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Find the Equation of the Plane Through the Line of Intersection of the Planes X + Y + Z = 1 and 2x + 3 + Y + 4 Z = 5 and Twice of Its Y -intercept is Equal to Three Times Its Z -intercept - Mathematics

Sum

Find the equation of the plane through the line of intersection of the planes \[x + y + z =\]1 and 2x \[+\] 3 \[+\] y \[+\] 4\[z =\] 5 and twice of its \[y\] -intercept is equal to three times its \[z\]-intercept

 
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Solution

The equation of the family of the planes passing through the intersection of the planes xy + z = 1 and 2x + 3y + 4z = 5 is

(x + y + z − 1) + k(2x + 3y + 4z − 5) = 0, where k is some constant

⇒ (2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1                .....(1)

\[\Rightarrow \frac{x}{\left( \frac{5k + 1}{2k + 1} \right)} + \frac{y}{\left( \frac{5k + 1}{3k + 1} \right)} + \frac{z}{\left( \frac{5k + 1}{4k + 1} \right)} = 1\]
\[\Rightarrow \frac{x}{\left( \frac{5k + 1}{2k + 1} \right)} + \frac{y}{\left( \frac{5k + 1}{3k + 1} \right)} + \frac{z}{\left( \frac{5k + 1}{4k + 1} \right)} = 1\]
It is given that twice of y-intercept is equal to three times its z-intercept.
\[\therefore 2\left( \frac{5k + 1}{3k + 1} \right) = 3\left( \frac{5k + 1}{4k + 1} \right)\]
\[ \Rightarrow \left( 5k + 1 \right)\left( 8k + 2 - 9k - 3 \right) = 0\]
\[ \Rightarrow \left( 5k + 1 \right)\left( - k - 1 \right) = 0\]
\[ \Rightarrow \left( 5k + 1 \right)\left( k + 1 \right) = 0\]
\[\Rightarrow 5k + 1 = 0 or k + 1 = 0\]
\[ \Rightarrow k = - \frac{1}{5} or k = - 1\]
Putting 
\[k = - \frac{1}{5}\]  in (1), we get
\[\left( - \frac{2}{5} + 1 \right)x + \left( - \frac{3}{5} + 1 \right)y + \left( - \frac{4}{5} + 1 \right)z = 5 \times \left( - \frac{1}{5} \right) + 1\]
\[ \Rightarrow 3x + 2y + z = 0\]

This plane passes through the origin. So, the intercepts made by the plane with the coordinate axes is 0. Hence, this equation of plane is not accepted as twice of y-intercept is not equal to three times its z-intercept.

Putting \[k = - 1\]  in (1), we get
\[\left( - 2 + 1 \right)x + \left( - 3 + 1 \right)y + \left( - 4 + 1 \right)z = 5 \times \left( - 1 \right) + 1\]
\[ \Rightarrow - x - 2y - 3z = - 4\]
\[ \Rightarrow x + 2y + 3z = 4\]
Here, twice of y-intercept is equal to three times its z-intercept.

Thus, the equation of the required plane is x + 2y + 3z = 4.
 
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 29 The Plane
Exercise 29.8 | Q 19 | Page 40
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