Question

Find the equation of the plane that contains the point A(2, 1, – 1) and is perpendicular to the line of intersection of the planes 2x + y – z = 3 and x + 2y + z = 2.​

Answer 1

Appropriate Question :-

Find the equation of the plane that contains the point A(2, 1, – 1) and through the line of intersection of the planes 2x + y – z = 3 and x + 2y + z = 2.

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

The required equation of plane which passes through the line of intersection of the planes 2x + y – z - 3 = 0 and x + 2y + z - 2 = 0 is given by

$\rm :\longmapsto\:2x + y - z - 3 + k(x + 2y + z - 2) = 0$

$\rm :\longmapsto\:2x + y - z - 3 + kx + 2ky + kz - 2k = 0$

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

Since, this plane contains the point (2, 1, - 1),

Thus,

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

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

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

$\bf\implies \:k \: = - \: 3$

Put value of k = - 3, in equation (1), we get

$\rm :\longmapsto\:( - 3 + 2)x + ( - 6 + 1)y + ( - 3 - 1)z - ( - 6 + 3) = 0$

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

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

Additional Information :-

Let us consider two planes,

$\rm :\longmapsto\: \vec{r}. \: \vec{n_1} = d$

and

$\rm :\longmapsto\: \vec{r}. \: \vec{n_2} = d$

Then,

1. Two planes are perpendicular iff

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

2. Two planes are parallel iff

$\rm :\longmapsto\:\vec{n_1} \: \times \: \vec{n_2} = 0 \: \: \: or \: \: \: \vec{n_1} = k \: \vec{n_2}$

3. Angle between two planes is given by

$\rm :\longmapsto\:cos\theta = \dfrac{\vec{n_1} \: . \: \vec{n_2}}{ |\vec{n_1}| \: |\vec{n_2}| }$