Question

If the distance of the point \[\left( {1,1,1} \right)\] from the origin is half its distance from the plane \[x + y + z + k = 0\], then what is the value of \[k\]?
A. \[ \pm 3\]
B. \[ \pm 6\]
C. \[ - 3,9\]
D. \[3, - 9\]

Answer 1

Hint: First, calculate the distance between the point \[\left( {1,1,1} \right)\] and the origin. Then, calculate the distance between the point \[\left( {1,1,1} \right)\] and the plane \[x + y + z + k = 0\]. After that, apply the given condition and solve it to get the required answer.

Formula used: The distance between the origin and a point \[\left( {{x_1},{y_1},{z_1}} \right)\] is: \[d = \sqrt {{x_1}^2 + {y_1}^2 + {z_1}^2} \]
The smallest distance between a point \[\left( {{x_1},{y_1},{z_1}} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]

Complete step by step solution: Given:
The distance of the point \[\left( {1,1,1} \right)\] from the origin is half its distance from the plane \[x + y + z + k = 0\].

Let’s calculate the distance between the origin and the point \[\left( {1,1,1} \right)\].
Apply the formula of the distance between the point and the origin \[d = \sqrt {{x_1}^2 + {y_1}^2 + {z_1}^2} \].
We get,
\[d = \sqrt {{1^2} + {1^2} + {1^2}} \]
\[ \Rightarrow d = \sqrt {1 + 1 + 1} \]
\[ \Rightarrow d = \sqrt 3 \] \[.....\left( 1 \right)\]
Now calculate the distance between the plane \[x + y + z + k = 0\] and the point \[\left( {1,1,1} \right)\].
So, use the formula of the smallest distance between a plane and a point \[D = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
Here, \[\left( {{x_1},{y_1},{z_1}} \right) = \left( {1,1,1} \right)\] and the equation of plane is \[x + y + z + k = 0\].
Substitute the values in the formula. We get
\[D = \left| {\dfrac{{\left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + k}}{{\sqrt {{1^2} + {1^2} + {1^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{1 + 1 + 1 + k}}{{\sqrt 3 }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{k + 3}}{{\sqrt 3 }}} \right|\]
\[ \Rightarrow D = \pm \dfrac{{k + 3}}{{\sqrt 3 }}\] \[.....\left( 2 \right)\]

It is given that the distance of the point \[\left( {1,1,1} \right)\] from the origin is half its distance from the plane \[x + y + z + k = 0\].
From the equations \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
\[d = \dfrac{1}{2}D\]
\[ \Rightarrow \sqrt 3 = \dfrac{1}{2}\left( { \pm \dfrac{{k + 3}}{{\sqrt 3 }}} \right)\]
\[ \Rightarrow 6 = \pm \left( {k + 3} \right)\]
\[ \Rightarrow 6 = k + 3\] or \[6 = - k - 3\]
\[ \Rightarrow k = 3\] or \[k = - 9\]
Thus, the values of \[k\] are \[3\] and \[ - 9\].

Thus, Option (D) is correct.

Note: Sometimes students use the formula of the smallest distance between a plane and a point as \[D = \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. In this type of question, they will get only one value. So, always remember to apply the modulus function.