Vardaan Learning Institute

Find the direction cosines of a line which makes equal angles with the positive coordinate axes.Solution: Let angles be $\alpha = \beta = \gamma$. Then $l=m=n$. Since $l^2 + m^2 + n^2 = 1 \Rightarrow 3l^2 = 1 \Rightarrow l = \pm 1/\sqrt{3}$. The direction cosines are $(\pm 1/\sqrt{3}, \pm 1/\sqrt{3}, \pm 1/\sqrt{3})$. For positive axes, it's $(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$.

Find the direction cosines of the line passing through the two points $(-2, 4, -5)$ and $(1, 2, 3)$.Solution: D.R.s $a = 1 - (-2) = 3$, $b = 2 - 4 = -2$, $c = 3 - (-5) = 8$. Magnitude $= \sqrt{3^2 + (-2)^2 + 8^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$. D.C.s are $(3/\sqrt{77}, -2/\sqrt{77}, 8/\sqrt{77})$.

If a line has direction ratios $2, -1, -2$, determine its direction cosines.Solution: Magnitude $= \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$. D.C.s are $(2/3, -1/3, -2/3)$.

Show that the points $A(2, 3, -4), B(1, -2, 3)$ and $C(3, 8, -11)$ are collinear using direction ratios.Solution: D.R. of AB $= (1-2, -2-3, 3-(-4)) = (-1, -5, 7)$. D.R. of BC $= (3-1, 8-(-2), -11-3) = (2, 10, -14)$. Since D.R. of BC is $-2$ times D.R. of AB, lines are parallel. Since point B is common, they are collinear.

A line makes angles $90^\circ, 135^\circ, 45^\circ$ with the x, y, and z axes respectively. Find its direction cosines.Solution: $l = \cos 90^\circ = 0$, $m = \cos 135^\circ = -1/\sqrt{2}$, $n = \cos 45^\circ = 1/\sqrt{2}$. D.C.s are $(0, -1/\sqrt{2}, 1/\sqrt{2})$.

Find the direction cosines of the line joining the points $(1, 0, 0)$ and $(0, 1, 1)$.Solution: D.R.s $= (0-1, 1-0, 1-0) = (-1, 1, 1)$. Magnitude $= \sqrt{1+1+1} = \sqrt{3}$. D.C.s are $(-1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$.

Find the vector equation of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3\hat{i} + 2\hat{j} - 8\hat{k}$.Solution: Position vector $\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}$. Parallel vector $\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$. Equation is $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$.

Find the Cartesian equation of the line passing through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.Solution: Direction ratios of parallel line are $3, 5, 6$. Equation is $\frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6} \Rightarrow \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}$.

The Cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$. Write its vector equation.Solution: Point is $(5, -4, 6)$ and D.R.s are $3, 7, 2$. Vector equation: $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$.

Find the equation of the line in vector and Cartesian form that passes through the origin and $(5, -2, 3)$.Solution: Vector: $\vec{a} = 0$, $\vec{b} = 5\hat{i} - 2\hat{j} + 3\hat{k}$. $\vec{r} = \lambda(5\hat{i} - 2\hat{j} + 3\hat{k})$. Cartesian: $\frac{x-0}{5-0} = \frac{y-0}{-2-0} = \frac{z-0}{3-0} \Rightarrow \frac{x}{5} = \frac{y}{-2} = \frac{z}{3}$.

Find the points on the line $\frac{x-1}{2} = \frac{y+2}{3} = \frac{z-3}{6}$ at a distance of $7$ units from the point $(1, -2, 3)$.Solution: General point $P(2\lambda+1, 3\lambda-2, 6\lambda+3)$. Distance from $A(1, -2, 3)$ is $\sqrt{(2\lambda)^2 + (3\lambda)^2 + (6\lambda)^2} = \sqrt{49\lambda^2} = 7|\lambda|$. Given distance $= 7 \Rightarrow 7|\lambda| = 7 \Rightarrow \lambda = \pm 1$. Points are $P_1(3, 1, 9)$ and $P_2(-1, -5, -3)$.

