Class 12: Three Dimensional Geometry – Miscellaneous Problems

$\displaystyle \textbf{Question 1: } \text{Prove that } (1,1,1) \text{ cannot be the direction cosine of the straight line.}$
$\displaystyle \text{} \quad \text{ISC Specimen 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know if }l,m,n\text{ are direction cosines, then }l^{2}+m^{2}+n^{2}=1$
$\displaystyle \text{But }1^{2}+1^{2}+1^{2}=3\neq1$
$\displaystyle \therefore (1,1,1)\text{ cannot be direction cosines of a line}$
$\\$

$\displaystyle \textbf{Question 2: } \text{Shown below is a cube. Find the equation of its diagonal which passes }$
$\displaystyle \text{through } \text{the centre of the cube and also the origin.} \quad \text{ISC Specimen 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Coordinates of }D=(0,0,0),\ F=(2,2,2)$
$\displaystyle \text{Centre lies on }DF\Rightarrow C=\left(\frac{2+0}{2},\frac{2+0}{2},\frac{2+0}{2}\right)=(1,1,1)$
$\displaystyle \text{Direction ratios of diagonal through origin and }C=(1,1,1)$
$\displaystyle \therefore \text{Equation: }\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}\Rightarrow x=y=z$
$\\$

$\displaystyle \textbf{Question 3: } \text{The equation of the line passing } (1,-1,0) \text{ and parallel to the line }$
$\displaystyle \text{} \frac{x-1}{1}=\frac{y+2}{-2}=\frac{z+1}{-1} \text{ is } \text{ISC Specimen Sem-I 2022}$
$\displaystyle \text{(a) } \frac{x-1}{1}=\frac{y+1}{-2}=\frac{z}{-1} \quad \text{(b) } \frac{x-1}{2}=\frac{y+2}{1}=\frac{z+1}{-3}$
$\displaystyle \text{(c) } \frac{x-6}{1}=\frac{y-2}{-2}=\frac{z+1}{3} \quad \text{(d) } \frac{x-2}{1}=\frac{y+2}{-2}=\frac{z+3}{-1}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given point }(1,-1,0)\text{ and line }\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z+1}{-1}$
$\displaystyle \text{Direction ratios }(1,-2,-1)$
$\displaystyle \therefore \text{Required line: }\frac{x-1}{1}=\frac{y+1}{-2}=\frac{z}{-1}$
$\\$

$\displaystyle \textbf{Question 4: } \text{What will be the angle between the two lines}$
$\displaystyle \frac{-x+2}{-2}=\frac{y-1}{7}=\frac{z+3}{-3} \text{ and } \frac{x+2}{-1}=\frac{2y-8}{4}=\frac{z-5}{4}?$
$\displaystyle \text{ISC Specimen Sem-I 2022}$
$\displaystyle \text{(a) } \frac{\pi}{2} \quad \text{(b) } \frac{\pi}{4} \quad \text{(c) } 0 \quad \text{(d) } \pi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Angle between lines: }$
$\displaystyle \cos\theta=\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\ \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$\displaystyle \text{Given lines: }\frac{x-2}{2}=\frac{y-1}{7}=\frac{z+3}{-3},\ \frac{x+2}{-1}=\frac{y-4}{2}=\frac{z-5}{4}$
$\displaystyle \cos\theta=\frac{2(-1)+7(2)+(-3)(4)}{\sqrt{2^{2}+7^{2}+(-3)^{2}}\sqrt{(-1)^{2}+2^{2}+4^{2}}}$
$\displaystyle =\frac{-2+14-12}{\sqrt{4+49+9}\sqrt{1+4+16}}=0$
$\displaystyle \therefore \theta=\cos^{-1}(0)=\frac{\pi}{2}$
$\\$

