Write an implicit equation for the plane passing through the point (-5, -4, 5) that is perpendicular to the line r(t) = (2 – 4t, 2 + t, 4t – 4). Plane equation:
The Correct Answer and Explanation is:
To find the implicit equation of a plane passing through the point (−5,−4,5)(-5, -4, 5)(−5,−4,5) and perpendicular to the line given by the parametric equation r(t)=(2−4t,2+t,4t−4)\mathbf{r}(t) = (2 – 4t, 2 + t, 4t – 4)r(t)=(2−4t,2+t,4t−4), we can follow these steps:
Step 1: Direction of the Line
The direction vector of the line r(t)=(2−4t,2+t,4t−4)\mathbf{r}(t) = (2 – 4t, 2 + t, 4t – 4)r(t)=(2−4t,2+t,4t−4) can be extracted by taking the coefficients of ttt in each component of the parametric equation.
- The xxx-component is 2−4t2 – 4t2−4t, so the coefficient of ttt is −4-4−4.
- The yyy-component is 2+t2 + t2+t, so the coefficient of ttt is 111.
- The zzz-component is 4t−44t – 44t−4, so the coefficient of ttt is 444.
Thus, the direction vector of the line is d=(−4,1,4)\mathbf{d} = (-4, 1, 4)d=(−4,1,4).
Step 2: Normal Vector to the Plane
The plane is perpendicular to the line, which means the normal vector to the plane is parallel to the direction vector of the line. Therefore, the normal vector n\mathbf{n}n of the plane is the same as the direction vector d\mathbf{d}d, i.e., n=(−4,1,4)\mathbf{n} = (-4, 1, 4)n=(−4,1,4).
Step 3: Plane Equation
The general equation of a plane is given by:a(x−x0)+b(y−y0)+c(z−z0)=0a(x – x_0) + b(y – y_0) + c(z – z_0) = 0a(x−x0)+b(y−y0)+c(z−z0)=0
where (x0,y0,z0)(x_0, y_0, z_0)(x0,y0,z0) is a point on the plane and (a,b,c)(a, b, c)(a,b,c) is the normal vector to the plane.
- The normal vector is (−4,1,4)(-4, 1, 4)(−4,1,4), so a=−4a = -4a=−4, b=1b = 1b=1, and c=4c = 4c=4.
- The point (−5,−4,5)(-5, -4, 5)(−5,−4,5) lies on the plane, so x0=−5x_0 = -5×0=−5, y0=−4y_0 = -4y0=−4, and z0=5z_0 = 5z0=5.
Substitute these values into the equation of the plane:−4(x+5)+1(y+4)+4(z−5)=0-4(x + 5) + 1(y + 4) + 4(z – 5) = 0−4(x+5)+1(y+4)+4(z−5)=0
Step 4: Simplify
Expand the terms:−4x−20+y+4+4z−20=0-4x – 20 + y + 4 + 4z – 20 = 0−4x−20+y+4+4z−20=0
Combine like terms:−4x+y+4z−36=0-4x + y + 4z – 36 = 0−4x+y+4z−36=0
Thus, the implicit equation of the plane is:−4x+y+4z=36-4x + y + 4z = 36−4x+y+4z=36
Conclusion:
The implicit equation of the plane passing through the point (−5,−4,5)(-5, -4, 5)(−5,−4,5) and perpendicular to the line r(t)=(2−4t,2+t,4t−4)\mathbf{r}(t) = (2 – 4t, 2 + t, 4t – 4)r(t)=(2−4t,2+t,4t−4) is:−4x+y+4z=36-4x + y + 4z = 36−4x+y+4z=36