Find the Cartesian equation of a line passing through $(1, 2, 3)$ and parallel to the x-axis.Solution: D.R.s of x-axis are $1, 0, 0$. Equation is $\frac{x-1}{1} = \frac{y-2}{0} = \frac{z-3}{0}$.

Find the angle between the lines $\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})$ and $\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$.Solution: $\vec{b_1} = 3\hat{i} + 2\hat{j} + 6\hat{k}$, $\vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k}$. $\cos\theta = \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}|} = \frac{3(1) + 2(2) + 6(2)}{\sqrt{9+4+36}\sqrt{1+4+4}} = \frac{3+4+12}{7 \times 3} = \frac{19}{21}$. $\theta = \cos^{-1}(19/21)$.

Find the angle between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$.Solution: D.R.s are $(2, 2, 1)$ and $(4, 1, 8)$. $\cos\theta = \frac{2(4) + 2(1) + 1(8)}{\sqrt{4+4+1}\sqrt{16+1+64}} = \frac{8+2+8}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$. $\theta = \cos^{-1}(2/3)$.

Find the value of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.Solution: Convert to standard form: $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$ and $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$. For perpendicular lines, dot product of D.R.s $= 0$. $-3(-3p/7) + (2p/7)(1) + 2(-5) = 0 \Rightarrow 9p/7 + 2p/7 = 10 \Rightarrow 11p = 70 \Rightarrow p = 70/11$.

Show that the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.Solution: D.R.s are $(7, -5, 1)$ and $(1, 2, 3)$. Dot product $= 7(1) + (-5)(2) + 1(3) = 7 - 10 + 3 = 0$. Hence, perpendicular.

Find the angle between the pair of lines whose direction ratios are $(1, 1, 2)$ and $(\sqrt{3}-1, -\sqrt{3}-1, 4)$.Solution: $\cos\theta = \frac{1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4)}{\sqrt{1+1+4}\sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 16}} = \frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6}\sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}+16}} = \frac{6}{\sqrt{6}\sqrt{24}} = \frac{6}{\sqrt{144}} = \frac{6}{12} = \frac{1}{2}$. $\theta = 60^\circ$ (or $\pi/3$).

Find the value of $\lambda$ if the lines $\frac{x-1}{-3} = \frac{y-2}{2\lambda} = \frac{z-3}{2}$ and $\frac{x-1}{3\lambda} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular.Solution: Dot product $= -3(3\lambda) + 2\lambda(1) + 2(-5) = 0 \Rightarrow -9\lambda + 2\lambda - 10 = 0 \Rightarrow -7\lambda = 10 \Rightarrow \lambda = -10/7$.

Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$ and $\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$.Solution: $\vec{a_2} - \vec{a_1} = 3\hat{i} + 3\hat{j} + 3\hat{k}$. $\vec{b_1} \times \vec{b_2} = (-3-6)\hat{i} - (1-4)\hat{j} + (3+6)\hat{k} = -9\hat{i} + 3\hat{j} + 9\hat{k}$. Numerator $= (-9)(3) + 3(3) + 9(3) = -27 + 9 + 27 = 9$. Denominator $= \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19}$. Distance $= \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}}$.

Find the shortest distance between the lines given by $\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}$ and $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$.Solution: Points: $A(-1, -1, -1), B(3, 5, 7) \Rightarrow AB = (4, 6, 8)$. D.R.s $\vec{b_1} = (7, -6, 1)$, $\vec{b_2} = (1, -2, 1)$. $\vec{b_1} \times \vec{b_2} = (-6+2)\hat{i} - (7-1)\hat{j} + (-14+6)\hat{k} = -4\hat{i} - 6\hat{j} - 8\hat{k}$. Num $= -4(4) - 6(6) - 8(8) = -16 - 36 - 64 = -116$. Denom $= \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}$. Dist $= \frac{116}{\sqrt{116}} = \sqrt{116} = 2\sqrt{29}$.