$\displaystyle \textbf{Question 5: } \text{What are the direction ratios of the line passing through two points }$
$\displaystyle \text{} (-2,4,5) \text{ and } (1,2,3)? \text{ ISC Specimen Sem-I 2022}$
$\displaystyle \text{(a) } <1,2,3> \quad \text{(b) } <-3,2,2>$
$\displaystyle \text{(c) } <2,4,5> \quad \text{(d) } <0,-1,4>$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Direction ratios of line }PQ$
$\displaystyle =\langle x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}\rangle$
$\displaystyle =\langle -2-1,\ 4-2,\ 5-3\rangle=\langle -3,2,2\rangle$
$\\$

$\displaystyle \textbf{Question 6: } \text{Show that } \frac{4-x}{-1}=\frac{y+3}{-4}=\frac{z+1}{7} \text{ and}$
$\displaystyle \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8} \text{ intersect each other. Also, find } \text{out the point of intersection.}$
$\displaystyle \quad \text{ISC Specimen 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } \frac{4-x}{-1}=\frac{y+3}{-4}=\frac{z+1}{7} \text{ and } \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$
$\displaystyle \text{Let } \frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}=\lambda$
$\displaystyle \Rightarrow x=\lambda+4,\; y=-4\lambda-3,\; z=7\lambda-1 \quad (i)$
$\displaystyle \text{Now, } \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\mu$
$\displaystyle \Rightarrow x=2\mu+1,\; y=-3\mu-1,\; z=8\mu-10 \quad (ii)$
$\displaystyle \text{Comparing } x: \lambda+4=2\mu+1 \Rightarrow 2\mu-\lambda=3 \quad (iii)$
$\displaystyle \text{Comparing } y: -4\lambda-3=-3\mu-1 \Rightarrow 3\mu-4\lambda=2 \quad (iv)$
$\displaystyle \text{Solving (iii) and (iv), } \lambda=1,\; \mu=2$
$\displaystyle \text{Substituting in (i), } x=5,\; y=-7,\; z=6$
$\displaystyle \text{Answer: Point of intersection is } (5,-7,6)$
$\\$

$\displaystyle \textbf{Question 7: } \text{Find the equation of the lines passing through the } \text{point } (2,1,3) \text{ and}$
$\displaystyle \text{perpendicular to the lines } \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} \text{ and } \frac{x}{-3}=\frac{y}{2}=\frac{z}{5}.$
$\displaystyle \quad \text{ISC Specimen 2021, 19}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Any line through the point } (2,1,3) \text{ is given by } \frac{x-2}{a}=\frac{y-1}{b}=\frac{z-3}{c}$
$\displaystyle \text{where } a,b,c \text{ are direction ratios}$
$\displaystyle \text{Now, the line is perpendicular to the lines } \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} \text{ and } \frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$
$\displaystyle \text{Hence, } (a,b,c)\cdot(1,2,3)=0 \text{ and } (a,b,c)\cdot(-3,2,5)=0$
$\displaystyle a+2b+3c=0 \quad \text{and} \quad -3a+2b+5c=0$
$\displaystyle \text{Solving, ratios of } a:b:c \text{ are obtained}$
$\displaystyle \text{Answer: Required direction ratios satisfy above equations}$
$\displaystyle \text{where direction ratios of these two lines are } (1,2,3) \text{ and } (-3,2,5), \text{ respectively.}$
$\displaystyle \therefore a+2b+3c=0 \quad ...(ii)$
$\displaystyle \text{and } -3a+2b+5c=0 \quad ...(iii)$
$\displaystyle \text{If two lines having DR's } (a_1,b_1,c_1) \text{ and } (a_2,b_2,c_2) \text{ are perpendicular, then } a_1a_2+b_1b_2+c_1c_2=0$
$\displaystyle \text{By cross-multiplication method, we get } \frac{a}{10-6}=\frac{-b}{5+9}=\frac{c}{2+6}$
$\displaystyle \Rightarrow \frac{a}{4}=\frac{b}{-14}=\frac{c}{8}$
$\displaystyle \Rightarrow \frac{a}{2}=\frac{b}{-7}=\frac{c}{4}=\lambda$
$\displaystyle \Rightarrow a=2\lambda,\; b=-7\lambda,\; c=4\lambda$
$\displaystyle \text{On putting values in Eq. (i), we get } \frac{x-2}{2\lambda}=\frac{y-1}{-7\lambda}=\frac{z-3}{4\lambda}$
$\displaystyle \Rightarrow \frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}$
$\displaystyle \text{which is the required cartesian equation of the line.}$
$\displaystyle \text{The required vector equation of line which passes through } (2,1,3) \text{ and parallel to } (2\hat{i}-7\hat{j}+4\hat{k}) \text{ is }$
$\displaystyle \overrightarrow{r}=(2\hat{i}+\hat{j}+3\hat{k})+\lambda(2\hat{i}-7\hat{j}+4\hat{k})$
$\displaystyle \text{Answer: } \overrightarrow{r}=(2\hat{i}+\hat{j}+3\hat{k}) +\lambda(2\hat{i}-7\hat{j}+4\hat{k})$
$\\$

