Question

Find the eq of plane passing through the line of intersection of the plane 2x+y-z=3 and 5x-3y+4z+9=0and parallel to line

Answer 1

Answer:

7x + 9y - 10z - 27 = 0

Step-by-step explanation :

The Accurate Question :

Find the equation of the plane passing through the line of intersection of the planes 2x + y - z = 3 and 5x - 3y + 4z + 9 = 0 and parallel to the line -

$\sf \implies \dfrac{x-1}{2}=\dfrac{y-3}{4} =\dfrac{z-5}{5}$

Solution :

As given,

Equation of the planes are :

• 2x + y - z = 3
• 5x - 3y + 4z + 9 = 0

Now,

The equation of the plane passing through the line of intersection of these planes are :

$\sf\\ \implies (2x+y-z-3)+ \lambda (5x-3y+4z+9)=0\\ \\ \implies x(2+5\lambda)+y(1-3\lambda)+2(4\lambda-1)+9\lambda-3=0\quad[Eq.1]$

Now,

The plane is parallel to the line ,

$\sf \implies \dfrac{x-1}{2}=\dfrac{y-3}{4} =\dfrac{z-5}{5}$

Therefore,

$\sf \implies 2(2+5\lambda)+4(1-3\lambda)+5(4\lambda-1)=0$

$\sf\\ \\\implies 18\lambda+3=0\\ \\ \implies \lambda=-16$

Now,

Put the value of λ in Equation,

$\sf\\ \\\implies x(2-56)+y(1+36)+z(-46-1)-96-3=0\\ \\ \implies 7x+9y-10z-27=0$

Hence,

The equation of the required plane :

=> 7x + 9y - 10z - 27 = 0

Remember :

Vector equation of a line passing through a point and parallel to a given vector is :

$\vec{r}=\vec{a}+\lambda\vec{b}$

$\rule{300}{1.5}$

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