Question

If the points \[\left( {1,1,k} \right)\] and \[\left( { - 3,0,1} \right)\] be equidistant from the plane \[3x + 4y - 12z + 13 = 0\], then \[k = \]
A) 0
B) 1
C) 2
D) None of these.

Answer 1

Hint: At first we find the distance from the given point and then they are equal, so write them by equal sign we get the value of \[k\].



Formula Used: The length of the perpendicular from a point P\[\left( {{x_1},{y_1},{z_1}} \right)\] to the plane \[ax + by + cz + d = 0\] is given by \[p = \dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\].



Complete step by step solution: Now the perpendicular distance of the plane \[3x + 4y - 12z + 13 = 0\] from the point \[\left( {1,1,k} \right)\] is
\[\dfrac{{\left| {3 \times 1 + 4 \times 1 - 12 \times k + 13} \right|}}{{\sqrt {{3^2} + {4^2} + {{\left( { - 12} \right)}^2}} }}\]
\[ = \dfrac{{\left| {3 + 4 - 12k + 13} \right|}}{{\sqrt {9 + 16 + 144} }}\]
\[ = \dfrac{{\left| {20 - 12k} \right|}}{{\sqrt {169} }}\]
\[ = \dfrac{{\left| {20 - 12k} \right|}}{{13}}\]unit.
Again the perpendicular distance of the plane \[3x + 4y - 12z + 13 = 0\] from the point \[\left( { - 3,0,1} \right)\] is \[\dfrac{{\left| {3 \times \left( { - 3} \right) + 4 \times 0 - 12 \times 1 + 13} \right|}}{{\sqrt {{3^2} + {4^2} + {{\left( { - 12} \right)}^2}} }}\]
\[ = \dfrac{{\left| { - 9 - 12 + 13} \right|}}{{\sqrt {9 + 16 + 144} }}\]
\[ = \dfrac{{\left| { - 8} \right|}}{{\sqrt {169} }}\]
\[ = \dfrac{8}{{13}}\]unit.
Now from the given condition,
\[\dfrac{{\left| {20 - 12k} \right|}}{{13}} = \dfrac{8}{{13}}\]
\[ \Rightarrow \left| {20 - 12k} \right| = 8\]
\[ \Rightarrow 20 - 12k = \pm 8\]
\[ \Rightarrow 20 - 12k = 8\] or, \[20 - 12k = - 8\]
\[ \Rightarrow 12k = 20 - 8 = 12\] 0r, \[12k = 20 + 8 = 28\]
\[ \Rightarrow k = \dfrac{{12}}{{12}} = 1\] or, \[k = \dfrac{{28}}{{12}} = \dfrac{7}{3}\]
\[ \Rightarrow k = 1\] or, \[k = \dfrac{7}{3}\]



Option ‘B’ is correct


Note: It is to be noted that the distance formula must be \[\dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\] and not \[\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|\]. Also when mod sign removed then the other sides of the equal sign must be give \[ \pm \] sign.