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**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.

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