Find the shortest distance between the parallel lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(2\hat{i} - \hat{j} + \hat{k})$.Solution: $\vec{a_2} - \vec{a_1} = \hat{i} - \hat{k}$. $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$. $\vec{b} \times (\vec{a_2}-\vec{a_1}) = (1-0)\hat{i} - (-2-1)\hat{j} + (0+1)\hat{k} = \hat{i} + 3\hat{j} + \hat{k}$ Wait: determinant $| \hat{i} \quad \hat{j} \quad \hat{k} | \atop | 2 \quad -1 \quad 1 | \atop | 1 \quad 0 \quad -1 |$ $= \hat{i}(1-0) - \hat{j}(-2-1) + \hat{k}(0 - (-1)) = \hat{i} + 3\hat{j} + \hat{k}$. Num $= \sqrt{1+9+1} = \sqrt{11}$. Denom $= \sqrt{4+1+1} = \sqrt{6}$. Dist $= \sqrt{11/6}$.

Show that the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-4}{5} = \frac{y-1}{2} = z$ intersect.Solution: $A(1, 2, 3), B(4, 1, 0) \Rightarrow AB = (3, -1, -3)$. $\vec{b_1} = (2, 3, 4)$, $\vec{b_2} = (5, 2, 1)$. Determinant for numerator $= 3(3-8) - (-1)(2-20) + (-3)(4-15) = 3(-5) + 1(-18) - 3(-11) = -15 - 18 + 33 = 0$. Since SD $= 0$, they intersect.

Find the shortest distance between the lines passing through points $A(1, 2, 3), B(2, 4, 5)$ and $C(-1, 0, 1), D(0, 1, 2)$.Solution: Line 1 dir: $AB = (1, 2, 2)$. Line 2 dir: $CD = (1, 1, 1)$. Vector connecting lines: $AC = (-2, -2, -2)$. Since $AB \times CD = (0, 1, -1)$. Num $= (-2)(0) + (-2)(1) + (-2)(-1) = -2 + 2 = 0$. SD $= 0$ (The lines intersect).

Find the coordinates of the foot of the perpendicular drawn from the point $P(0, 2, 3)$ to the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$.Solution: Gen pt $F(5\lambda-3, 2\lambda+1, 3\lambda-4)$. D.R.s of PF: $(5\lambda-3, 2\lambda-1, 3\lambda-7)$. Perpendicular condition: $5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0 \Rightarrow 25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0 \Rightarrow 38\lambda = 38 \Rightarrow \lambda = 1$. Foot is $F(2, 3, -1)$.

Find the perpendicular distance from the point $(1, 0, 0)$ to the line $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8}$.Solution: Gen pt $F(2\lambda+1, -3\lambda-1, 8\lambda-10)$. PF D.R.s: $(2\lambda, -3\lambda-1, 8\lambda-10)$. Dot product: $2(2\lambda) - 3(-3\lambda-1) + 8(8\lambda-10) = 0 \Rightarrow 4\lambda + 9\lambda + 3 + 64\lambda - 80 = 0 \Rightarrow 77\lambda = 77 \Rightarrow \lambda = 1$. Foot $F(3, -4, -2)$. Distance $PF = \sqrt{(3-1)^2 + (-4-0)^2 + (-2-0)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.

Find the image of the point $(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.Solution: Gen pt $F(\lambda, 2\lambda+1, 3\lambda+2)$. PF D.R.s: $(\lambda-1, 2\lambda-5, 3\lambda-1)$. Dot product: $1(\lambda-1) + 2(2\lambda-5) + 3(3\lambda-1) = 0 \Rightarrow \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0 \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$. Foot $F(1, 3, 5)$. Image $I(a,b,c)$: $\frac{a+1}{2} = 1 \Rightarrow a=1$; $\frac{b+6}{2} = 3 \Rightarrow b=0$; $\frac{c+3}{2} = 5 \Rightarrow c=7$. Image is $(1, 0, 7)$.

Find the coordinates of the foot of the perpendicular drawn from the point $(2, 4, -1)$ to the line $\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9}$.Solution: Gen pt $F(\lambda-5, 4\lambda-3, -9\lambda+6)$. PF D.R.s: $(\lambda-7, 4\lambda-7, -9\lambda+7)$. Dot product: $1(\lambda-7) + 4(4\lambda-7) - 9(-9\lambda+7) = 0 \Rightarrow \lambda - 7 + 16\lambda - 28 + 81\lambda - 63 = 0 \Rightarrow 98\lambda = 98 \Rightarrow \lambda = 1$. Foot $F(-4, 1, -3)$.

