Find the equation of the planes that passes through three points.
(a) (1,1,-1),(6,4,-5),(-4,-2,-3)
(b) (1,1,0),(1,2,1),(-2,2,-1).
(a)The given points are A (1,1,-1), B (6,4,-5), and C (-4,-2,3).
\(\begin{vmatrix}1&1&-1\\6&4&-5\\-4&-2&3\end{vmatrix}\)=(12-10)-(18-20)-(-12+16)
=2+2-4 =0
Since, A, B, C are collinear points, there will be infinite number of planes passing through the given points.
(b)The given points are A (1,1,0), B (1,2,1), and C (-2,2,-1).
\(\begin{vmatrix}1&1&0\\1&2&1\\-2&-2&1\end{vmatrix}\)=(-2-2)-(2+2)
=-8≠0
Therefore, a plane will pass through the points A, B, and C.
It is known that the equation of the plane through the points, (x1,y1,z1), (x2,y2,z2), and (x3,y3,z3), is
\(\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}\)=0
\(\Rightarrow \begin{vmatrix}x-1&y-1&z\\0&1&1\\-3&1&-1\end{vmatrix}\)=0
\(\Rightarrow\) (-2)(x-1)-3(y-1)+3z=0
\(\Rightarrow\)-2x-3y+3z+2+3=0
\(\Rightarrow\)- 2x-3y+3z=-5
\(\Rightarrow\) 2x+3y-3z=5
This is the cartesian equation of the required plane.
List - I | List - II | ||
(P) | γ equals | (1) | \(-\hat{i}-\hat{j}+\hat{k}\) |
(Q) | A possible choice for \(\hat{n}\) is | (2) | \(\sqrt{\frac{3}{2}}\) |
(R) | \(\overrightarrow{OR_1}\) equals | (3) | 1 |
(S) | A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is | (4) | \(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\) |
(5) | \(\sqrt{\frac{2}{3}}\) |
Let \(\alpha x+\beta y+y z=1\) be the equation of a plane passing through the point\((3,-2,5)\)and perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,5)\) Then the value of \(\alpha \beta y\)is equal to ____
What is the Planning Process?
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely: