Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1,4,2) and(2,3,5).Also find the perpendicular distance of the plane from the origin.
Answers
Answer:
4x+5y+7z-58=0
4x+5y+7z-58=0
The required plane is
Step-by-step explanation:
The normal vector to the plane 2x-3y+z-2=0 is
2\vec{i}-3\vec{j}+\vec{k}
2\vec{i}-3\vec{j}+\vec{k}
The normal vector to the plane 4x+y-3z+1=0 is
4\vec{i}+\vec{j}-3\vec{k}
4\vec{i}+\vec{j}-3\vec{k}
Hence the required plane passes through the point (2,3,.5) and parallel to the vectors
2\vec{i}-3\vec{j}+\vec{k}
2\vec{i}-3\vec{j}+\vec{k}
4\vec{i}+\vec{j}-3\vec{k}
4\vec{i}+\vec{j}-3\vec{k}
and
Equation of the required plane is
\begin{gathered}|\begin{array}{ccc}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{array}|=0\end{gathered}
\begin{gathered}|\begin{array}{ccc}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{array}|=0\end{gathered}
\begin{gathered}|\begin{array}{ccc}x-2&y-3&z-5\\2&-3&1\\4&1&-3\end{array}|=0\end{gathered}
\begin{gathered}|\begin{array}{ccc}x-2&y-3&z-5\\2&-3&1\\4&1&-3\end{array}|=0\end{gathered}