Plane Questions in English

(C) The equation of the plane is $6x - 3y + 2z = 6$. Dividing by $6$,we get $\frac{x}{1} + \frac{y}{-2} + \frac{z}{3} = 1$. Thus,the intercepts are $a = 1, b = -2, c = 3$.
The normal vector to the plane is $\vec{n} = 6\hat{i} - 3\hat{j} + 2\hat{k}$. The magnitude is $|\vec{n}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
The direction cosines are $l = \frac{6}{7}, m = -\frac{3}{7}, n = \frac{2}{7}$.
The perpendicular distance $p$ from the origin to the plane $Ax + By + Cz + D = 0$ is $p = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. Here $p = \frac{|-6|}{7} = \frac{6}{7}$.
Now,calculate $|al + bm + cn| = |(1)(\frac{6}{7}) + (-2)(-\frac{3}{7}) + (3)(\frac{2}{7})| = |\frac{6}{7} + \frac{6}{7} + \frac{6}{7}| = |\frac{18}{7}|$.
Since $p = \frac{6}{7}$,we have $|al + bm + cn| = 3 \times \frac{6}{7} = 3p$.

(A) Let the intercepts of the plane on the coordinate axes be $a, b, c$. Thus,the points are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.
The centroid of $\triangle ABC$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.
Given that the centroid is $(1, r, r^2)$,we have:
$\frac{a}{3} = 1 \Rightarrow a = 3$
$\frac{b}{3} = r \Rightarrow b = 3r$
$\frac{c}{3} = r^2 \Rightarrow c = 3r^2$
The equation of the plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Substituting the values of $a, b, c$,we get $\frac{x}{3} + \frac{y}{3r} + \frac{z}{3r^2} = 1$.
Since the plane passes through $(5, 5, -12)$,we have:
$\frac{5}{3} + \frac{5}{3r} - \frac{12}{3r^2} = 1$
Multiplying by $3r^2$,we get $5r^2 + 5r - 12 = 3r^2$,which simplifies to $2r^2 + 5r - 12 = 0$.
Factoring the quadratic equation: $(2r - 3)(r + 4) = 0$.
Thus,$r = \frac{3}{2}$ or $r = -4$.

(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Substituting the point $(2, -1, -3)$,we get $a(x - 2) + b(y + 1) + c(z + 3) = 0$.
Since the plane is parallel to the lines with direction ratios $(3, 2, -4)$ and $(2, -3, 2)$,the normal vector $(a, b, c)$ must be perpendicular to these direction vectors.
Thus,$3a + 2b - 4c = 0$ and $2a - 3b + 2c = 0$.
Using the cross product to find the normal vector $(a, b, c) = (3, 2, -4) \times (2, -3, 2) = \begin{vmatrix} i & j & k \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = i(4 - 12) - j(6 + 8) + k(-9 - 4) = -8i - 14j - 13k$.
Taking the normal vector as $(8, 14, 13)$,the equation of the plane is $8(x - 2) + 14(y + 1) + 13(z + 3) = 0$.
Expanding this,we get $8x - 16 + 14y + 14 + 13z + 39 = 0$,which simplifies to $8x + 14y + 13z + 37 = 0$.

(B) Given point $P(a, b, c)$.
Perpendicular $PA$ is drawn to the $YZ$-plane. The coordinates of $A$ are $(0, b, c)$.
Perpendicular $PB$ is drawn to the $ZX$-plane. The coordinates of $B$ are $(a, 0, c)$.
The origin $O$ is $(0, 0, 0)$.
The plane passes through $O(0, 0, 0)$,$A(0, b, c)$,and $B(a, 0, c)$.
The normal vector $\vec{n}$ to the plane is given by $\vec{OA} \times \vec{OB}$.
$\vec{OA} = 0\hat{i} + b\hat{j} + c\hat{k}$
$\vec{OB} = a\hat{i} + 0\hat{j} + c\hat{k}$
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & b & c \\ a & 0 & c \end{vmatrix} = \hat{i}(bc - 0) - \hat{j}(0 - ac) + \hat{k}(0 - ab) = bc\hat{i} + ac\hat{j} - ab\hat{k}$.
The equation of the plane passing through the origin is $bcx + acy - abz = 0$.

