Find the equation of the plane passing through the point (2, 3, 4), (-3, 5, 1) and (4, -1, 2)

Asked by Aaryan | 1 year ago |  70

##### Solution :-

Given:

The three points are:

(2, 3, 4), (-3, 5, 1) and (4, -1, 2)

By using the formula, equation of plane passing through three points is given as:

$$\begin{vmatrix} x-2 & y-3 & z-4 \\[0.3em] -5& 2 & -3 \\[0.3em] 2 & -4 & -2 \end{vmatrix}$$

-16x – 16y + 16z + 16 = 0

Divide by -16, we get

Hence, the equation of plane is x + y – z – 1 = 0.

Answered by Sakshi | 1 year ago

### Related Questions

#### A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (1, -2, 3).

A 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.

#### Write the equation of the plane whose intercepts on the coordinate axes are x/a+yb+z/c = 1

Write the equation of the plane whose intercepts on the coordinate axes are $$\dfrac{x}{a}+ \dfrac{y}{b}+ \dfrac{z}{c}=1$$

#### A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (α,β,γ) .

A plane meets the coordinate axes in points A, B, C and the centroid of the triangle ABC is (α,β,γ) .Find the equation of the plane