$\displaystyle \textbf{Question 8: } \text{Find the shortest distance between the lines}$
$\displaystyle \frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{7} \text{ and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}.$
$\displaystyle \quad \text{ISC Specimen 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, lines are } \frac{x-8}{3}=\frac{y+9}{-16}=\frac{z-10}{7} \text{ and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$
$\displaystyle \text{Here, } x_1=8,\; y_1=-9,\; z_1=10,\; a_1=3,\; b_1=-16,\; c_1=7$
$\displaystyle \text{and } x_2=15,\; y_2=29,\; z_2=5,\; a_2=3,\; b_2=8,\; c_2=-5$
$\displaystyle \text{Shortest distance between two lines is given by }$
$\displaystyle \frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \end{vmatrix}} {\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}$
$\displaystyle =\frac{\begin{vmatrix} 7 & 38 & -5\\ 3 & -16 & 7\\ 3 & 8 & -5 \end{vmatrix}} {\sqrt{(80-56)^2+(21+15)^2+(24+48)^2}}$
$\displaystyle =\frac{7((-16)(-5)-7\cdot 8)-38(3(-5)-7\cdot 3) -5(3\cdot 8-(-16)\cdot 3)}{\sqrt{(24)^2+(36)^2+(72)^2}}$
$\displaystyle =\frac{7(80-56)-38(-15-21)-5(24+48)} {\sqrt{576+1296+5184}}$
$\displaystyle =\frac{7(24)+38(36)-5(72)}{\sqrt{7056}}$
$\displaystyle =\frac{168+1368-360}{84}=\frac{1176}{84}=14$
$\displaystyle \text{Answer: Shortest distance }=14$
$\\$

$\displaystyle \textbf{Question 9: } \text{Find the intercepts on the coordinate axes cut off by}$
$\displaystyle \text{the plane } \overrightarrow{r}\cdot(2\overrightarrow{i}-4\overrightarrow{j}+3\overrightarrow{k})=12. \quad \text{ISC Specimen 2024}$
$\displaystyle \text{(a) } a=6, b=-3, c=4 \quad \text{(b) } a=6, b=3, c=4$
$\displaystyle \text{(c) } a=-6, b=-3, c=4 \quad \text{(d) } a=6, b=-3, c=-4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, } \overrightarrow{r}\cdot(2\hat{i}-4\hat{j}+3\hat{k})=12$
$\displaystyle \Rightarrow (x\hat{i}+y\hat{j}+z\hat{k})\cdot(2\hat{i}-4\hat{j}+3\hat{k})=12$
$\displaystyle \Rightarrow 2x-4y+3z=12$
$\displaystyle \Rightarrow \frac{x}{6}+\frac{y}{-3}+\frac{z}{4}=1$
$\displaystyle \text{Intercepts on the coordinate axes are } a=6,\; b=-3,\; c=4$
$\displaystyle \text{Answer: } a=6,\; b=-3,\; c=4$
$\\$