(A) Let the equation of the plane after rotation be $P_{3}: \ell x+my+nz=0$.
The line of intersection of the planes $P_{1}: \ell x+my=0$ and $P_{2}: z=0$ is the line where $\ell x+my=0$ and $z=0$.
The normal vectors are $\vec{n}_{1} = (\ell, m, 0)$ and $\vec{n}_{3} = (\ell, m, n)$.
The angle $\alpha$ between the planes $P_{1}$ and $P_{3}$ is given by $\cos \alpha = \frac{|\vec{n}_{1} \cdot \vec{n}_{3}|}{|\vec{n}_{1}| |\vec{n}_{3}|}$.
$\cos \alpha = \frac{|\ell^{2}+m^{2}|}{\sqrt{\ell^{2}+m^{2}} \sqrt{\ell^{2}+m^{2}+n^{2}}} = \sqrt{\frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}}$.
Squaring both sides,$\cos^{2} \alpha = \frac{\ell^{2}+m^{2}}{\ell^{2}+m^{2}+n^{2}}$.
$\Rightarrow \cos^{2} \alpha (\ell^{2}+m^{2}+n^{2}) = \ell^{2}+m^{2}$.
$\Rightarrow n^{2} \cos^{2} \alpha = (\ell^{2}+m^{2})(1 - \cos^{2} \alpha) = (\ell^{2}+m^{2}) \sin^{2} \alpha$.
$\Rightarrow n^{2} = (\ell^{2}+m^{2}) \tan^{2} \alpha$.
$\Rightarrow n = \pm \sqrt{\ell^{2}+m^{2}} \tan \alpha$.
Substituting $n$ into the equation of $P_{3}$,we get $\ell x+my \pm z \sqrt{\ell^{2}+m^{2}} \tan \alpha = 0$.

(C) Let the points be $A(1, 2, 3)$ and $B(3, 4, 5)$.
The midpoint $M$ of the line segment $AB$ is given by $M = \left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{3+5}{2}\right) = (2, 3, 4)$.
The direction ratios of the line segment $AB$ are $(3-1, 4-2, 5-3) = (2, 2, 2)$.
Since the plane bisects $AB$ at right angles,the line $AB$ is normal to the plane. Thus,the normal vector is $\vec{n} = 2\hat{i} + 2\hat{j} + 2\hat{k}$,which can be simplified to $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The equation of a plane passing through a point $\vec{a}$ with normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Here,$\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
Substituting these values,we get: $((x\hat{i} + y\hat{j} + z\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k})) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.
$(x-2)\hat{i} + (y-3)\hat{j} + (z-4)\hat{k} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$.
$(x-2) + (y-3) + (z-4) = 0$.
$x + y + z - 9 = 0$,or $x + y + z = 9$.

(D) The plane passes through $(1, 2, -3)$ and $(2, -2, 1)$. The vector connecting these two points is $\vec{v} = (2-1)\hat{i} + (-2-2)\hat{j} + (1-(-3))\hat{k} = \hat{i} - 4\hat{j} + 4\hat{k}$.
Since the plane is parallel to the $X$-axis,its normal is perpendicular to the unit vector $\hat{i} = (1, 0, 0)$.
The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{v}$ and $\hat{i}$:
$\vec{n} = \vec{v} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -4 & 4 \\ 1 & 0 & 0 \end{vmatrix} = \hat{i}(0) - \hat{j}(0-4) + \hat{k}(0-(-4)) = 4\hat{j} + 4\hat{k}$.
We can simplify the normal vector to $\vec{n}' = (0, 1, 1)$.
The equation of the plane passing through $(1, 2, -3)$ with normal $(0, 1, 1)$ is:
$0(x-1) + 1(y-2) + 1(z+3) = 0$
$y - 2 + z + 3 = 0$
$y + z + 1 = 0$.

(C) The equations of the given planes are $x+y+2z-6=0$ and $2x-y+z-9=0$.
Comparing these with the general form $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$,we get the normal vectors $\vec{n_1} = (1, 1, 2)$ and $\vec{n_2} = (2, -1, 1)$.
The angle $\theta$ between the two planes is given by the formula:
$\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}$.
Substituting the values:
$\cos \theta = \frac{|(1)(2) + (1)(-1) + (2)(1)|}{\sqrt{1^2+1^2+2^2} \sqrt{2^2+(-1)^2+1^2}}$.
$\cos \theta = \frac{|2 - 1 + 2|}{\sqrt{1+1+4} \sqrt{4+1+1}} = \frac{3}{\sqrt{6} \times \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(A) The equation of the plane is given by $2x - y + 2z - 1 = 0$.
The normal vector $\vec{n}$ to this plane is given by the coefficients of $x, y,$ and $z$,which is $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$.
The direction vector of the $X$-axis is $\vec{a} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.
The angle $\theta$ between the normal vector $\vec{n}$ and the $X$-axis is given by the formula $\cos \theta = \frac{|\vec{n} \cdot \vec{a}|}{|\vec{n}| |\vec{a}|}$.
Calculating the dot product: $\vec{n} \cdot \vec{a} = (2)(1) + (-1)(0) + (2)(0) = 2$.
Calculating the magnitudes: $|\vec{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ and $|\vec{a}| = \sqrt{1^2 + 0^2 + 0^2} = 1$.
Therefore,$\cos \theta = \frac{2}{3 \times 1} = \frac{2}{3}$.
Thus,$\theta = \cos^{-1} \frac{2}{3}$.