Does this line belong to this plane?

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To determine if the line defined by the equation [x,y,z]=[-3,-6,-11]+k[22,1,-11] belongs to the plane formed by points A(2,5,6), B(-7,1,4), and C(6,-2,-9), one must find the equation of the plane using the cross product of vectors AB and AC to establish a normal vector. The line does not need to be collinear with the vectors between the points to be in the plane; instead, it must satisfy the plane's equation. Calculating the plane's equation allows for checking if the line intersects or lies within the plane. The discussion emphasizes that collinearity is not a requirement for the line to belong to the plane. Ultimately, verifying the line against the plane's equation is essential for confirmation.
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i am given an equation of a line: [x,y,z]=[-3,-6,-11]+k[22,1,-11]
and i am being asked if it belongs to the plane defined by these three points A(2,5,6) B(-7,1,4) and C(6,-2,-9)
so first i calculated all three vectors between the points so i have AB=[-9,-4,2] AC=[4,-7,-15] BC=[13,-3,-13] and clearly none are collinear with [22,-1,11] so clearly the line is not part of the plane yet the book claims that it is part of the plane. Am i doing something wrong?
 
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pharoh123 said:
i am given an equation of a line: [x,y,z]=[-3,-6,-11]+k[22,1,-11]
and i am being asked if it belongs to the plane defined by these three points A(2,5,6) B(-7,1,4) and C(6,-2,-9)
so first i calculated all three vectors between the points so i have AB=[-9,-4,2] AC=[4,-7,-15] BC=[13,-3,-13] and clearly none are collinear with [22,-1,11] so clearly the line is not part of the plane yet the book claims that it is part of the plane. Am i doing something wrong?
Your given line can be in the plane without having to be collinear with any of the three lines you found. You can determine whether the given line is in the plane by finding the equation of the plane, and then determining whether the given line satisfies that plane equation.

Do you know how to find the equation of a plane given three points in the plane?
 
Mark44 said:
Your given line can be in the plane without having to be collinear with any of the three lines you found. You can determine whether the given line is in the plane by finding the equation of the plane, and then determining whether the given line satisfies that plane equation.

Do you know how to find the equation of a plane given three points in the plane?

yeah i could just put [x,y,z]=A + s*AB + t*AC where t and s are coefficients of the vectors AC and AB
but i don't know how that helps
 
No, I mean find the equation of the plane. The cross product of AB and AC will give you a normal to the plane, say <n1, n2, n3>. Then use any of the points on the plane and the normal to find the equation of the plane.

When you have the equation of the plane, determine whether your line satisfies the plane equation.
 
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