$\displaystyle \textbf{Question 10: } \text{The planes } 2x-y+4z=5 \text{ and } 5x-2.5y+10z=6 \text{ are}$
$\displaystyle \text{ISC Specimen 2023}$
$\displaystyle \text{(a) parallel} \quad \text{(b) intersect on } Y\text{-axis}$
$\displaystyle \text{(c) perpendicular} \quad \text{(d) pass through } (0,0,\frac{5}{4})$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, planes } 2x-y+4z=5 \text{ and } 5x-2.5y+10z=6$
$\displaystyle \text{Multiplying first equation by 2: } 10x-5y+20z=10$
$\displaystyle \text{Dividing second equation by 5: } 2x-y+4z=\frac{12}{5}$
$\displaystyle \text{Since coefficients of } x,y,z \text{ are proportional, planes are parallel}$
$\displaystyle \text{Answer: Planes are parallel}$
$\\$

$\displaystyle \textbf{Question 11: } \text{Find the equation of the plane with intercept } 3 \text{ on the}$
$\displaystyle Y\text{-axis and parallel to } XZ\text{-plane.} \quad \text{ISC Specimen 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of plane } ZOX \text{ is } y=0$
$\displaystyle \text{Any plane parallel to it is of the form } y=a$
$\displaystyle \text{Since } y\text{-intercept is } 3$
$\displaystyle \Rightarrow a=3$
$\displaystyle \text{Required plane is } y=3$
$\displaystyle \text{Answer: } y=3$
$\\$

$\displaystyle \textbf{Question 12: } \text{If the intercept form of the equation of the plane}$
$\displaystyle 2x-3y+4z=12 \text{ is } \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, \text{ then the values of}$
$\displaystyle a,b,c \text{ are respectively.} \quad \text{ISC Specimen Sem-II 2022}$
$\displaystyle \text{(a) } a=6, b=-4, c=3 \quad \text{(b) } a=-6, b=-4, c=3$
$\displaystyle \text{(c) } a=6, b=4, c=3 \quad \text{(d) } a=6, b=4, c=-3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Equation of plane } 2x-3y+4z=12$
$\displaystyle \Rightarrow \frac{2x}{12}-\frac{3y}{12}+\frac{4z}{12}=1$
$\displaystyle \Rightarrow \frac{x}{6}+\frac{y}{-4}+\frac{z}{3}=1$
$\displaystyle \text{Hence } a=6,\; b=-4,\; c=3$
$\displaystyle \text{Answer: } a=6,\; b=-4,\; c=3$
$\\$

$\displaystyle \textbf{Question 13: } \text{The distance of the plane whose equation is given by}$
$\displaystyle 3x-4y+12z=3, \text{ from the origin will be} \quad \text{ISC Specimen Sem-II 2022}$
$\displaystyle \text{(a) } \frac{3}{13} \quad \text{(b) } -\frac{2}{13} \quad \text{(c) } -3 \quad \text{(d) } \frac{13}{19}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, equation of plane is } 3x-4y+12z=3.$
$\displaystyle \text{Distance of } (x_1,y_1,z_1) \text{ from the plane } Ax+By+Cz+D=0 \text{ is}$
$\displaystyle D'=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$
$\displaystyle \text{Here, } (x_1,y_1,z_1)=(0,0,0)$
$\displaystyle A=3,\; B=-4,\; C=12,\; D=-3$
$\displaystyle \text{Distance }(D')=\left|\frac{3\times0+(-4)\times0+12\times0+(-3)}{\sqrt{3^2+(-4)^2+12^2}}\right|$
$\displaystyle =\frac{|-3|}{\sqrt{9+16+144}}=\frac{3}{\sqrt{169}}=\frac{3}{13}\text{ units}$
$\displaystyle \text{Answer: } \frac{3}{13}\text{ units}$
$\\$

