)
Using the equation of the line D:
x= 7+12t, y= -1-6t, z= 2-2t,
Find the equation of a line that's entirely part of the plane P and that is perpendicular to the line D at point Q(1,2,3)
Thank you
To find the equation of a line that is perpendicular to the given line and passes through point Q(1,2,3), you need to find the direction vector of the plane P first.
The direction vector of the plane P can be determined by finding two vectors that lie on the plane. One way to do this is by subtracting the coordinates of two points on the plane.
Point A(1,0,-7) and B(-2,-1,0) are on the plane P.
The direction vector AB can be found by subtracting the coordinates of B from A:
AB = (1 - (-2), 0 - (-1), -7 - 0)
= (3, 1, -7)
Similarly, you can find the direction vector AC:
AC = (0 - 1, 0 - 2, -3 - (-7))
= (-1, -2, 4)
Now, you can find the cross product of AB and AC to obtain a vector that is orthogonal (perpendicular) to the plane P.
Let's calculate the cross product:
N = AB x AC
N = (3, 1, -7) x (-1, -2, 4)
N = (1*(-2) - (-2)*1, (-7)*(-1) - 4*3, 3*(-2) - 1*(-7))
N = (-2 + 2, 7 + 12, -6 + 7)
N = (0, 19, 1)
The vector N (0, 19, 1) is the normal vector of the plane P. Any line that lies entirely in the plane P needs to have a direction vector perpendicular to N.
Now, let's find the equation of the line that is perpendicular to the line D at point Q(1,2,3).
We can use the point-normal form of a line equation, which is:
(x - x0) / a = (y - y0) / b = (z - z0) / c
where (a, b, c) is the vector perpendicular to the line, and (x0, y0, z0) is a point on the line.
We have the direction vector of the line D, which is (12, -6, -2). Since the line we want is perpendicular to D, the direction vector of the new line would be perpendicular to (12, -6, -2).
The dot product of the direction vector of the new line and (12, -6, -2) should be zero.
Let the direction vector of the new line be (m, n, p).
(m, n, p) · (12, -6, -2) = 0
12m - 6n - 2p = 0
There are infinitely many solutions to this equation. So we can choose any two values of (m, n, p) that satisfy the equation.
For simplicity, we can start by setting m = 1.
12(1) - 6n - 2p = 0
12 - 6n - 2p = 0
-6n - 2p = -12
3n + p = 6
Now, choose an arbitrary value for n (let's say n = 2) and solve for p:
3(2) + p = 6
6 + p = 6
p = 0
So we have m = 1, n = 2, and p = 0.
Now, we can use the point Q(1,2,3) and the direction vector (1, 2, 0) to write the equation of the line:
(x - 1) / 1 = (y - 2) / 2 = (z - 3) / 0
Note that (z - 3) / 0 does not provide any information about z, which means z can be any value (it is a free variable). We can represent it as a parameter t.
So the equation of the line that lies in the plane P and is perpendicular to line D at point Q(1,2,3) is:
x - 1 = 1t
y - 2 = 2t
z = t
Simplifying the equations, we can write:
x = 1 + t
y = 2 + 2t
z = t
This is the equation of the line you are looking for.