Find the length of the perpendicular drawn from the origin to the line $\frac{x+2}{2} = \frac{y-3}{-2} = \frac{z}{-1}$.Solution: Gen pt $F(2\lambda-2, -2\lambda+3, -\lambda)$. OF D.R.s are same as F coords. Dot product: $2(2\lambda-2) - 2(-2\lambda+3) - 1(-\lambda) = 0 \Rightarrow 4\lambda - 4 + 4\lambda - 6 + \lambda = 0 \Rightarrow 9\lambda = 10 \Rightarrow \lambda = 10/9$. Foot $F(2/9, 7/9, -10/9)$. Dist $= \sqrt{(2/9)^2 + (7/9)^2 + (-10/9)^2} = \frac{\sqrt{4+49+100}}{9} = \frac{\sqrt{153}}{9}$.

Find the vector equation of a plane which is at a distance of $7$ units from the origin and normal to the vector $3\hat{i} + 5\hat{j} - 6\hat{k}$.Solution: Normal vector $\vec{n} = 3\hat{i} + 5\hat{j} - 6\hat{k}$. Unit normal $\hat{n} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{9+25+36}} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}}$. Equation: $\vec{r} \cdot \hat{n} = d \Rightarrow \vec{r} \cdot \left( \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}} \right) = 7$.

Find the Cartesian equation of the plane passing through the point $(2, 3, 4)$ and perpendicular to the vector $3\hat{i} - 2\hat{j} + 6\hat{k}$.Solution: Normal D.R.s are $3, -2, 6$. Point $(2, 3, 4)$. Eq: $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0 \Rightarrow 3(x-2) - 2(y-3) + 6(z-4) = 0 \Rightarrow 3x - 6 - 2y + 6 + 6z - 24 = 0 \Rightarrow 3x - 2y + 6z = 24$.

Find the equation of the plane passing through the non-collinear points $(1, 1, 0), (1, 2, 1),$ and $(-2, 2, -1)$.Solution: Eq: $\begin{vmatrix} x-1 & y-1 & z-0 \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = 0 \Rightarrow (x-1)(-1-1) - (y-1)(0+3) + z(0+3) = 0 \Rightarrow -2(x-1) - 3(y-1) + 3z = 0 \Rightarrow -2x + 2 - 3y + 3 + 3z = 0 \Rightarrow 2x + 3y - 3z = 5$.

Find the intercepts cut off by the plane $2x - 3y + 4z = 12$ on the coordinate axes.Solution: Divide by 12: $\frac{x}{6} - \frac{y}{4} + \frac{z}{3} = 1 \Rightarrow \frac{x}{6} + \frac{y}{-4} + \frac{z}{3} = 1$. Intercepts are $a=6, b=-4, c=3$.

Find the equation of the plane with intercepts $2, 3,$ and $-4$ on the $x, y,$ and $z$ axes respectively.Solution: $\frac{x}{2} + \frac{y}{3} + \frac{z}{-4} = 1$. Multiply by LCM 12: $6x + 4y - 3z = 12$.

Reduce the equation of the plane $x - 2y + 2z = 6$ to normal form, and hence find the perpendicular distance from the origin.Solution: Divide by $\sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3$. Eq becomes $\frac{1}{3}x - \frac{2}{3}y + \frac{2}{3}z = 2$. The perpendicular distance from origin is $2$.

Find the equation of the plane through $(1, 0, -2)$ and parallel to the plane $2x + 3y - z = 5$.Solution: Parallel plane has same normal D.R.s: $2x + 3y - z = k$. Passes through $(1, 0, -2) \Rightarrow 2(1) + 3(0) - (-2) = k \Rightarrow 2 + 2 = 4 \Rightarrow k = 4$. Eq: $2x + 3y - z = 4$.

Find the angle between the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$.Solution: Normals: $\vec{n_1} = (2, 1, -2)$, $\vec{n_2} = (3, -6, -2)$. $\cos\theta = \frac{|2(3) + 1(-6) + (-2)(-2)|}{\sqrt{4+1+4}\sqrt{9+36+4}} = \frac{|6 - 6 + 4|}{3 \times 7} = \frac{4}{21}$. $\theta = \cos^{-1}(4/21)$.