$\displaystyle \textbf{Question 14: } \text{The equation of the plane which is parallel to}$
$\displaystyle 2x-3y+z=0 \text{ and which passes through } (1,-1,2) \text{ is}$
$\displaystyle \quad \text{ISC Specimen Sem-II 2022}$
$\displaystyle \text{(a) } 2x-3y+z-7=0 \quad \text{(b) } 2x-3y+z+7=0$
$\displaystyle \text{(c) } 2x-3y+z-8=0 \quad \text{(d) } 2x-3y+z+6=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Any plane parallel to the given plane is } 2x-3y+z+k=0$
$\displaystyle \text{If it passes through the point } (1,-1,2), \text{ then}$
$\displaystyle 2(1)-3(-1)+2+k=0$
$\displaystyle \Rightarrow 2+3+2+k=0$
$\displaystyle \Rightarrow k=-7$
$\displaystyle \text{Hence, required equation of plane is } 2x-3y+z-7=0$
$\displaystyle \text{Answer: } 2x-3y+z-7=0$
$\\$

$\displaystyle \textbf{Question 15: } \text{The intercepts made on the coordinate axes by the}$
$\displaystyle \text{plane } 2x+y-2z=3 \text{ are} \quad \text{ISC Specimen Sem-II 2022}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, } 2x+y-2z=3$
$\displaystyle \Rightarrow \frac{2x}{3}+\frac{y}{3}-\frac{2z}{3}=1$
$\displaystyle \Rightarrow \frac{x}{\frac{3}{2}}+\frac{y}{3}+\frac{z}{-\frac{3}{2}}=1$
$\displaystyle \therefore \text{Intercept on X-axis }=\frac{3}{2}$
$\displaystyle \text{Intercept on Y-axis }=3$
$\displaystyle \text{Intercept on Z-axis }=-\frac{3}{2}$
$\displaystyle \text{Answer: } \frac{3}{2},\;3,\;-\frac{3}{2}$
$\\$

$\displaystyle \textbf{Question 16: } \text{The intercepts made by the plane } 3x-2y+4z=12 \text{ on}$
$\displaystyle \text{the coordinate axes are} \quad \text{ISC Specimen 2021}$
$\displaystyle \text{(a) } 6,-4,3 \quad \text{(b) } 4,-6,3 \quad \text{(c) } 2,-3,4 \quad \text{(d) } \frac{1}{4},-\frac{1}{6},\frac{1}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given that, } 3x-2y+4z=12$
$\displaystyle \Rightarrow \frac{3x}{12}+\frac{-2y}{12}+\frac{4z}{12}=1$
$\displaystyle \Rightarrow \frac{x}{4}+\frac{y}{-6}+\frac{z}{3}=1$
$\displaystyle \text{Hence, intercepts are } 4,-6,3$
$\displaystyle \text{Answer: } 4,-6,3$
$\\$

$\displaystyle \textbf{Question 17: } \text{Find the equation of the plane passing through}$
$\displaystyle (-2,1,3) \text{ and perpendicular to the line having direction}$
$\displaystyle \text{ratios } <3,1,5>. \quad \text{ISC Specimen 2021}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, } x_1=-2,\; y_1=1,\; z_1=3$
$\displaystyle \text{and } a=3,\; b=1,\; c=5$
$\displaystyle \text{So, required equation of plane is }$
$\displaystyle a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
$\displaystyle \Rightarrow 3(x+2)+1(y-1)+5(z-3)=0$
$\displaystyle \Rightarrow 3x+6+y-1+5z-15=0$
$\displaystyle \Rightarrow 3x+y+5z-10=0$
$\displaystyle \Rightarrow 3x+y+5z=10$
$\displaystyle \text{Answer: } 3x+y+5z=10$
$\\$

