Question

Find the equation of the plane passing through the following points: (2,3,4), (−3,5,1) and (4,−1,2).

Answer 1

Hint: In this question, we use the concept of the equation of Plane. Equation of plane passing through three non collinear points $\left( {{x_1},{y_1},{z_1}} \right),\left( {{x_2},{y_2},{z_2}} \right),\left( {{x_3},{y_3},{z_3}} \right)$ is

{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{array}} \right| = 0$$ Complete step-by-step answer: Given, we have three points (2,3,4), (−3,5,1) and (4,−1,2). When we have three non collinear points so the equation plane passing through three non collinear points is $$\left| {\begin{array}{*{20}{c}} {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{array}} \right| = 0$$ Now, $\left( {{x_1},{y_1},{z_1}} \right) = \left( {2,3,4} \right)$ , $\left( {{x_2},{y_2},{z_2}} \right) = \left( { - 3,5,1} \right)$ and $\left( {{x_3},{y_3},{z_3}} \right) = \left( {4, - 1,2} \right)$ .

\Rightarrow \left| {\begin{array}{{20}{c}}
{x - 2}&{y - 3}&{z - 4} \\
{ - 3 - 2}&{5 - 3}&{1 - 4} \\
{4 - 2}&{ - 1 - 3}&{2 - 4}
\end{array}} \right| = 0 \\
\Rightarrow \left| {\begin{array}{{20}{c}}
{x - 2}&{y - 3}&{z - 4} \\
{ - 5}&2&{ - 3} \\
2&{ - 4}&{ - 2}
\end{array}} \right| = 0 \\

Now solve the determinant, $ \Rightarrow \left( {x - 2} \right)\left( { - 4 - 12} \right) - \left( {y - 3} \right)\left( {10 + 6} \right) + \left( {z - 4} \right)\left( {20 - 4} \right) = 0 \\\ \Rightarrow \left( {x - 2} \right)\left( { - 16} \right) - \left( {y - 3} \right)\left( {16} \right) + \left( {z - 4} \right)\left( {16} \right) = 0 \\\ $ Divide by 16 in above equation, $ \Rightarrow \left( {x - 2} \right)\left( { - 1} \right) - \left( {y - 3} \right) + \left( {z - 4} \right) = 0 \\\ \Rightarrow - x + 2 - y + 3 + z - 4 = 0 \\\ \Rightarrow - x - y + z + 1 = 0 \\\ \Rightarrow x + y - z - 1 = 0 \\\ \Rightarrow x + y - z = 1 \\\ $ So, the equation of the plane passing through the points (2,3,4), (−3,5,1) and (4,−1,2) is $x + y - z = 1$. Note: Whenever we face such types of problems we can use two methods, one method we already mention above and in second method we can find vector equation of plane passing through three points with position vector $\mathop a\limits^ \to ,\mathop b\limits^ \to ,\mathop c\limits^ \to $ is $\left( {\mathop r\limits^ \to - \mathop a\limits^ \to } \right).\left[ {\left( {\mathop b\limits^ \to - \mathop a\limits^ \to } \right) \times \left( {\mathop c\limits^ \to - \mathop a\limits^ \to } \right)} \right] = 0$ then put $\mathop r\limits^ \to = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge $ . So, we will get the required answer.