Find the scalar equation for the plane passing through the point P=(0, 2, −1) and containing the line L defined by
x = 2+t
y = 4+t
z = −6+7t
Thank you Arora, It works first find the two vectors with the above points:
(0,2,-1)-(2,4,-6) = (-2,-2,5)
[3,5,1]-[2,4,-6] = [1,1,7]
cross multiply the points (to get normal vector) and we get:
19i+19j+0k
plug in the numbers: 19x+19y+0z=d
find d: 19(0)+19(2)=d ; d = 38
final solution: 19x+19y=38
To find the scalar equation for the plane passing through the point P=(0, 2, −1) and containing the line L defined by x = 2+t, y = 4+t, z = −6+7t, we need to find the normal vector of the plane.
Step 1: Find the direction vector of the line L:
The direction vector of the line L is given by the coefficients of t in the parametric equations. So, the direction vector of L is <1, 1, 7>.
Step 2: Find two other points on the plane:
Since the plane contains the line L, we can choose two other points on the line. Let's choose t = 1 and t = 0.
When t = 1, the point Q on the line is (2+1, 4+1, -6+7(1)) = (3, 5, 1).
When t = 0, the point R on the line is (2+0, 4+0, -6+7(0)) = (2, 4, -6).
Step 3: Find two vectors from P to Q and P to R:
The vector from P to Q is <3-0, 5-2, 1-(-1)> = <3, 3, 2>.
The vector from P to R is <2-0, 4-2, -6-(-1)> = <2, 2, -5>.
Step 4: Find the cross product of the two vectors:
The cross product of the vectors <3, 3, 2> and <2, 2, -5> is <19, -19, -6>.
Step 5: Write the equation of the plane:
The equation of the plane passing through the point P=(0, 2, -1) can be written as:
19(x - 0) - 19(y - 2) - 6(z - (-1)) = 0,
which simplifies to:
19x - 19y - 6z + 59 = 0.
So, the scalar equation for the plane passing through the point P=(0, 2, −1) and containing the line L defined by x = 2+t, y = 4+t, z = −6+7t is 19x - 19y - 6z + 59 = 0.
To find the scalar equation for the plane passing through the point P and containing the given line L, we can use the normal vector of the plane.
Step 1: Determine the normal vector
Since the plane contains the line L, then the normal vector of the plane should be perpendicular to the direction vector of the line. The direction vector of the line L is given by <1, 1, 7>. To find the normal vector, we can take the cross product of the direction vector of the line with a vector from any point on the line to the given point P.
Let's choose the point Q on the line L when t = 0. Plugging in t = 0 into the parametric equations of the line, we get Q = (2, 4, -6).
Now, we can find the normal vector N by taking the cross product of the direction vector of the line L and the vector PQ = P - Q.
PQ = P - Q = <0, 2, -1> - <2, 4, -6> = <-2, -2, 5>
N = <1, 1, 7> x <-2, -2, 5>
To compute the cross product, we can use the determinant method:
N = (1 * -2 - 1 * -2)i - (1 * -2 - 1 * 5)j + (1 * -2 - 2 * -2)k
= (-4)i + (3)j + (-2)k
= <-4, 3, -2>
Therefore, the normal vector of the plane is N = <-4, 3, -2>.
Step 2: Write the scalar equation
The scalar equation of a plane can be written as Ax + By + Cz + D = 0, where A, B, C, and D are constants.
To find these constants, we can substitute the coordinates of the point P into the equation and use the normal vector as coefficients:
A(0) + B(2) + C(-1) + D = 0
Simplifying, we find that B - C + D = 0.
Now, let's substitute the coordinates of the line into the equation:
A(2 + t) + B(4 + t) + C(-6 + 7t) + D = 0
Expanding and simplifying, we get:
(2A + 4B - 6C) + (At + Bt + 7Ct) + D = 0
Since this equation holds for any value of t, the coefficients of t must be zero:
A + B + 7C = 0.
Now we have two equations:
B - C + D = 0
A + B + 7C = 0
Solving this system of equations, we can express A, B, C, and D in terms of one of the variables. Let's solve for B in terms of C:
B = C - D
A + (C - D) + 7C = 0
A + 8C - D = 0
D = A + 8C
Substituting D in terms of A and C into the first equation:
(C - D) - C + (A + 8C) = 0
A + 8C - C + A + 8C = 0
2A + 15C = 0
2A = -15C
A = -15C/2
Substituting A in terms of C into the second equation:
(-15C/2) + B + 7C = 0
B = 15C/2 - 7C
B = C/2
Now we have the expressions:
A = -15C/2
B = C/2
D = A + 8C = -15C/2 + 8C = -7C/2
Therefore, the scalar equation of the plane passing through the point P and containing the line L is:
-15Cx/2 + Cy/2 - 7Cz/2 + C = 0
Since t is a variable, you can obtain three different points on the plane:
a) (0,2,-1) Given in the question
b) (2,4,-6) by setting t=0
c) (3,5,1) by setting t=1
If you know three points in a plane, you can use them to find a unique equation for the plane. Try it out