$\displaystyle \textbf{Question 18: } \text{Find the angle between the line}$
$\displaystyle \frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} \text{ and the plane } 2x+3y-5z=4. \quad \text{ISC Specimen 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, the line } \frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$
$\displaystyle \text{and plane } 2x+3y-5z=4$
$\displaystyle \text{Here, } \overrightarrow{b}=2\hat{i}+3\hat{j}+6\hat{k} \text{ and } \overrightarrow{n}=2\hat{i}+3\hat{j}-5\hat{k}$
$\displaystyle \text{Angle between line and plane is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot\overrightarrow{n}}{|\overrightarrow{b}||\overrightarrow{n}|}$
$\displaystyle =\frac{(2\hat{i}+3\hat{j}+6\hat{k})\cdot(2\hat{i}+3\hat{j}-5\hat{k})}{|2\hat{i}+3\hat{j}+6\hat{k}||2\hat{i}+3\hat{j}-5\hat{k}|}$
$\displaystyle \sin\theta=\frac{4+9-30}{\sqrt{4+9+36}\sqrt{4+9+25}}=\frac{-17}{7\sqrt{38}}$
$\displaystyle \Rightarrow \theta=\sin^{-1}\left(\frac{-17}{7\sqrt{38}}\right)$
$\displaystyle \text{Answer: } \theta=\sin^{-1}\left(\frac{-17}{7\sqrt{38}}\right)$
$\\$

$\displaystyle \textbf{Question 19: } \text{Find the equation of the plane passing through the}$
$\displaystyle \text{points } (-2,6,6), (1,-1,0) \text{ and } (1,2,-1). \quad \text{ISC Specimen Sem-II 2022}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } A(-2,6,6),\; B(1,-1,0) \text{ and } C(1,2,-1).$
$\displaystyle \text{Here, DR's of } AB \text{ are } (3,-7,-6)$
$\displaystyle \text{DR's of } BC \text{ are } (0,3,-1)$
$\displaystyle \therefore \text{The points are non-collinear.}$
$\displaystyle \text{Since, cartesian equation of plane passing through three points } A(x_1,y_1,z_1),$
$\displaystyle B(x_2,y_2,z_2) \text{ and } C(x_3,y_3,z_3) \text{ is given by}$
$\displaystyle \begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x+2&y-6&z-6\\3&-7&-6\\3&-4&-7\end{vmatrix}=0$
$\displaystyle \Rightarrow (x+2)[49-24]-(y-6)[-21+18]+(z-6)[-12+21]=0$
$\displaystyle \Rightarrow (x+2)(25)+3(y-6)+9(z-6)=0$
$\displaystyle \Rightarrow 25x+3y+9z=22$
$\displaystyle \text{Answer: } 25x+3y+9z=22$
$\\$

$\displaystyle \textbf{Question 20: } \text{Find the equation of the plane passing through the}$
$\displaystyle \text{point } (1,1,1) \text{ and is perpendicular to the line}$
$\displaystyle \frac{x-1}{3}=\frac{y-2}{0}=\frac{z-3}{4}. \text{ Also, find the distance of this}$
$\displaystyle \text{plane from the origin.} \quad \text{ISC Specimen Sem-II 2022}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{ The required plane is perpendicular to the line } \frac{x-1}{3}=\frac{y-2}{0}=\frac{z-3}{4}$
$\displaystyle \therefore \text{Its direction ratios are } (3,0,4)$
$\displaystyle \text{Now, equation of plane passing through } (1,1,1) \text{ and having DR's } 3,0,4$
$\displaystyle \text{is given by } 3(x-1)+0(y-1)+4(z-1)=0$
$\displaystyle \Rightarrow 3x-3+4z-4=0$
$\displaystyle \Rightarrow 3x+4z-7=0$
$\displaystyle \therefore \text{Distance of this plane from the origin is}$
$\displaystyle \left|\frac{3(0)+4(0)-7}{\sqrt{9+16}}\right|$
$\displaystyle =\frac{7}{5}\text{ units}$
$\displaystyle \text{Answer: } \frac{7}{5}\text{ units}$
$\\$