Find the angle between the line $\frac{x+1}{2} = \frac{y}{3} = \frac{z-3}{6}$ and the plane $10x + 2y - 11z = 3$.Solution: Line vector $\vec{b} = (2, 3, 6)$. Plane normal $\vec{n} = (10, 2, -11)$. $\sin\theta = \frac{|2(10) + 3(2) + 6(-11)|}{\sqrt{4+9+36}\sqrt{100+4+121}} = \frac{|20 + 6 - 66|}{7 \times 15} = \frac{|-40|}{105} = \frac{8}{21}$. $\theta = \sin^{-1}(8/21)$.

Find the coordinates of the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{1}$ and the plane $x - y + z = 5$.Solution: Gen pt $(3\lambda+2, 4\lambda-1, \lambda+2)$. Substitute into plane: $(3\lambda+2) - (4\lambda-1) + (\lambda+2) = 5 \Rightarrow 3\lambda + 2 - 4\lambda + 1 + \lambda + 2 = 5 \Rightarrow 5 = 5$. This identity means the line lies completely on the plane (infinite intersection points).

Find the coordinates of the point where the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ crosses the plane $2x + y + z = 7$.Solution: Line Eq: $\frac{x-3}{-1} = \frac{y+4}{1} = \frac{z+5}{6} = \lambda$. Gen pt $(-\lambda+3, \lambda-4, 6\lambda-5)$. Sub in plane: $2(-\lambda+3) + (\lambda-4) + (6\lambda-5) = 7 \Rightarrow -2\lambda+6+\lambda-4+6\lambda-5 = 7 \Rightarrow 5\lambda - 3 = 7 \Rightarrow 5\lambda = 10 \Rightarrow \lambda = 2$. Point is $(1, -2, 7)$.

Find the value of $k$ if the plane $x - 2y + kz = 4$ is parallel to the line $\frac{x}{2} = \frac{y-1}{3} = \frac{z+1}{4}$.Solution: For parallel, dot product of line vector and plane normal = 0. $2(1) + 3(-2) + 4(k) = 0 \Rightarrow 2 - 6 + 4k = 0 \Rightarrow 4k = 4 \Rightarrow k = 1$.

Show that the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) $ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$ are coplanar.Solution: $\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (0-1)\hat{j} + (-1+1)\hat{k} = 3\hat{i} - \hat{j}$. $\vec{b_1} = (3, -1, 0)$, $\vec{b_2} = (2, 0, 3)$. Determinant $|3 \quad -1 \quad 0| \atop |3 \quad -1 \quad 0| \atop |2 \quad \;\; 0 \quad 3|$ $= 0$ (Two identical rows). Thus, SD = 0, lines intersect, so they are coplanar.

Find the distance of the point $(2, 5, -3)$ from the plane $6x - 3y + 2z - 4 = 0$.Solution: $D = \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{36+9+4}} = \frac{|12 - 15 - 6 - 4|}{7} = \frac{|-13|}{7} = \frac{13}{7}$.

Find the distance between the parallel planes $2x - y + 3z + 4 = 0$ and $6x - 3y + 9z - 3 = 0$.Solution: Divide 2nd eq by 3: $2x - y + 3z - 1 = 0$. Now $d_1 = 4, d_2 = -1$. Distance $= \frac{|4 - (-1)|}{\sqrt{4+1+9}} = \frac{5}{\sqrt{14}}$.

Find the distance from the origin to the plane $3x - 4y + 12z = 26$.Solution: Eq: $3x - 4y + 12z - 26 = 0$. Distance $= \frac{|-26|}{\sqrt{9+16+144}} = \frac{26}{13} = 2$.

Find the perpendicular distance of the point $(1, 2, 3)$ from the plane $x + 2y - 3z + 5 = 0$.Solution: $D = \frac{|1(1) + 2(2) - 3(3) + 5|}{\sqrt{1+4+9}} = \frac{|1 + 4 - 9 + 5|}{\sqrt{14}} = \frac{1}{\sqrt{14}}$.