Vardaan Learning Institute

A variable plane is at a constant perpendicular distance $p$ from the origin and meets the coordinate axes at points $A, B,$ and $C$. Prove analytically that the locus of the centroid of the tetrahedron $OABC$ is given by the equation $x^{-2} + y^{-2} + z^{-2} = 16p^{-2}$. Solution: Let the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Its distance from the origin $(0,0,0)$ is $p$, so $\frac{|0+0+0-1|}{\sqrt{(1/a)^2 + (1/b)^2 + (1/c)^2}} = p \implies \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$.
The intercepts yield the vertices: $A(a,0,0), B(0,b,0), C(0,0,c)$, and $O(0,0,0)$.
Let the centroid of the tetrahedron be $(x_1, y_1, z_1)$.
$x_1 = \frac{a+0+0+0}{4} \implies a = 4x_1$.
$y_1 = \frac{0+b+0+0}{4} \implies b = 4y_1$.
$z_1 = \frac{0+0+c+0}{4} \implies c = 4z_1$.
Substitute $a, b, c$ into the distance relation: $\frac{1}{(4x_1)^2} + \frac{1}{(4y_1)^2} + \frac{1}{(4z_1)^2} = \frac{1}{p^2}$.
$\frac{1}{16x_1^2} + \frac{1}{16y_1^2} + \frac{1}{16z_1^2} = \frac{1}{p^2} \implies \frac{1}{x_1^2} + \frac{1}{y_1^2} + \frac{1}{z_1^2} = \frac{16}{p^2}$.
Thus, the locus is $x^{-2} + y^{-2} + z^{-2} = 16p^{-2}$.

Find the magnitude and the equations of the line of shortest distance between the lines $L_1: \frac{x-8}{3} = \frac{y+9}{-16} = \frac{z-10}{7}$ and $L_2: \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$. Solution: $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 - (-48)) = 24\hat{i} + 36\hat{j} + 72\hat{k}$.
Dividing by 12, proportional direction ratios for the shortest distance line are $(2, 3, 6)$. Magnitude $|\vec{b}| = \sqrt{4+9+36} = 7$.
$\vec{a_2} - \vec{a_1} = (15-8, 29-(-9), 5-10) = (7, 38, -5)$.
Shortest Distance $d = \frac{|(7)(2) + (38)(3) + (-5)(6)|}{7} = \frac{|14 + 114 - 30|}{7} = \frac{98}{7} = 14$ units.
Equation of the line of SD is given by the intersection of two planes: Plane containing $L_1$ and SD vector, and Plane containing $L_2$ and SD vector.
Plane 1: $\begin{vmatrix} x-8 & y+9 & z-10 \\ 3 & -16 & 7 \\ 2 & 3 & 6 \end{vmatrix} = 0 \implies -117(x-8) - 4(y+9) + 41(z-10) = 0 \implies 117x + 4y - 41z - 490 = 0$.
Plane 2: $\begin{vmatrix} x-15 & y-29 & z-5 \\ 3 & 8 & -5 \\ 2 & 3 & 6 \end{vmatrix} = 0 \implies 63(x-15) - 28(y-29) - 7(z-5) = 0 \implies 9x - 4y - z - 14 = 0$.
The equations of the line of SD are $117x + 4y - 41z - 490 = 0 = 9x - 4y - z - 14$.

A variable plane passes through a fixed point $(a, b, c)$ and meets the coordinate axes at the points $A, B,$ and $C$. Show that the locus of the point common to the three planes drawn through $A, B,$ and $C$ parallel to the coordinate planes is $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 1$. Solution: Let the plane be $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 1$. Since it passes through $(a,b,c)$, we have $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 1$.
The intercepts are $A(p, 0, 0), B(0, q, 0), C(0, 0, r)$.
The plane through $A$ parallel to the $yz$-plane is $x = p$.
The plane through $B$ parallel to the $zx$-plane is $y = q$.
The plane through $C$ parallel to the $xy$-plane is $z = r$.
Let $(x_1, y_1, z_1)$ be the point common to these three planes. Then $x_1 = p, y_1 = q, z_1 = r$.
Substitute $p, q, r$ into the initial constraint: $\frac{a}{x_1} + \frac{b}{y_1} + \frac{c}{z_1} = 1$.
Hence, the locus is $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 1$.

Find the exact equation of the line which represents the reflection of the line $\frac{x-1}{9} = \frac{y-2}{-1} = \frac{z+3}{-3}$ in the plane $3x - 3y + 10z = 26$. Solution: Let's check if the line is parallel to the plane: dot product of direction vector $(9, -1, -3)$ and normal $(3, -3, 10)$ is $9(3) + (-1)(-3) + (-3)(10) = 27 + 3 - 30 = 0$. Yes, they are parallel.
Since the line is parallel, its reflection will also be parallel to it. So, the new direction ratios remain $9, -1, -3$.
Take a point on the line: $P(1, 2, -3)$. We find its image in the plane.
Equation of normal from $P$: $\frac{x-1}{3} = \frac{y-2}{-3} = \frac{z+3}{10} = k$. Foot $F(3k+1, -3k+2, 10k-3)$.
$F$ lies on the plane: $3(3k+1) - 3(-3k+2) + 10(10k-3) = 26 \implies 9k+3 + 9k-6 + 100k-30 = 26 \implies 118k - 33 = 26 \implies 118k = 59 \implies k = 1/2$.
For the image $I$, use $k' = 2k = 1$. Image point is $I(3(1)+1, -3(1)+2, 10(1)-3) = (4, -1, 7)$.
The reflected line passes through $(4, -1, 7)$ and is parallel to the original line.
Equation: $\frac{x-4}{9} = \frac{y+1}{-1} = \frac{z-7}{-3}$.

Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$. Solution: Let the line drawn from $P(3, 8, 2)$ parallel to the plane intersect the given line at $Q$.
The general coordinates of $Q$ on the given line are $(2\lambda+1, 4\lambda+3, 3\lambda+2)$.
The vector $\vec{PQ} = (2\lambda+1-3, 4\lambda+3-8, 3\lambda+2-2) = (2\lambda-2, 4\lambda-5, 3\lambda)$.
Since $PQ$ is parallel to the plane, $\vec{PQ}$ is perpendicular to the normal of the plane $(3, 2, -2)$.
$3(2\lambda-2) + 2(4\lambda-5) - 2(3\lambda) = 0 \implies 6\lambda - 6 + 8\lambda - 10 - 6\lambda = 0 \implies 8\lambda - 16 = 0 \implies \lambda = 2$.
Coordinates of $Q$ are $(2(2)+1, 4(2)+3, 3(2)+2) = (5, 11, 8)$.
Distance $PQ = \sqrt{(5-3)^2 + (11-8)^2 + (8-2)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ units.

Using the properties of scalar triple products, prove that the lines given by $\frac{x-a+d}{\alpha-\delta} = \frac{y-a}{\alpha} = \frac{z-a-d}{\alpha+\delta}$ and $\frac{x-b+c}{\beta-\gamma} = \frac{y-b}{\beta} = \frac{z-b-c}{\beta+\gamma}$ are strictly coplanar. Solution: Points on lines: $P_1(a-d, a, a+d)$ and $P_2(b-c, b, b+c)$.
Vector $\vec{P_1 P_2} = (b-c-a+d, b-a, b+c-a-d)$.
Direction vectors: $\vec{b_1} = (\alpha-\delta, \alpha, \alpha+\delta)$ and $\vec{b_2} = (\beta-\gamma, \beta, \beta+\gamma)$.
Coplanarity condition requires the determinant of these three vectors to be zero.
$D = \begin{vmatrix} b-a-c+d & b-a & b-a+c-d \\ \alpha-\delta & \alpha & \alpha+\delta \\ \beta-\gamma & \beta & \beta+\gamma \end{vmatrix}$.
Apply column operation: $C_1 \to C_1 + C_3$.
Column 1 becomes: $(b-a-c+d) + (b-a+c-d) = 2(b-a)$.
Row 2, Column 1 becomes: $(\alpha-\delta) + (\alpha+\delta) = 2\alpha$.
Row 3, Column 1 becomes: $(\beta-\gamma) + (\beta+\gamma) = 2\beta$.
Now $C_1$ is $2(b-a), 2\alpha, 2\beta$, which is exactly $2 \times C_2$.
Since Column 1 and Column 2 are proportional ($C_1 = 2C_2$), the determinant $D = 0$. Thus, the lines are coplanar.