$\displaystyle \textbf{Question 121: } \text{Find the cartesian equation of the line passing through}$
$\displaystyle \text{the points } (-1,0,2) \text{ and } (3,4,6). \quad \text{ISC Specimen 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of line passing through } (-1,0,2) \text{ and } (3,4,6) \text{ is }$
$\displaystyle \frac{x+1}{3+1}=\frac{y-0}{4-0}=\frac{z-2}{6-2}$
$\displaystyle \Rightarrow \frac{x+1}{4}=\frac{y}{4}=\frac{z-2}{4}$
$\displaystyle \Rightarrow x+1=y=z-2$
$\displaystyle \text{Answer: } x+1=y=z-2$
$\\$

$\displaystyle \textbf{Question 22: } \text{Shown below are equations of two planes}$
$\displaystyle \text{Plane 1: } \frac{x}{-2}+\frac{3y}{4}-\frac{5z}{4}=1$
$\displaystyle \text{Plane 2: } \frac{x}{3}-\frac{y}{4}-\frac{z}{1}=1$
$\displaystyle \text{Is the point } P(2,-3,4) \text{ closer to plane 1 or plane 2?}$
$\displaystyle \text{Find the angle between the planes.} \quad \text{ISC Specimen 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, equation of two planes can be rewritten as }$
$\displaystyle -2x+3y-5z=4 \text{ and } 4x-3y-12z=12$
$\displaystyle \text{Given point is } (2,-3,4)$
$\displaystyle \text{Required distance }=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}$
$\displaystyle =\frac{|-2(2)+3(-3)-5(4)-4|}{\sqrt{4+9+25}}$
$\displaystyle =\frac{|-4-9-20-4|}{\sqrt{38}}=\frac{37}{\sqrt{38}}$
$\displaystyle \text{and for second plane }$
$\displaystyle =\frac{|4(2)+(-3)(-3)-12(4)-12|}{\sqrt{16+9+144}}$
$\displaystyle =\frac{|8+9-48-12|}{13}=\frac{43}{13}$
$\displaystyle \text{Answer: } \frac{37}{\sqrt{38}},\; \frac{43}{13}$
$\displaystyle \text{Hence, the point } P \text{ is closer to plane 2.}$
$\displaystyle \text{Now, given equations of planes are } -2x+3y-5z-4=0 \text{ and } 4x-3y-12z-12=0$
$\displaystyle \text{Comparing with } a_1x+b_1y+c_1z+d_1=0 \text{ and } a_2x+b_2y+c_2z+d_2=0,$
$\displaystyle a_1=-2,\; b_1=3,\; c_1=-5 \text{ and } a_2=4,\; b_2=-3,\; c_2=-12$
$\displaystyle \text{Angle between two planes is given by }$
$\displaystyle \theta=\cos^{-1}\left(\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right)$
$\displaystyle =\cos^{-1}\left(\frac{(-2)(4)+(3)(-3)+(-5)(-12)}{\sqrt{4+9+25}\sqrt{16+9+144}}\right)$
$\displaystyle =\cos^{-1}\left(\frac{-8-9+60}{\sqrt{38}\sqrt{169}}\right)$
$\displaystyle =\cos^{-1}\left(\frac{43}{13\sqrt{38}}\right)$
$\displaystyle \text{Answer: } \theta=\cos^{-1}\left(\frac{43}{13\sqrt{38}}\right)$
$\\$

