The plane containing the lines r1(t)=⟨1,−4,−1⟩+t⟨1,2,−2⟩ and r2(t)=⟨1,−4,−1⟩+t⟨1,0,3⟩ has scalar equation
To find the scalar equation of the plane containing the given lines, we need to find the normal vector to the plane. Since the two lines are in the plane, the direction vectors of the lines lie in the plane as well.
Let's find the normal vector by taking the cross product of the direction vectors of the lines, denoted by "v" and "w":
v = <1, 2, -2>
w = <1, 0, 3>
To find the cross product, we can use the following formula:
v x w = <(2 * 3) - (-2 * 0), (-2 * 1) - (3 * 1), (1 * 0) - (1 * 2)>
= <6, -5, -2>
This cross product vector, <6, -5, -2>, is the normal vector to the plane. Now we can use this normal vector to determine the scalar equation of the plane.
Let (x, y, z) be any point on the plane. Then the vector from this point to the point (1, -4, -1) on the plane, denoted as "u", should be orthogonal (perpendicular) to the normal vector we found earlier.
u = <x - 1, y - (-4), z - (-1)>
= <x - 1, y + 4, z + 1>
The dot product of "u" with the normal vector should be zero:
<6, -5, -2> · <x - 1, y + 4, z + 1> = 0
Expanding the dot product:
6(x - 1) - 5(y + 4) - 2(z + 1) = 0
Simplifying:
6x - 6 - 5y - 20 - 2z - 2 = 0
6x - 5y - 2z - 28 = 0
Therefore, the scalar equation of the plane containing the given lines is 6x - 5y - 2z - 28 = 0.
To find the scalar equation of the plane containing the given lines, we need to find two vectors that lie in the plane. These two vectors can be obtained by taking the direction vectors of the two lines.
The direction vector of the first line, r1(t), is ⟨1, 2, -2⟩.
The direction vector of the second line, r2(t), is ⟨1, 0, 3⟩.
Now, let's find the normal vector of the plane by taking the cross product of the two direction vectors.
Normal vector, n = ⟨1, 2, -2⟩ × ⟨1, 0, 3⟩
Using the cross product formula, we have:
n = ⟨(2 * 3) - (0 * -2), (-2 * 1) - (3 * 1), (1 * 0) - (1 * 2)⟩
= ⟨6 + 0, -2 - 3, 0 - 2⟩
= ⟨6, -5, -2⟩
Now, we have the normal vector of the plane, which is ⟨6, -5, -2⟩. To write the scalar equation of the plane, we equate it to the dot product of the normal vector and a point on the plane (we can choose any point that lies on the plane).
Let's take the point (1, -4, -1) on the plane.
The scalar equation of the plane is:
6(x - 1) - 5(y + 4) - 2(z + 1) = 0
Simplifying it further:
6x - 6 - 5y - 20 - 2z - 2 = 0
6x - 5y - 2z - 28 = 0
So, the scalar equation of the plane containing the given lines is 6x - 5y - 2z - 28 = 0.
the normal is r1×r2 = 6i-5j-2k
So, since the normal contains the point (1,-4,-1) the plane is
6x-5y-2z = 28