Find the equations of the lines which bisect the angles between the two lines given by $\frac{x-1}{2} = \frac{y+1}{-2} = \frac{z}{1}$ and $\frac{x-1}{1} = \frac{y+1}{2} = \frac{z}{-2}$. Solution: The lines intersect at $P(1, -1, 0)$. Direction vectors are $\vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}$.
Their magnitudes are both $\sqrt{4+4+1} = 3$.
Unit vectors are $\hat{u} = \frac{1}{3}(2, -2, 1)$ and $\hat{v} = \frac{1}{3}(1, 2, -2)$.
The direction vectors of the angle bisectors are parallel to $\hat{u} + \hat{v}$ and $\hat{u} - \hat{v}$.
$\vec{d_1} \parallel \hat{u} + \hat{v} = \frac{1}{3}(3, 0, -1) \sim (3, 0, -1)$.
$\vec{d_2} \parallel \hat{u} - \hat{v} = \frac{1}{3}(1, -4, 3) \sim (1, -4, 3)$.
The bisecting lines pass through $P(1, -1, 0)$.
Equations: $B_1: \frac{x-1}{3} = \frac{y+1}{0} = \frac{z}{-1}$ and $B_2: \frac{x-1}{1} = \frac{y+1}{-4} = \frac{z}{3}$.

A line $L$ passes through the origin and makes angles $\alpha, \beta, \gamma$ with the coordinate planes $x=0, y=0, z=0$ respectively. Prove that $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 2$. Solution: Let the direction cosines of line $L$ be $l, m, n$. Thus $l^2 + m^2 + n^2 = 1$.
The normal to the $yz$-plane ($x=0$) is the $x$-axis (with direction cosines $1, 0, 0$).
The angle $\theta_1$ between the line and the normal to the $yz$-plane is given by $\cos\theta_1 = l(1) + m(0) + n(0) = l$.
The angle $\alpha$ the line makes with the plane is complementary to $\theta_1$, so $\alpha = 90^\circ - \theta_1 \implies \sin\alpha = \cos\theta_1 = l$.
Similarly, the angle $\beta$ with the $zx$-plane ($y=0$) gives $\sin\beta = m$.
And the angle $\gamma$ with the $xy$-plane ($z=0$) gives $\sin\gamma = n$.
Squaring and adding: $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = l^2 + m^2 + n^2 = 1$.
We know $\sin^2 x = 1 - \cos^2 x$. Substituting this:
$(1 - \cos^2\alpha) + (1 - \cos^2\beta) + (1 - \cos^2\gamma) = 1 \implies 3 - (\cos^2\alpha + \cos^2\beta + \cos^2\gamma) = 1$.
Therefore, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 2$. Proved.

Find the geometric condition under which the three planes $x=cy+bz$, $y=az+cx$, and $z=bx+ay$ intersect exactly in a single line. Solution: Rewrite the planes as a homogeneous system of linear equations:
$x - cy - bz = 0$
$cx - y + az = 0$
$bx + ay - z = 0$
For the planes to intersect in a line (infinitely many solutions), the determinant of the coefficient matrix must be zero.
$\begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix} = 0$
Expanding the determinant: $1(1 - a^2) - (-c)(-c - ab) + (-b)(ac + b) = 0$
$1 - a^2 - (c^2 + abc) - (abc + b^2) = 0$
$1 - a^2 - c^2 - abc - abc - b^2 = 0 \implies 1 - a^2 - b^2 - c^2 - 2abc = 0$.
The required condition is $a^2 + b^2 + c^2 + 2abc = 1$.

