Question

If \[P\left( {2,6,3} \right)\] be the point, then what is the equation of the plane through \[P\] at right angle to\[OP\], \[O\] being the origin?
A. \[2x + 6y + 3z = 7\]
B. \[2x - 6y + 3z = 7\]
C. \[2x + 6y - 3z = 49\]
D. \[2x + 6y + 3z = 49\]

Answer 1

Hint: First, use the formula of the plane passing through a point and calculate the equation of the required plane. Then, calculate the direction ratios of the line joining the points \[P\left( {2,6,3} \right)\] and \[O\left( {0,0,0} \right)\]. After that, substitute the direction ratios in the equation of plane and solve it to get the required answer.

Formula used: The equation of plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\] is: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\] .
The direction ratios of the line joining the points \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\] is: \[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]


Complete step by step solution: Given:
The plane passes through the point \[P\left( {2,6,3} \right)\].
The plane is perpendicular to the line \[OP\], where \[O\] is an origin.

Let’s calculate the equation of the required plane.
Apply the formula of the equation of plane passing through the point.
We get,
The equation of the required plane with direction ratios \[\left( {a,b,c} \right)\] is:
\[a\left( {x - 2} \right) + b\left( {y - 6} \right) + c\left( {z - 3} \right) = 0\] \[.....\left( 1 \right)\]

The given line joins the two points \[P\left( {2,6,3} \right)\] and \[O\left( {0,0,0} \right)\].
Apply the formula of the direction ratios of a line.
We get,
Direction ratios: \[\left( {2 - 0,6 - 0,3 - 0} \right)\]
\[ \Rightarrow \] Direction ratios: \[\left( {2,6,3} \right)\]

Since the plane is perpendicular to the line \[OP\].
So, the direction ratios of the line and the plane are equal.
We get,
\[a = 2,b = 6,c = 3\]
Now substitute these values in the equation \[\left( 1 \right)\].
\[2\left( {x - 2} \right) + 6\left( {y - 6} \right) + 3\left( {z - 3} \right) = 0\]
\[ \Rightarrow 2x - 4 + 6y - 36 + 3z - 9 = 0\]
\[ \Rightarrow 2x + 6y + 3z - 49 = 0\]
\[ \Rightarrow 2x + 6y + 3z = 49\]
Thus, the equation of the plane is \[2x + 6y + 3z = 49\].

Thus, Option (D) is correct.

Note: Students often get confused between the direction ratios and direction cosines. Remember the following formulas:
For the line joining the points \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\]:
Direction ratios: \[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]
Direction cosines: \[\left( {\dfrac{{{x_2} - {x_1}}}{w},\dfrac{{{y_2} - {y_1}}}{w},\dfrac{{{z_2} - {z_1}}}{w}} \right)\], where \[w = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]