We have to prove that points (0, -1, -1), (4, 5, 1), (3, 9, 4) and (-4, 4, 4) are coplanar.
Now let us find the equation of plane passing through three point’s i.e.
(0, -1, -1), (4, 5, 1), (3, 9, 4)
By using the formula, equation of plane passing through three points is given as:
\( \begin{vmatrix} x & y+1 & z+1 \\[0.3em] 4& 6 & 2 \\[0.3em] 3 & 10 & 5 \end{vmatrix}\)
10x – 14y + 22z + 8 = 0
Divide by 2, we get
5x – 7y + 11z + 4 = 0 ……. (1)
By using the fourth point (-4, 4, 4),
Substitute the values as x = -4, y =4, z = 4 in equation (1), we get
5x – 7y + 11z + 4 = 0
5(-4) – 7(4) + 11(4) + 4 = 0
-48 + 48 = 0
0 = 0
LHS = RHS
Since, fourth point satisfies the equation of plane passing through three points. So, all the points are coplanar.
Hence, the equation of common plane is 5x – 7y + 11z + 4 = 0.
Answered by Sakshi | 1 year agoA plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (1, -2, 3). Find the equation of the plane.
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