A plane passes through a fixed point $(p, q, r)$ and cuts the coordinate axes in points $A, B,$ and $C$. Find the locus of the center of the sphere passing through the origin $O$ and the points $A, B, C$. Solution: Let the intercepts of the variable plane be $a, b, c$. Equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
It passes through $(p,q,r)$, so $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$.
The vertices of the tetrahedron $OABC$ are $(0,0,0), (a,0,0), (0,b,0), (0,0,c)$.
The center $(x_1, y_1, z_1)$ of the sphere passing through these points is the midpoint of the principal diagonal of the bounding box, i.e., $x_1 = a/2, y_1 = b/2, z_1 = c/2$.
Thus, $a = 2x_1, b = 2y_1, c = 2z_1$.
Substitute these into the initial constraint: $\frac{p}{2x_1} + \frac{q}{2y_1} + \frac{r}{2z_1} = 1$.
Multiply by 2: $\frac{p}{x_1} + \frac{q}{y_1} + \frac{r}{z_1} = 2$.
The locus is $\frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 2$.

Find the equation of the line passing through the origin which intersects the two lines $L_1: \frac{x-3}{1} = \frac{y-5}{2} = \frac{z-7}{3}$ and $L_2: \frac{x+1}{2} = \frac{y+2}{1} = \frac{z+3}{4}$. Solution: The required line lies in the plane $P_1$ formed by the origin and $L_1$, and also in the plane $P_2$ formed by the origin and $L_2$.
Normal to $P_1$: $\vec{n_1} = \vec{OA} \times \vec{b_1} = (3, 5, 7) \times (1, 2, 3) = \hat{i}(15-14) - \hat{j}(9-7) + \hat{k}(6-5) = \hat{i} - 2\hat{j} + \hat{k}$. Equation of $P_1$: $x - 2y + z = 0$.
Normal to $P_2$: $\vec{n_2} = \vec{OB} \times \vec{b_2} = (-1, -2, -3) \times (2, 1, 4) = \hat{i}(-8 - (-3)) - \hat{j}(-4 - (-6)) + \hat{k}(-1 - (-4)) = -5\hat{i} - 2\hat{j} + 3\hat{k}$. Equation of $P_2$: $-5x - 2y + 3z = 0$.
The required line is the intersection of these two planes. Its direction vector is $\vec{n_1} \times \vec{n_2}$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ -5 & -2 & 3 \end{vmatrix} = \hat{i}(-6 - (-2)) - \hat{j}(3 - (-5)) + \hat{k}(-2 - 10) = -4\hat{i} - 8\hat{j} - 12\hat{k} \sim (1, 2, 3)$.
Since it passes through the origin, the equation is $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$.

A tetrahedron has vertices at $O(0,0,0), A(1,2,1), B(2,1,3),$ and $C(-1,1,2)$. Find the exact acute angle between the two triangular faces $OAB$ and $ABC$. Solution: The angle between the faces is the angle between their normal vectors.
Normal to face OAB: $\vec{n_1} = \vec{OA} \times \vec{OB} = (1, 2, 1) \times (2, 1, 3) = \hat{i}(6-1) - \hat{j}(3-2) + \hat{k}(1-4) = 5\hat{i} - \hat{j} - 3\hat{k}$.
Normal to face ABC: Vectors are $\vec{AB} = (2-1, 1-2, 3-1) = (1, -1, 2)$ and $\vec{AC} = (-1-1, 1-2, 2-1) = (-2, -1, 1)$.
$\vec{n_2} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = \hat{i}(-1 - (-2)) - \hat{j}(1 - (-4)) + \hat{k}(-1 - 2) = \hat{i} - 5\hat{j} - 3\hat{k}$.
Angle: $\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} = \frac{|5(1) + (-1)(-5) + (-3)(-3)|}{\sqrt{25+1+9}\sqrt{1+25+9}} = \frac{|5 + 5 + 9|}{\sqrt{35}\sqrt{35}} = \frac{19}{35}$.
$\theta = \cos^{-1}\left(\frac{19}{35}\right)$.

