Question

What is the equation of the plane through the three points \[\left( {1,1,1} \right),\left( {1, - 1,1} \right)\] and \[\left( { - 7, - 3, - 5} \right)\]?
A. \[3x - 4z + 1 = 0\]
B. \[3x - 4y + 1 = 0\]
C. \[3x + 4y + 1 = 0\]
D. None of these

Answer 1

Hint: Here, it is given that the plane is passing through three points. First, find the general equation of the plane passing through the point \[\left( {1,1,1} \right)\]. Also, find the equations of the plane on the basis of the other two points \[\left( {1, - 1,1} \right)\] and \[\left( { - 7, - 3, - 5} \right)\]. In the end, solve all the equations and find the required equation of the plane.

Formula used: The general equation of the plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with the direction ratio \[\left( {a,b,c} \right)\] is \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\].

Complete step by step solution: Given: The plane passes through the points \[\left( {1,1,1} \right),\left( {1, - 1,1} \right)\] and \[\left( { - 7, - 3, - 5} \right)\].

Let’s calculate the general equation of the plane passing through the point \[\left( {1,1,1} \right)\].
The general equation is:
\[a\left( {x - 1} \right) + b\left( {y - 1} \right) + c\left( {z - 1} \right) = 0\] \[.....\left( 1 \right)\]

The plane will pass through the points \[\left( {1, - 1,1} \right)\] and \[\left( { - 7, - 3, - 5} \right)\] if,
\[a\left( {1 - 1} \right) + b\left( { - 1 - 1} \right) + c\left( {1 - 1} \right) = 0\]
\[ \Rightarrow - 2b = 0\]
\[ \Rightarrow b = 0\] \[.....\left( 2 \right)\]
And
\[a\left( { - 7 - 1} \right) + b\left( { - 3 - 1} \right) + c\left( { - 5 - 1} \right) = 0\]
\[ \Rightarrow - 8a - 4b - 6c = 0\]
\[ \Rightarrow 4a + 2b + 3c = 0\] \[.....\left( 3 \right)\]

Now solve the equations \[\left( 2 \right)\] and \[\left( 3 \right)\].
\[4a + 2\left( 0 \right) + 3c = 0\]
\[ \Rightarrow 4a + 3c = 0\]
\[ \Rightarrow a = - \dfrac{3}{4}c\] \[.....\left( 4 \right)\]

Substitute the equations \[\left( 2 \right)\] and \[\left( 4 \right)\] in the equation \[\left( 1 \right)\].
\[ - \dfrac{3}{4}c\left( {x - 1} \right) + 0\left( {y - 1} \right) + c\left( {z - 1} \right) = 0\]
\[ \Rightarrow - \dfrac{3}{4}c\left( {x - 1} \right) + c\left( {z - 1} \right) = 0\]
\[ \Rightarrow 3\left( {x - 1} \right) - 4\left( {z - 1} \right) = 0\]
\[ \Rightarrow 3x - 3 - 4z + 4 = 0\]
\[ \Rightarrow 3x - 4z + 1 = 0\]
Thus, the equation of the plane passing through the points \[\left( {1,1,1} \right),\left( {1, - 1,1} \right)\] and \[\left( { - 7, - 3, - 5} \right)\] is \[3x - 4z + 1 = 0\].

Thus, Option (A) is correct.

Note: We can solve this problem in another way. To solve this question directly use the following formula.
The equation of the plane passing through the points \[\left( {{x_1},{y_1},{z_1}} \right)\], \[\left( {{x_2},{y_2},{z_2}} \right)\] and \[\left( {{x_3},{y_3},{z_3}} \right)\] is: \[\left| {\begin{array}{*{20}{c}}{x - {x_1}}&{y - {y_1}}&{z - {z_1}}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right| = 0\]
Substitute the values and solve the determinant to get the equation of the plane.