Find the equation of the plane that contains the line x = 1 - t, y = 2 + t,
z = -3t and the point P(2, 0, -6). [Hint: A vector from
any point on the line to P will lie in the plane]
a few seconds with google can almost always help. Try this:
math.stackexchange.com/questions/640463/equation-of-a-plane-containing-a-point-and-a-line
Why did the plane go to therapy? Because it couldn't find its equation! But don't worry, I'm here to help.
To find the equation of the plane that contains the given line and the point P(2, 0, -6), we can start by finding two vectors that lie in the plane. One vector can be found using any point on the line and the point P.
Let's choose t = 0 on the line, we get the point (1, 2, 0). The vector from this point to P is given by:
(P - (1, 2, 0)) = (2, 0, -6) - (1, 2, 0) = (2 - 1, 0 - 2, -6 - 0) = (1, -2, -6).
Now, we can take another vector that lies along the direction of the line. We already have the parametric equations of the line, so we can take the direction vector as:
(1, 1, -3) * t.
Since we want the equation of the plane to contain the line and the point P, any vector in the plane can be written as a linear combination of these two vectors.
Let's call the equation of the plane Ax + By + Cz + D = 0.
The vector (1, -2, -6) lies in the plane, so it satisfies the equation of the plane:
A(1) + B(-2) + C(-6) + D = 0.
The direction vector (1, 1, -3) * t also lies in the plane, so it satisfies the equation of the plane:
A(1 - t) + B(1 + t) + C(-3t) + D = 0.
Now, we have two equations to solve for A, B, C, and D:
A - 2B - 6C + D = 0,
A - At + B + Bt - 3Ct + D = 0.
To solve these equations, we can substitute the coordinates of the point P(2, 0, -6) into them:
A - 2B - 6C + D = 0,
2A + D = 0.
Substituting A = -D/2 into the first equation:
-D/2 - 2B - 6C + D = 0,
2(-D/2) + D = 0.
Simplifying these equations, we get:
-D - 4B - 12C + 2D = 0,
-D + 2D = 0.
Combining like terms, we get:
3D - 4B - 12C = 0,
D = 0.
Substituting D = 0 into the first equation, we get:
-4B - 12C = 0.
Dividing both sides by -4, we get:
B + 3C = 0.
Now, we have two variables, B and C, that are dependent on each other. We can solve for A by substituting B = -3C into the equation:
-4B - 12C = 0,
-4(-3C) - 12C = 0,
12C - 12C = 0.
This means that A can be any value since it cancels out. Let's choose A = 1 for simplicity.
So, the equation of the plane that contains the given line and the point P(2, 0, -6) is:
x - 3y = 0.
I hope this equation makes your day a bit brighter! If you have any more questions, feel free to ask!
To find the equation of the plane that contains the line and point given, we need to find two vectors that lie on the plane.
First, let's calculate the direction vector of the line. We can find this by taking the differences in the x, y, and z coordinates between two points on the line.
Let's choose t = 0 and t = 1 for convenience.
For t = 0, we have:
P1(1 - 0, 2 + 0, -3(0)) = P1(1, 2, 0)
For t = 1, we have:
P2(1 - 1, 2 + 1, -3(1)) = P2(0, 3, -3)
Now we can calculate the direction vector by subtracting the coordinates of P1 from P2.
Direction vector D = P2 - P1 = (0 - 1, 3 - 2, -3 - 0) = (-1, 1, -3)
Next, we need to find another vector lying on the plane. We will use the vector from any point on the line (x = 1 - t, y = 2 + t, z = -3t) to the point P(2, 0, -6).
Let's substitute the coordinates of the line equation into the coordinates of point P to find this vector.
Vector P1P = P - P1 = (2 - 1, 0 - 2, -6 - 0) = (1, -2, -6)
Now we have two vectors that lie on the plane: D = (-1, 1, -3) and P1P = (1, -2, -6).
We can find the normal vector of the plane by taking the cross product of these two vectors.
N = D x P1P = (1 * (-6) - (-2) * (-3), (-1) * (-6) - 1 * (-3), (-1) * (-2) - 1 * 1) = (-12 - 6, 6 - 3, 2 + 1) = (-18, 3, 3)
The equation of the plane can be written in the form Ax + By + Cz + D = 0, where (A, B, C) is the normal vector.
So, the equation of the plane that contains the line x = 1 - t, y = 2 + t, z = -3t and the point P(2, 0, -6) is:
-18x + 3y + 3z + D = 0
To find the value of D, substitute the coordinates of P into the equation:
-18(2) + 3(0) + 3(-6) + D = 0
-36 - 18 + D = 0
D = 54
Therefore, the equation of the plane is:
-18x + 3y + 3z + 54 = 0
To find the equation of the plane that contains the given line and point, we can use the fact that any two non-parallel vectors in the plane will be orthogonal (perpendicular) to the normal vector of the plane.
Step 1: Find two non-parallel vectors in the plane.
Let's take two arbitrary points on the given line, say A and B, where A has coordinates (1, 2, 0) and B has coordinates (0, 3, -3). Therefore, the vector AB will lie in the plane.
AB = (0 - 1, 3 - 2, -3 - 0) = (-1, 1, -3)
Step 2: Find a vector from any point on the line to the given point P.
Let's consider the point P(2, 0, -6) on the plane. We can calculate the vector AP.
AP = (2 - 1, 0 - 2, -6 - 0) = (1, -2, -6)
Step 3: Calculate the cross product of the vectors AB and AP to find the normal vector to the plane.
The normal vector, n, of the plane is the cross product of AB and AP.
n = AB × AP = (-1, 1, -3) × (1, -2, -6)
To find the cross product, we can use the following formula:
n = ((y1 * z2 - z1 * y2), (z1 * x2 - x1 * z2), (x1 * y2 - y1 * x2))
where (x1, y1, z1) and (x2, y2, z2) are the components of AB and AP, respectively.
Calculating the cross product, we get:
n = ((1 * (-6) - (-3) * (-2)), ((-3) * 1 - (-1) * (-6)), ((-1) * (-2) - 1 * (-3)))
= (-12 - 6, (-3 + 6), (2 + 3))
= (-18, 3, 5)
So, the normal vector to the plane is n = (-18, 3, 5).
Step 4: Write the equation of the plane.
The equation of a plane can be written as ax + by + cz = d, where (a, b, c) is the normal vector and (x, y, z) are the coordinates of any point on the plane.
Using the point P(2, 0, -6) and the normal vector n = (-18, 3, 5), we can write the equation of the plane as:
-18x + 3y + 5z = d
To find the value of d, substitute the coordinates of P into the equation:
-18 * 2 + 3 * 0 + 5 * (-6) = d
-36 + 0 - 30 = d
-66 = d
Therefore, the equation of the plane that contains the given line and point P is:
-18x + 3y + 5z = -66.