Help me find my error!
a)Write down the equation of the plane passing through the point (1,5,7) and parallel to the lines.
Line A: Line B: Point
x = 3- 2t x = 2 - 2t <1,5,7>
y = 1 - 5t y = 3 - 2t
z = 2 - t z = 1 + 4t
Parametric equation of the plane
<x,y,z> = <1,5,7> + t1 <-2,-5,-1> + t2 <-2,-2,4>
Equation of the plane
<-2,-5,-1> x <-2,-2,4> = -18i - 6j - 6k
-18(x-1) -6 (y-5) -6 (z-7)
-18x +18 -6y + 30 - 6z + 42
-18x - 6y - 6z + 90 = 0 <---My answer
-11x + 5y - 3z +7 = 0 <--- The textbook solution
Where have I gone wrong?
To find the equation of the plane passing through the point (1,5,7) and parallel to the given lines, we need to find the normal vector of the plane. The normal vector can be obtained by taking the cross product of the direction vectors of the two given lines, Line A and Line B.
Let's calculate the cross product of the direction vectors <-2,-5,-1> and <-2,-2,4>:
(-5 * 4 - (-1) * (-2))i - ((-2) * 4 - (-1) * (-2))j + ((-2) * (-2) - (-5) * (-2))k
(-20 + 2)i - (-8 - 2)j - (4 + 10)k
-18i + 10j - 14k
So, the normal vector of the plane is <-18, 10, -14>.
Now, let's find the equation of the plane using the point-normal form of the equation of a plane: Ax + By + Cz + D = 0.
Substituting the values into the equation:
-18(x-1) + 10(y-5) - 14(z-7) + D = 0
Expanding and simplifying the equation:
-18x + 18 + 10y - 50 - 14z + 98 + D = 0
-18x + 10y - 14z + 66 + D = 0
Comparing this equation with the textbook solution (-11x + 5y - 3z + 7 = 0), we see that the coefficients of x, y, and z are different. So, your answer (-18x - 6y - 6z + 90 = 0) is incorrect.
It seems that the calculation of the normal vector may have been incorrect. Please double-check your cross product calculation to obtain the correct normal vector, and then use it to find the correct equation of the plane.