
If the product of distances of the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] from the origin and the plane \[x - y + z + k = 0\] be 5. Then what is the value of \[k\]?
A. \[ - 2\]
B. \[ - 3\]
C. \[4\]
D. \[7\]
Answer
160.8k+ views
Hint: Here, an equation of plane and a point is given. First, calculate the distance between the given point and the origin. Then, by using the formula of the distance between the point and a plane calculate the distance between the given point and plane. After that, solve the given condition of the product of the both distances and get the required answer.
Formula Used:The smallest distance between a point \[\left( {p,q,r} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]
The distance between any point \[\left( {x,y,z} \right)\] and an origin is: \[\sqrt {{x^2} + {y^2} + {z^2}} \]
Complete step by step solution:Given:
The product of distances of the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] from the origin and the plane \[x - y + z + k = 0\] is 5.
We know the formula of the distance between an origin and a point \[\left( {x,y,z} \right)\] is \[\sqrt {{x^2} + {y^2} + {z^2}} \].
Apply this formula to calculate the distance.
So, the distance between the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] and the origin is:
\[{D_1} = \sqrt {{1^2} + {1^2} + {1^2}} \]
\[ \Rightarrow {D_1} = \sqrt {1 + 1 + 1} \]
\[ \Rightarrow {D_1} = \sqrt 3 \] \[.....\left( 1 \right)\]
Now we have to calculate the distance between the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] and the plane \[x - y + z + k = 0\].
So, use the formula of the smallest distance between a plane and a point.
Substitute the values in the formula \[D = \left| {\dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
We get,
\[D = \left| {\dfrac{{\left( 1 \right)\left( 1 \right) + \left( { - 1} \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + k}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {1^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{1 + - 1 + 1 + k}}{{\sqrt {1 + 1 + 1} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{1 + k}}{{\sqrt 3 }}} \right|\] \[.....\left( 2 \right)\]
It is given that, the product of distances of the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] from the origin and the plane \[x - y + z + k = 0\] is 5.
So, multiply the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[\left| {\dfrac{{1 + k}}{{\sqrt 3 }}} \right| \times \sqrt 3 = 5\]
\[ \Rightarrow \left| {1 + k} \right| = 5\]
\[ \Rightarrow 1 + k = 5\] or \[1 + k = - 5\]
\[ \Rightarrow k = 4\] or \[k = - 6\]
Option ‘C’ is correct
Note: Sometimes students use the formula of the smallest distance between a plane and a point as \[D = \dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. In this type of questions, they will get only one value. So, always remember to apply the modulus function.
Formula Used:The smallest distance between a point \[\left( {p,q,r} \right)\] and a plane \[ax + by + cz = d\] is: \[D = \left| {\dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\]
The distance between any point \[\left( {x,y,z} \right)\] and an origin is: \[\sqrt {{x^2} + {y^2} + {z^2}} \]
Complete step by step solution:Given:
The product of distances of the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] from the origin and the plane \[x - y + z + k = 0\] is 5.
We know the formula of the distance between an origin and a point \[\left( {x,y,z} \right)\] is \[\sqrt {{x^2} + {y^2} + {z^2}} \].
Apply this formula to calculate the distance.
So, the distance between the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] and the origin is:
\[{D_1} = \sqrt {{1^2} + {1^2} + {1^2}} \]
\[ \Rightarrow {D_1} = \sqrt {1 + 1 + 1} \]
\[ \Rightarrow {D_1} = \sqrt 3 \] \[.....\left( 1 \right)\]
Now we have to calculate the distance between the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] and the plane \[x - y + z + k = 0\].
So, use the formula of the smallest distance between a plane and a point.
Substitute the values in the formula \[D = \left| {\dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\].
We get,
\[D = \left| {\dfrac{{\left( 1 \right)\left( 1 \right) + \left( { - 1} \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + k}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2} + {1^2}} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{1 + - 1 + 1 + k}}{{\sqrt {1 + 1 + 1} }}} \right|\]
\[ \Rightarrow D = \left| {\dfrac{{1 + k}}{{\sqrt 3 }}} \right|\] \[.....\left( 2 \right)\]
It is given that, the product of distances of the point \[\left( {1,{\rm{ }}1,{\rm{ }}1} \right)\] from the origin and the plane \[x - y + z + k = 0\] is 5.
So, multiply the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[\left| {\dfrac{{1 + k}}{{\sqrt 3 }}} \right| \times \sqrt 3 = 5\]
\[ \Rightarrow \left| {1 + k} \right| = 5\]
\[ \Rightarrow 1 + k = 5\] or \[1 + k = - 5\]
\[ \Rightarrow k = 4\] or \[k = - 6\]
Option ‘C’ is correct
Note: Sometimes students use the formula of the smallest distance between a plane and a point as \[D = \dfrac{{ap + bq + cr + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. In this type of questions, they will get only one value. So, always remember to apply the modulus function.
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