Four points (0, -1, 0), (2, 1, -1), (1, 1, 1) and (3, 3, 0) are coplanar. Find the equation of the plane containing them.

Asked by Aaryan | 1 year ago |  55

##### Solution :-

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 ago

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