The equation of a plane passing through the line of intersection of the planes \[x + 2y + 3z = 2\] and \[x - y + z = 3\] and at a distance \[2\sqrt 3 \] from the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] is
A.\[5x - 11y + z = 17\]
B.\[\sqrt 2 x + y = 3\sqrt 2 - 1\]
C.\[x + y + z = \sqrt 3 \]
D.\[x - \sqrt 2 y = 1 - \sqrt 2 \]
Answer
279.9k+ views
Hint: In order to determine the required equation of a plane passing through the line intersection of the plane \[x + 2y + 3z = 2\] and \[x - y + z = 3\].First, we compare this equation with the two plane \[{{\rm P}_1} + \lambda {{\rm P}_2} = 0,\lambda \in \mathbb{R}\] then we get the equation as \[ax + by + cz + d = 0\] with the points
.The distance between the line is \[2\sqrt 3 \]. Adding the two planes does not yield their line of intersection. In fact, it is the equation of a plane passing through their line of intersection, rather than a line with the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] as \[(x,y,z)\]. We use the formula for the plane distance is \[D = \dfrac{{\left| {ax + by + cz + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. We need to solve the question to get the required solution.
Complete step-by-step answer:
In this given problem,
We are given the line of intersection of planes \[x + 2y + 3z = 2\]and \[x - y + z = 3\] at a distance \[2\sqrt 3 \] from the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\].
In this question we are supposed to find out the equation of a plane passing through the line of intersection of the planes
Let us consider the two plane equation as\[{P_1}:x + 2y + 3z = 2\]and \[{P_2}:x - y + z = 3\],
Let \[D\] be the distance,\[D = \dfrac{2}{{\sqrt 3 }}\]
For this we have to first determine the plane equation \[{{\rm P}_1} + \lambda {{\rm P}_2} = 0\] as \[(x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0\] to simplify the this equation as the form \[ax + by + cz + d = 0\] with the point \[(x,y,z)\] as \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\].
Now, we have to simplify the equation \[(x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0\]
\[
(x + 2y + 3z - 2 + \lambda x - \lambda y + \lambda z - 3\lambda ) = 0 \\
(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0 \to (3) \\
\]
Here, directly compare the equation\[(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0\] with\[ax + by + cz + d = 0\]. Where, \[a = (1 + \lambda ),b = (2 - \lambda ),c = (3 + \lambda ),d = (2 + 3\lambda )\]
The formula for the plane distance is \[D = \dfrac{{\left| {ax + by + cz + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
We have the values \[a = (1 + \lambda ),b = (2 - \lambda ),c = (3 + \lambda ),d = (2 + 3\lambda )\]and the given distance \[D = \dfrac{2}{{\sqrt 3 }}\] By substitute all the values and points \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] into the equation to find out the value\[\lambda \]
\[ \Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| {3(1 + \lambda ) + (2 - \lambda ) - (3 + \lambda ) - (2 + 3\lambda )} \right|}}{{\sqrt {{{(1 + \lambda )}^2} + {{(2 - \lambda )}^2} + {{(3 + \lambda )}^2}} }}\].
Comparing the denominator\[{(1 + \lambda )^2},({(2 - \lambda )^2},{(3 + \lambda )^2}\] with the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab,{(a - b)^2} = {a^2} + {b^2} - 2ab\] by expanding the bracket on RHS, we get
\[
\Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| {3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda } \right|}}{{\sqrt {1 + {\lambda ^2} + 2\lambda + 4 + {\lambda ^2} - 4\lambda + 9 + {\lambda ^2} + 6\lambda } }} \\
\Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| { - 2\lambda } \right|}}{{\sqrt {3{\lambda ^2} + 4\lambda + 14} }} \;
\]
Take square on both sides of numerator and denominator, we get
\[
\Rightarrow \dfrac{{{{(2)}^2}}}{{{{(\sqrt 3 )}^2}}} = \dfrac{{{{( - 2\lambda )}^2}}}{{{{(\sqrt {3{\lambda ^2} + 4\lambda + 14} )}^2}}} \\
\Rightarrow \dfrac{4}{3} = \dfrac{{4{\lambda ^2}}}{{(3{\lambda ^2} + 4\lambda + 14)}}\]
By simplify in further step to solving the fraction on both sides, we get
\[
4(3{\lambda ^2} + 4\lambda + 14) = 4(3){\lambda ^2} \\
3{\lambda ^2} + 4\lambda + 14 = 3{\lambda ^2} \;
\]
Combining all the terms together to find out the value,\[\lambda \], we get
\[
3{\lambda ^2} + 4\lambda + 14 - 3{\lambda ^2} = 0 \Rightarrow 4\lambda + 14 = 0 \\
4\lambda = - 14 \Rightarrow \lambda = - \dfrac{{14}}{4} \;
\]
Therefore, the value of \[\lambda = - \dfrac{7}{2}\]. Then
Let the equation \[(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0\] on comparing with \[ax + by + cz + d = 0\] and substitute the \[\lambda = - \dfrac{7}{2}\] into plane equation.
