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 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.
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) .Find the equation of the plane
Reduce the equations of the planes in the intercept form and find its intercepts on the coordinate axes: 2x – y + z = 5
Reduce the equations of the following planes in the intercept form and find its intercepts on the coordinate axes: 2x + 3y – z = 6