Answer :
Answer:
5x +9y +3z = 52
Step-by-step explanation:
You want the equation of the plane that passes through points 2i+4j+2k and 2i+3j+5k and is parallel to the vector 3i−2j+k.
Direction
The direction vector of the plane will be perpendicular to both the vector between the given points, and the given vector. It can be found as the cross product of those two vectors.
u = vector between points
u = (2i +3j +5k) -(2i +4j +2k) = (2-2)i +(3-4)j +(5-2)k = -j +3k
Then the normal to the plane is ...
n = (-j +3k) × (3i -2j +k) = (5i +9j +3k)
Equation
The constant in the equation of the plane is the dot product of one of the points in the plane and the normal vector:
n • p1 = (5i +9j +3k) • (2i +4j +2k) = 5·2 +9·4 +3·2 = 52
The equation of the plane is ...
5x +9y +3z = 52
