1
Question
A plane passes through the points P (1,1,1) , Q (3, -1, 2) and R (-3, 5 , -4). The equation of the plane is
A plane passes through the points P (1,1,1) , Q (3, -1, 2) and R (-3, 5 , -4). The equation of the plane is
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Solution
The correct option is C 6x+6y+16z−28=0
∣∣
∣∣x−x1y−y1z−z1x2−x1y2−y1z2−z1x3−x1y3−y1z3−z1∣∣
∣∣=0
Where (x,y,z) are the coordinates of any general point in the plane. And (x1,y1,z1),(x2,y2,z2),(x3,y3,z3) are the given points. Basically it’s nothing but the locus of any point (x,y,z) such that the volume of the tetrahedron made with the help of three fixed points is zero.
We’ll do the appropriate substitution.
∣∣
∣∣x−1y−1z−13−1−1−12−1−3−15−1−4−1∣∣
∣∣=0
(x-1)(10 - 4) - (y -1)(-10 + 4) + (z -1)(8+8) = 0
6x + 6y +16z - 28 = 0
∣∣ ∣∣x−x1y−y1z−z1x2−x1y2−y1z2−z1x3−x1y3−y1z3−z1∣∣ ∣∣=0
Where (x,y,z) are the coordinates of any general point in the plane. And (x1,y1,z1),(x2,y2,z2),(x3,y3,z3) are the given points. Basically it’s nothing but the locus of any point (x,y,z) such that the volume of the tetrahedron made with the help of three fixed points is zero.
We’ll do the appropriate substitution.
∣∣ ∣∣x−1y−1z−13−1−1−12−1−3−15−1−4−1∣∣ ∣∣=0
(x-1)(10 - 4) - (y -1)(-10 + 4) + (z -1)(8+8) = 0
6x + 6y +16z - 28 = 0