Given:
Four points are:
(0, -1, 0), (2, 1, -1), (1, 1, 1) and (3, 3, 0)
Now let us find the equation of plane passing through three point’s i.e.
(0, -1, 0), (2, 1, -1), (1, 1, 1)
By using the formula, equation of plane passing through three points is given as:
\( \begin{vmatrix} x & y+1 & z \\[0.3em] 2& 2 & -1 \\[0.3em] 1 & 2& 1 \end{vmatrix}\)
4x – 3y – 3 + 2z = 0
4x – 3y + 2z – 3 = 0 ……. (1)
By using the fourth point (3, 3, 0),
Substitute the values as x = 3, y =3, z = 0 in equation (1), we get
4x – 3y – 3 + 2z = 0
= 12- 12 = 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 4x – 3y + 2z – 3 = 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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