Question

(1 point) Find an equation for the plane that is perpendicular to the plane 2x – 5y – 3z = -5 and passes through the points P(5,3,-5) and P(2,4,1)​

Answer 1

$\sf \: Let \: the \: equation \: of \: plane \: containing \: the \: point \: (5, 3, -5)\: be \:$

$\sf \: a(x - 5) + b(y - 3) + c(z + 5) = 0 - - - (1)$

Where,

• (a, b, c) are direction ratios of plane.

Now,

• It is given that plane (1) passes through (2, 4, 1),

So,

$\rm :\longmapsto\:\sf \: a(2 - 5) + b(4 - 3) + c(1 + 5) = 0$

$\rm :\longmapsto\:\sf \: - 3a+ b + 6c = 0 - - - (2)$

Also,

Given that,

Plane (1) is perpendicular to the plane 2x - 5y - 3z = - 5

Now,

Direction ratios of plane 2x - 5y - 3z = - 5 is (2, - 5, - 3).

Thus,

By using condition of perpendicular of 2 planes, we have

$\rm :\longmapsto\:\sf \: 2a - 5 b - 3c = 0 - - - (3)$

Now,

Solving equation (2) and (3), we have

$\rm :\longmapsto\:\dfrac{a}{ - 3 + 30} = \dfrac{b}{12 - 9} = \dfrac{c}{15 - 2}$

$\rm :\longmapsto\:\dfrac{a}{ 27} = \dfrac{b}{3} = \dfrac{c}{13} = k$

$\rm :\longmapsto\:a = 27k \\\rm:\longmapsto\:b = 3k \\ \rm :\longmapsto\:c = 13k$

Substituting these values of a, b, c in equation (1), we get

$\rm :\longmapsto\:\sf \: 27k(x - 5) + 3k(y - 3) + 13k(z + 5) = 0$

$\rm :\longmapsto\:\sf \: 27(x - 5) + 3(y - 3) + 13(z + 5) = 0$

$\rm :\longmapsto\:\sf \: 27x - 135 + 3y - 9 + 13z + 65 = 0$

$\rm :\longmapsto\:\sf \: 27x + 3y + 13z - 79 = 0$

Additional Information :-

Let us consider two planes,

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

and

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

Then,

1. 2 planes are perpendicular iff

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

2. 2 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

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