Question

If the image of the point \[A(1,2, - 3)\] relative to the plane \[\pi \] is \[B( - 3,6,4)\] then equation of plane \[ax + by + cz = k\] is

(A) \[8x + 8y + 14z - 47 = 0\]

(B) \[8x - 8y - 14z + 47 = 0\]

(C) \[8x - 8y - 14z - 47 = 0\]

(D) \[8x + 8y - 14z + 47 = 0\]

Answer 1

Hint: We use the formula to find the mid-point of points A and B with the help of their coordinates and since the point lies on the plane it will satisfy the equation of a plane. Using the formula for slope using two endpoints we find the slope of the line AB. Substituting the point obtained from the slope in the equation of the plane we find the value of k, which then gives us the final value of the plane.

Formula used:

* The formula for midpoint of the two-point \[A({x_1},{y_1},{z_1})\]and \[B({x_2},{y_2},{z_2})\] is

\[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]

* The formula for slope of a line having two points \[A({x_1},{y_1},{z_1})\]and \[B({x_2},{y_2},{z_2})\] is

\[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]

Complete Step-by-step Solution

Let us assume the equation of plane is \[ax + by + cz = k\] … (1)

Now find the mid-point of the two points \[A(1,2, - 3)\] and \[B( - 3,6,4)\]

We know mid-point of a line joining two points \[A({x_1},{y_1},{z_1})\] and \[B({x_2},{y_2},{z_2})\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]

Substitute\[{x_1} = 1,{x_2} = - 3,{y_1} = 2,{y_2} = 6,{z_1} = - 3,{z_2} = 4\] in the formula for mid-point.

\[ \Rightarrow \left( {\dfrac{{1 + ( - 3)}}{2},\dfrac{{2 + 6}}{2},\dfrac{{ - 3 + 4}}{2}} \right)\]

\[ \Rightarrow \left( { - 1,4,\dfrac{1}{2}} \right)\]

Now since the point lies on the plane, it will satisfy equation of the plane. Substituting the value \[x = - 1,y = 4,z = \dfrac{1}{2}\] in the equation of the plane, we get

\[ \Rightarrow a( - 1) + b(4) + c\left( {\dfrac{1}{2}} \right) = k\]

\[ \Rightarrow - a + 4b + \dfrac{c}{2} = k\] … (2)

We know that slope of a line joining two points \[A({x_1},{y_1},{z_1})\] and \[B({x_2},{y_2},{z_2})\] is given by

\[\left( {{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right)\]

Substitute the value of \[{x_1} = 1,{x_2} = - 3,{y_1} = 2,{y_2} = 6,{z_1} = - 3,{z_2} = 4\] in the formula

\[ \Rightarrow ( - 3 - 1,6 - 2,4 - ( - 3))\]

\[ \Rightarrow ( - 4,4,7)\]

Now find the k value by putting the point of slope in equation of plane from equation (2).

\[ \Rightarrow - 1( - 4) + 4(4) + \dfrac{1}{2}(7) = k\]

\[ \Rightarrow k = 4 + 16 + \dfrac{7}{2}\]

Take LCM on RHS of the equation

\[\Rightarrow k = \dfrac{{8 + 32 + 7}}{2} \]

\[\Rightarrow k = \dfrac{{47}}{2} \]

Now put the value of k and slope value in equation (1) so we get the equation of the plane.

\[ \Rightarrow - 4x + 4y + 7z = \dfrac{{47}}{2}\]

Multiply both sides of the equation by 2

\[ \Rightarrow 2 \times ( - 4x + 4y + 7z) = 2 \times \dfrac{{47}}{2}\]

Cancel the same terms from RHS of the equation.

\[ \Rightarrow - 8x + 8y + 14z = 47\]

Shift all values to one side of the equation

\[ \Rightarrow - 8x + 8y + 14z - 47 = 0\]

So, this is not available in option by taking common from -1 we get the same equation that in option shown so the plane equation is:

\[8x - 8y - 14z + 47 = 0\]

$\therefore $ The plane equation is \[8x - 8y - 14z + 47 = 0\]. Thus, option (B) is correct.

Note:

Students are likely to make mistake in the calculation part where you have to change the sign from positive to negative and vice-versa when shifting the values from one side of the equation to another side. After getting an equation if the answer is not available in the option then convert that equation by multiplying some constant on both sides or take a constant common to form one of the options.