In a cube of edge length '$a$', find the shortest distance between the main diagonal of the cube and a face diagonal that does not intersect the main diagonal. Solution: Let the vertices of the cube be at $O(0,0,0)$ to $G(a,a,a)$.
Main diagonal $OG$ passes through $(0,0,0)$ and $(a,a,a)$. Vector eq: $\vec{r_1} = \lambda(a, a, a) \sim \lambda(1, 1, 1)$.
Face diagonal on the XY plane not intersecting $OG$ is $AC$, connecting $(a,0,0)$ and $(0,a,0)$. Vector eq: $\vec{r_2} = (a, 0, 0) + \mu(-a, a, 0) \sim (a, 0, 0) + \mu(-1, 1, 0)$.
$\vec{a_2} - \vec{a_1} = (a, 0, 0) - (0, 0, 0) = (a, 0, 0)$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 1 & 0 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(0 - (-1)) + \hat{k}(1 - (-1)) = -\hat{i} - \hat{j} + 2\hat{k}$.
$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{1+1+4} = \sqrt{6}$.
Numerator $= |(-\hat{i} - \hat{j} + 2\hat{k}) \cdot (a, 0, 0)| = |-a| = a$.
Shortest distance $= \frac{a}{\sqrt{6}}$.

Find the coordinates of the two possible points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ that are situated at an exact distance of $3\sqrt{2}$ units from the point $(1, 2, 3)$. Solution: Let the point on the line be $F(3\lambda-2, 2\lambda-1, 2\lambda+3)$.
Distance from $P(1, 2, 3)$ squared is $(3\sqrt{2})^2 = 18$.
$(3\lambda-2-1)^2 + (2\lambda-1-2)^2 + (2\lambda+3-3)^2 = 18$
$(3\lambda-3)^2 + (2\lambda-3)^2 + (2\lambda)^2 = 18$
$9\lambda^2 - 18\lambda + 9 + 4\lambda^2 - 12\lambda + 9 + 4\lambda^2 = 18$
$17\lambda^2 - 30\lambda + 18 = 18 \implies 17\lambda^2 - 30\lambda = 0 \implies \lambda(17\lambda - 30) = 0$.
Thus, $\lambda = 0$ or $\lambda = 30/17$.
For $\lambda = 0$, the point is $(-2, -1, 3)$.
For $\lambda = 30/17$, the point is $(3(30/17)-2, 2(30/17)-1, 2(30/17)+3) = \left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$.

Find the equation of the plane which contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$, and which is perpendicular to the plane $5x+3y-6z+8=0$. Solution: Family of planes: $(x+2y+3z-4) + \lambda(2x+y-z+5) = 0$
$(1+2\lambda)x + (2+\lambda)y + (3-\lambda)z - 4 + 5\lambda = 0$.
This is perpendicular to $5x+3y-6z+8=0$, so the dot product of their normals is zero:
$5(1+2\lambda) + 3(2+\lambda) - 6(3-\lambda) = 0$
$5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0 \implies 19\lambda - 7 = 0 \implies \lambda = 7/19$.
Substitute $\lambda=7/19$ back into the family equation:
$\left(1 + \frac{14}{19}\right)x + \left(2 + \frac{7}{19}\right)y + \left(3 - \frac{7}{19}\right)z - 4 + \frac{35}{19} = 0$
$\frac{33}{19}x + \frac{45}{19}y + \frac{50}{19}z - \frac{41}{19} = 0 \implies 33x + 45y + 50z = 41$.