Question

find the equation of plane passing through intersection of Plane x+y+z=1, 2x+ 3y+4z=5 to perpendicular to plane X-y+z=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us assume that,

$\rm :\longmapsto\:P_1 : x + y + z - 1 = 0$

and

$\rm :\longmapsto\:P_2 : 2x + 3y + 4z - 5 = 0$

So, required equation of plane which passes through the line of intersection of planes x + y + z - 1 = 0 and 2x + 3y + 4z - 5 = 0 is

$\sf :\longmapsto\:x + y + z - 1 + k(2x + 3y + 4z - 5) = 0$

$\sf :\longmapsto\:x + y + z - 1 + 2kx + 3ky + 4kz - 5k = 0$

$\red{\sf :\longmapsto\:(1 + 2k)x + (1 + 3k)y + (1 + 4k)z -(1 + 5k) = 0 - - (1)}$

Now, this plane is perpendicular to the plane x - y + z = 0.

We know,

Two planes ax + by + cz + d = 0 and px + qy + rz + s = 0 are perpendicular iff ap + bq + cr = 0.

So, using this, we get

$\rm :\longmapsto\:(1 + 2k)(1) + (1 + 3k)( - 1) + (1 + 4k)(1) = 0$

$\rm :\longmapsto\:1 + 2k - 1 - 3k + 1 + 4k = 0$

$\rm :\longmapsto\:3k + 1 = 0$

$\rm :\longmapsto\:3k = - \: 1$

$\bf\implies \:k \: = \: - \: \dfrac{1}{3}$

Now, on substituting this value of k in equation (1), we get

$\rm :\longmapsto\:\bigg[1 - \dfrac{2}{3} \bigg]x + \bigg[1 - \dfrac{3}{3} \bigg]y + \bigg[1 - \dfrac{4}{3} \bigg]z - \bigg[1 - \dfrac{5}{3} \bigg] = 0$

$\rm :\longmapsto\:\bigg[\dfrac{1}{3} \bigg]x + \bigg[1 - 1 \bigg]y - \bigg[ \dfrac{1}{3} \bigg]z + \bigg[ \dfrac{2}{3} \bigg] = 0$

$\rm :\longmapsto\:\dfrac{x}{3} - \dfrac{z}{3} + \dfrac{2}{3} = 0$

$\bf\implies \:x - z + 2 = 0$

So,

Required equation of plane passing through line of intersection of plane x+y+z=1 and 2x+ 3y+4z=5 and perpendicular to plane x-y+z=0 is x - z + 2 = 0.

Additional Information :-

1. Equation of plane is

$\boxed{ \tt{ \: \vec{r}.\vec{n} \: = \: d \: }}$

Let us consider two planes

$\boxed{ \tt{ \: \vec{r}.\vec{n_1} \: = \: d \: }}$

and

$\boxed{ \tt{ \: \vec{r}.\vec{n_2} \: = \: d \: }}$

2. Two planes are perpendicular iff

$\rm :\longmapsto\:\boxed{ \tt{ \: \vec{n_1}.\vec{n_2} = 0 \: }}$

3. Two planes are parallel iff

$\rm :\longmapsto\:\boxed{ \tt{ \: \vec{n_1} \times \vec{n_2} = 0 \: }}$

4. Angle between two planes is given by

$\rm :\longmapsto\:\boxed{ \tt{ \: cos \theta \: = \: \frac{\vec{n_1}.\vec{n_2}}{ |\vec{n_1}| \: |\vec{n_2}| } \: }}$

Answer 2

Answer:

Step-by-step explanation:

Given equations of plane :

$\tt{x+y+z=1}\\\tt{2x+3y+4z=5}$

Any plane passing through the intersection of the given planes is

$\sf{(x+y+z-1)+\lambda(2x+3y+4z-5)=0}$

$\sf{\implies\,x+y+z-1+2\lambda\,x+3\lambda\,y+4\lambda\,z-5\lambda\,=0}$

$\sf{\implies\,\boxed{\blue{(1+2\lambda)x+(1+3\lambda)y+(1+4\lambda)z-(1+5\lambda)=0}}\,\,\,\,\,\,\,\,...(1)}$

Now, this plane is perpendicular to the plane $\sf{x-y+z=0}$

So,

$\sf{(1+2\lambda)(1)+(1+3\lambda)(-1)+(1+4\lambda)(1)=0}$

$\sf{\implies1+2\lambda-1-3\lambda+1+4\lambda=0}$

$\sf{\implies2\lambda-3\lambda+1+4\lambda=0}$

$\sf{\implies3\lambda+1=0}$

$\sf{\implies\lambda=-\dfrac{1}{3}}$

Put the value of λ in (1), we get,

$\sf{\implies\,\bigg(1-\dfrac{2}{3}\bigg)x+\bigg(1-\dfrac{3}{3}\bigg)y+\bigg(1-\dfrac{4}{3}\bigg)z-\bigg(1-\dfrac{5}{3}\bigg)=0}$

$\sf{\implies\,\bigg(\dfrac{3-2}{3}\bigg)x+\bigg(1-1\bigg)y+\bigg(\dfrac{3-4}{3}\bigg)z-\bigg(\dfrac{3-5}{3}\bigg)=0}$

$\sf{\implies\,\bigg(\dfrac{1}{3}\bigg)x+\bigg(\dfrac{-1}{3}\bigg)z-\bigg(\dfrac{-2}{3}\bigg)=0}$

$\sf{\implies\,x-z+2=0}$