\[
\lambda = - \dfrac{7}{2} \Rightarrow \left( {1 - \dfrac{7}{2}} \right)x + \left( {2 + \dfrac{7}{2}} \right)y + \left( {3 - \dfrac{7}{2}} \right)z - \left( {2 + 3\left( {\dfrac{{ - 7}}{2}} \right)} \right) = 0 \;
\]
Take LCM on the above equation, so we get
\[
\left( {\dfrac{{2 - 7}}{2}} \right)x + \left( {\dfrac{{4 + 7}}{2}} \right)y + \left( {\dfrac{{6 - 7}}{2}} \right)z - \left( {\dfrac{{4 - 21}}{2}} \right) = 0 \\
- \dfrac{5}{2}x + \dfrac{{11}}{2}y - \dfrac{1}{2}z + \dfrac{{17}}{2} = 0 \;
\]
We perform multiplication on both sides by \[2\] to simplify the fraction, we can get
\[ - 5x + 11y - z + 17 = 0\]
Therefore, The equation of a plane passing through the line of intersection of the planes \[x + 2y + 3z = 2\] and \[x - y + z = 3\] and at a distance \[2\sqrt 3 \]from the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] is \[ - 5x + 11y - z + 17 = 0\]
So, the final answer is option A: \[5x - 11y + z = 17\]
So, the correct answer is “Option A”.
Note: In this problem, always try to understand the mathematical statement .when using the distance formula, use the notation for the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] as \[(x,y,z)\]. This will help to find out the final equation by substituting the values into the formula.
.The distance between the line is \[2\sqrt 3 \]. Adding the two planes does not yield their line of intersection. In fact, it is the equation of a plane passing through their line of intersection, rather than a line with the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] as \[(x,y,z)\]. We use the formula for the plane distance is \[D = \dfrac{{\left| {ax + by + cz + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]. We need to solve the question to get the required solution.
Complete step-by-step answer:
In this given problem,
We are given the line of intersection of planes \[x + 2y + 3z = 2\]and \[x - y + z = 3\] at a distance \[2\sqrt 3 \] from the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\].
In this question we are supposed to find out the equation of a plane passing through the line of intersection of the planes
Let us consider the two plane equation as\[{P_1}:x + 2y + 3z = 2\]and \[{P_2}:x - y + z = 3\],
Let \[D\] be the distance,\[D = \dfrac{2}{{\sqrt 3 }}\]
For this we have to first determine the plane equation \[{{\rm P}_1} + \lambda {{\rm P}_2} = 0\] as \[(x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0\] to simplify the this equation as the form \[ax + by + cz + d = 0\] with the point \[(x,y,z)\] as \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\].
Now, we have to simplify the equation \[(x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0\]
\[
(x + 2y + 3z - 2 + \lambda x - \lambda y + \lambda z - 3\lambda ) = 0 \\
(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0 \to (3) \\
\]
Here, directly compare the equation\[(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0\] with\[ax + by + cz + d = 0\]. Where, \[a = (1 + \lambda ),b = (2 - \lambda ),c = (3 + \lambda ),d = (2 + 3\lambda )\]
The formula for the plane distance is \[D = \dfrac{{\left| {ax + by + cz + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
We have the values \[a = (1 + \lambda ),b = (2 - \lambda ),c = (3 + \lambda ),d = (2 + 3\lambda )\]and the given distance \[D = \dfrac{2}{{\sqrt 3 }}\] By substitute all the values and points \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] into the equation to find out the value\[\lambda \]
\[ \Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| {3(1 + \lambda ) + (2 - \lambda ) - (3 + \lambda ) - (2 + 3\lambda )} \right|}}{{\sqrt {{{(1 + \lambda )}^2} + {{(2 - \lambda )}^2} + {{(3 + \lambda )}^2}} }}\].
