Question

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Answer 1

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

Given points are

$\rm :\longmapsto\: - 2\hat{i} + 6\hat{j} - 6\hat{k}$

$\rm :\longmapsto\: - 3\hat{i} + 10\hat{j} - 9\hat{k}$

$\rm :\longmapsto\: - 5\hat{i} - 6\hat{j} - 6\hat{k}$

So, the points in cartesian form are

$\rm :\longmapsto\:( - 2,6, - 6), \: ( - 3,10, - 9), \: ( - 5, - 6, - 6)$

Now,

We have to find the equation of plane passes through

$\rm :\longmapsto\:( - 2,6, - 6), \: ( - 3,10, - 9), \: ( - 5, - 6, - 6)$

Let us assume that direction ratios of the plane be (a, b, c).

So, equation of plane which passes through the point (-2, 6, - 6) and having direction ratios (a, b, c) is given by

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

Now,

Plane (1) passes through (-3, 10, - 9), So

$\rm :\longmapsto\:a( - 3 + 2) + b(10 - 6) + c( - 9 + 6) = 0$

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

Again,

Plane (1) passes through (- 5, - 6, - 6), So

$\rm :\longmapsto\:a( - 5 + 2) + b( - 6 - 6) + c( - 6 + 6) = 0$

$\rm :\longmapsto\: - 3a - 12b = 0$

$\rm :\longmapsto\: - 3a = 12b$

$\rm :\longmapsto\:a = - \: 4b - - (3)$

Substituting the value of a in equation (2), we get

$\rm :\longmapsto\: - ( - 4b) + 4b - 3c = 0$

$\rm :\longmapsto\: 4b+ 4b - 3c = 0$

$\rm :\longmapsto\: 8b - 3c = 0$

$\rm :\longmapsto\: 8b = 3c$

$\rm :\implies\:c = \dfrac{8b}{3} - - - (4)$

On substituting the values from equation (3) and (4) in equation (1), we get

$\rm :\longmapsto\:( - 4b)(x + 2) + b(y - 6) + \dfrac{8b}{3} (z + 6) = 0$

On cancel 'b' , we get

$\rm :\longmapsto\: - 4(x + 2) + (y - 6) + \dfrac{8}{3} (z + 6) = 0$

$\rm :\longmapsto\: - 12(x + 2) + 3(y - 6) + 8 (z + 6) = 0$

$\rm :\longmapsto\: - 12x - 24 + 3y - 18 + 8 z + 48 = 0$

$\rm :\longmapsto\: - 12x + 3y + 8 z + 6= 0$

$\rm :\longmapsto\: 12x - 3y - 8 z - 6= 0$

$\rm :\longmapsto\: 12x - 3y - 8 z = 6$

So, in vector form, the equation of plane is

$\rm :\longmapsto\: \vec{r} \: . \: (12\hat{i} - 3\hat{j} - 8\hat{k}) = 6$

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. Two planes are parallel iff

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

or

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

2. Two planes are perpendicular, iff

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

3. Angle 'e' between two planes is

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