- #1
ineedhelpnow
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find the equation of the plane through the origin and perpendicular to the vector (1,-2,5). this is the only relevant equation i have found $a(x-x_0)+b(y-y_0)+c(x-x_0)=0$
The equation of the plane can be written as x - 2y + 5z = 0. This is because the normal vector of the plane is (1,-2,5), and any point on the plane must satisfy the equation a(x - x0) + b(y - y0) + c(z - z0) = 0, where (x0, y0, z0) is a point on the plane and (a,b,c) is the normal vector.
Since the equation of the plane is x - 2y + 5z = 0, it can be seen that when all variables are equal to 0, the equation is satisfied. Therefore, the point (0,0,0), which is the origin, is a point on the plane, indicating that the plane passes through the origin.
Yes, the equation of the plane can be graphed in three dimensions. The plane will appear as a flat surface passing through the origin, perpendicular to the vector (1,-2,5).
The perpendicularity of the plane and the vector is determined by the dot product of the normal vector of the plane and the given vector. If the dot product is equal to 0, then the two vectors are perpendicular.
Yes, there is a unique equation for a plane passing through the origin and perpendicular to a given vector. This is because the normal vector of a plane is unique, and any point on the plane can be used to determine the equation using the formula mentioned in the answer to the first question.