Comparing the denominator\[{(1 + \lambda )^2},({(2 - \lambda )^2},{(3 + \lambda )^2}\] with the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab,{(a - b)^2} = {a^2} + {b^2} - 2ab\] by expanding the bracket on RHS, we get
\[
\Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| {3 + 3\lambda + 2 - \lambda - 3 - \lambda - 2 - 3\lambda } \right|}}{{\sqrt {1 + {\lambda ^2} + 2\lambda + 4 + {\lambda ^2} - 4\lambda + 9 + {\lambda ^2} + 6\lambda } }} \\
\Rightarrow \dfrac{2}{{\sqrt 3 }} = \dfrac{{\left| { - 2\lambda } \right|}}{{\sqrt {3{\lambda ^2} + 4\lambda + 14} }} \;
\]
Take square on both sides of numerator and denominator, we get
\[
\Rightarrow \dfrac{{{{(2)}^2}}}{{{{(\sqrt 3 )}^2}}} = \dfrac{{{{( - 2\lambda )}^2}}}{{{{(\sqrt {3{\lambda ^2} + 4\lambda + 14} )}^2}}} \\
\Rightarrow \dfrac{4}{3} = \dfrac{{4{\lambda ^2}}}{{(3{\lambda ^2} + 4\lambda + 14)}}\]
By simplify in further step to solving the fraction on both sides, we get
\[
4(3{\lambda ^2} + 4\lambda + 14) = 4(3){\lambda ^2} \\
3{\lambda ^2} + 4\lambda + 14 = 3{\lambda ^2} \;
\]
Combining all the terms together to find out the value,\[\lambda \], we get
\[
3{\lambda ^2} + 4\lambda + 14 - 3{\lambda ^2} = 0 \Rightarrow 4\lambda + 14 = 0 \\
4\lambda = - 14 \Rightarrow \lambda = - \dfrac{{14}}{4} \;
\]
Therefore, the value of \[\lambda = - \dfrac{7}{2}\]. Then
Let the equation \[(1 + \lambda )x + (2 - \lambda )y + (3 + \lambda )z - (2 + 3\lambda ) = 0\] on comparing with \[ax + by + cz + d = 0\] and substitute the \[\lambda = - \dfrac{7}{2}\] into plane equation.
\[
\lambda = - \dfrac{7}{2} \Rightarrow \left( {1 - \dfrac{7}{2}} \right)x + \left( {2 + \dfrac{7}{2}} \right)y + \left( {3 - \dfrac{7}{2}} \right)z - \left( {2 + 3\left( {\dfrac{{ - 7}}{2}} \right)} \right) = 0 \;
\]
Take LCM on the above equation, so we get
\[
\left( {\dfrac{{2 - 7}}{2}} \right)x + \left( {\dfrac{{4 + 7}}{2}} \right)y + \left( {\dfrac{{6 - 7}}{2}} \right)z - \left( {\dfrac{{4 - 21}}{2}} \right) = 0 \\
- \dfrac{5}{2}x + \dfrac{{11}}{2}y - \dfrac{1}{2}z + \dfrac{{17}}{2} = 0 \;
\]
We perform multiplication on both sides by \[2\] to simplify the fraction, we can get
\[ - 5x + 11y - z + 17 = 0\]
Therefore, The equation of a plane passing through the line of intersection of the planes \[x + 2y + 3z = 2\] and \[x - y + z = 3\] and at a distance \[2\sqrt 3 \]from the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] is \[ - 5x + 11y - z + 17 = 0\]
So, the final answer is option A: \[5x - 11y + z = 17\]
So, the correct answer is “Option A”.
Note: In this problem, always try to understand the mathematical statement .when using the distance formula, use the notation for the point \[\left( {3,{\text{ }}1,{\text{ }} - 1} \right)\] as \[(x,y,z)\]. This will help to find out the final equation by substituting the values into the formula.
Recently Updated Pages
Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How many meters are there in a kilometer And how many class 8 maths CBSE

What is pollution? How many types of pollution? Define it

Change the following sentences into negative and interrogative class 10 english CBSE

What were the major teachings of Baba Guru Nanak class 7 social science CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a labelled sketch of the human eye class 12 physics CBSE
