
Find the Cartesian equation of the plane, passing through the line of intersection of the planes: \[r.(2\hat i + 3\hat j - 4\hat k) + 5 = 0\] and \[r.(\hat i - 5\hat j + 7\hat k) + 2 = 0\] and intersecting y-axis at \[\left( {0,3} \right)\]
Answer
524.1k+ views
Hint: In this problem we will approach with a direct vector approach. We will first find the equation of the plane of the intersection of two planes with a lambda form. Then as it intersects y axis at some point, we will find another condition to proceed and find the value of lambda.
Complete step-by-step answer:
We have two planes as, \[r.(2\hat i + 3\hat j - 4\hat k) + 5 = 0\] and \[r.(\hat i - 5\hat j + 7\hat k) + 2 = 0\] .
As, any plane passing through the intersection of two planes \[r.(a\hat i + b\hat j + c\hat k) + d = 0\] and \[r.(e\hat i + f\hat j + g\hat k) + h = 0\] given as, \[[r.(a\hat i + b\hat j + c\hat k) + d] + \lambda [r.(e\hat i + f\hat j + g\hat k) + h] = 0\] .
Now, Any plane passing through the intersection of the planes \[r.(2\hat i + 3\hat j - 4\hat k) + 5 = 0\] and \[r.(\hat i - 5\hat j + 7\hat k) + 2 = 0\] is,
\[[r.(2\hat i + 3\hat j - 4\hat k) + 5] + \lambda [r.(\hat i - 5\hat j + 7\hat k) + 2] = 0\]
Now, on simplifying, we get,
\[r[(2 + \lambda )\hat i + (3 - 5\lambda )\hat j + ( - 4 + 7\lambda )\hat k] = - 5 - 2\lambda \] ……(i)
Again, If equation (i) intersects the y-axis at \[\left( {0,3} \right)\] , then the point \[\left( {0,3} \right)\] will satisfy the equation. In other words the point \[\left( {0,3} \right)\] will be on that line.
So, now the point, \[r = 0\hat i + 0\hat j + 3\hat k\] ,
As it is on the line, the dot product with the given intersection of planes will be zero.
So, we get, \[(0\hat i + 0\hat j + 3\hat k).[(2 + \lambda )\hat i + (3 - 5\lambda )\hat j + ( - 4 + 7\lambda )\hat k] = 0\]
\[ \Rightarrow 0(2 + \lambda ) + 0(3 - \lambda ) + 3( - 4 + 7\lambda ) = 0\]
\[ \Rightarrow - 12 + 21\lambda = 0\]
\[ \Rightarrow 21\lambda = 12\]
\[ \Rightarrow \lambda = \dfrac{{12}}{{21}} = \dfrac{4}{7}\]
∴ The required plane is obtained after substituting the value of \[\lambda \]
\[[(2 + \dfrac{4}{7})\hat i + (3 - 5.\dfrac{4}{7})\hat j + ( - 4 + 7.\dfrac{4}{7})\hat k] = - 5 - 2.\dfrac{4}{7}\]
\[ \Rightarrow (\dfrac{{18}}{7})\hat i + (\dfrac{{21 - 20}}{7})\hat j + ( - 4 + 4)\hat k] = \dfrac{{ - 35 - 8}}{7}\]
\[ \Rightarrow (\dfrac{{18}}{7})\hat i + (\dfrac{1}{7})\hat j = \dfrac{{ - 43}}{7}\]
Multiplying both sides by 7, we get,
\[ \Rightarrow 18\hat i + \hat j = - 43\]
So, we have, the equation of the plane, as,
\[18x + y = - 43\] or \[18x + y + 43 = 0\;\] which is the required equation.
Note: Planes are either parallel, or they're perpendicular, otherwise they intersect each other at some other angle. Planes are said to be parallel if the ratio equality is true, whereas they are perpendicular if the dot product of their normal vectors is 0.
Complete step-by-step answer:
We have two planes as, \[r.(2\hat i + 3\hat j - 4\hat k) + 5 = 0\] and \[r.(\hat i - 5\hat j + 7\hat k) + 2 = 0\] .
As, any plane passing through the intersection of two planes \[r.(a\hat i + b\hat j + c\hat k) + d = 0\] and \[r.(e\hat i + f\hat j + g\hat k) + h = 0\] given as, \[[r.(a\hat i + b\hat j + c\hat k) + d] + \lambda [r.(e\hat i + f\hat j + g\hat k) + h] = 0\] .
Now, Any plane passing through the intersection of the planes \[r.(2\hat i + 3\hat j - 4\hat k) + 5 = 0\] and \[r.(\hat i - 5\hat j + 7\hat k) + 2 = 0\] is,
\[[r.(2\hat i + 3\hat j - 4\hat k) + 5] + \lambda [r.(\hat i - 5\hat j + 7\hat k) + 2] = 0\]
Now, on simplifying, we get,
\[r[(2 + \lambda )\hat i + (3 - 5\lambda )\hat j + ( - 4 + 7\lambda )\hat k] = - 5 - 2\lambda \] ……(i)
Again, If equation (i) intersects the y-axis at \[\left( {0,3} \right)\] , then the point \[\left( {0,3} \right)\] will satisfy the equation. In other words the point \[\left( {0,3} \right)\] will be on that line.
So, now the point, \[r = 0\hat i + 0\hat j + 3\hat k\] ,
As it is on the line, the dot product with the given intersection of planes will be zero.
So, we get, \[(0\hat i + 0\hat j + 3\hat k).[(2 + \lambda )\hat i + (3 - 5\lambda )\hat j + ( - 4 + 7\lambda )\hat k] = 0\]
\[ \Rightarrow 0(2 + \lambda ) + 0(3 - \lambda ) + 3( - 4 + 7\lambda ) = 0\]
\[ \Rightarrow - 12 + 21\lambda = 0\]
\[ \Rightarrow 21\lambda = 12\]
\[ \Rightarrow \lambda = \dfrac{{12}}{{21}} = \dfrac{4}{7}\]
∴ The required plane is obtained after substituting the value of \[\lambda \]
\[[(2 + \dfrac{4}{7})\hat i + (3 - 5.\dfrac{4}{7})\hat j + ( - 4 + 7.\dfrac{4}{7})\hat k] = - 5 - 2.\dfrac{4}{7}\]
\[ \Rightarrow (\dfrac{{18}}{7})\hat i + (\dfrac{{21 - 20}}{7})\hat j + ( - 4 + 4)\hat k] = \dfrac{{ - 35 - 8}}{7}\]
\[ \Rightarrow (\dfrac{{18}}{7})\hat i + (\dfrac{1}{7})\hat j = \dfrac{{ - 43}}{7}\]
Multiplying both sides by 7, we get,
\[ \Rightarrow 18\hat i + \hat j = - 43\]
So, we have, the equation of the plane, as,
\[18x + y = - 43\] or \[18x + y + 43 = 0\;\] which is the required equation.
Note: Planes are either parallel, or they're perpendicular, otherwise they intersect each other at some other angle. Planes are said to be parallel if the ratio equality is true, whereas they are perpendicular if the dot product of their normal vectors is 0.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