$\displaystyle \textbf{Question 23: } \text{A line with positive direction cosines passes through}$
$\displaystyle \text{the point } P(-2,3,2) \text{ and makes equal angles with the}$
$\displaystyle \text{coordinate axes. The line passes through the plane}$
$\displaystyle 3x-4y+2z=11 \text{ at point } Q.$
$\displaystyle \text{Find the length of the line segment } PQ. \quad \text{ISC Specimen 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, a line with position direction cosines passes through } P(-2,3,2)$
$\displaystyle \therefore l=\cos\alpha,\; m=\cos\beta,\; n=\cos\gamma$
$\displaystyle \text{We know that } l^2+m^2+n^2=1$
$\displaystyle \Rightarrow \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\displaystyle \text{Since all are equal, } 3\cos^2\alpha=1$
$\displaystyle \Rightarrow \cos^2\alpha=\frac{1}{3} \Rightarrow \cos\alpha=\frac{1}{\sqrt{3}}$
$\displaystyle \therefore \text{Equation of line is } \frac{x+2}{\frac{1}{\sqrt{3}}} =\frac{y-3}{\frac{1}{\sqrt{3}}}=\frac{z-2}{\frac{1}{\sqrt{3}}}$
$\displaystyle \Rightarrow \frac{x+2}{1}=\frac{y-3}{1}=\frac{z-2}{1}=K$
$\displaystyle \text{Any point on the line is } Q(K-2,K+3,K+2)$
$\displaystyle \text{Point } Q \text{ lies on plane } 3x-4y+2z=11$
$\displaystyle \Rightarrow 3(K-2)-4(K+3)+2(K+2)=11$
$\displaystyle \Rightarrow 3K-6-4K-12+2K+4=11$
$\displaystyle \Rightarrow K-14=11 \Rightarrow K=25$
$\displaystyle \therefore Q=(23,28,27)$
$\displaystyle \text{Length of } PQ=\sqrt{(23+2)^2+(28-3)^2+(27-2)^2}$
$\displaystyle =\sqrt{25^2+25^2+25^2}=\sqrt{3\times25^2}=25\sqrt{3}$
$\displaystyle \text{Answer: } 25\sqrt{3}$
$\\$

$\displaystyle \textbf{Question 24: } \text{Find the equation of the plane passing through the}$
$\displaystyle \text{points } (0,4,-3) \text{ and } (6,-4,3), \text{ if the sum of their}$
$\displaystyle \text{intercepts on three axes is } 0. \quad \text{ISC Specimen 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let equation of plane be } \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\displaystyle \text{As } (0,4,-3) \text{ lies on it, } \frac{4}{b}-\frac{3}{c}=1 \quad ...(i)$
$\displaystyle \text{and } (6,-4,3) \text{ lies on it, } \frac{6}{a}-\frac{4}{b}+\frac{3}{c}=1 \quad ...(ii)$
$\displaystyle \text{On adding (i) and (ii), we get } \frac{6}{a}=2 \Rightarrow a=3$
$\displaystyle \text{Also } a+b+c=0 \Rightarrow b+c=-3 \Rightarrow c=-3-b$
$\displaystyle \text{Substituting in (i), } \frac{4}{b}+\frac{3}{3+b}=1$
$\displaystyle \Rightarrow 4(3+b)+3b=b(3+b)$
$\displaystyle \Rightarrow 12+4b+3b=3b+b^2$
$\displaystyle \Rightarrow b^2-4b-12=0$
$\displaystyle \Rightarrow (b-6)(b+2)=0$
$\displaystyle \Rightarrow b=6 \text{ or } b=-2$
$\displaystyle \text{Answer: } a=3,\; b=6 \text{ or } -2$
$\displaystyle \text{Therefore, } b=6 \text{ or } b=-2$
$\displaystyle \text{Hence, } c=-9 \text{ or } c=-1$
$\displaystyle \text{Thus, equations of planes are } \frac{x}{3}+\frac{y}{6}+\frac{z}{-9}=1 \text{ or } \frac{x}{3}+\frac{y}{-2}+\frac{z}{-1}=1$
$\displaystyle \text{Answer: } \frac{x}{3}+\frac{y}{6}-\frac{z}{9}=1 \text{ or } \frac{x}{3}-\frac{y}{2}-z=1$
$\\$

$\displaystyle \textbf{Question 25: } \text{Find the distance of the point } A(3,4,4) \text{ from the point,}$
$\displaystyle \text{where the line joining points } P(3,-4,-5) \text{ and } Q(2,-3,1)$
$\displaystyle \text{intersects the plane } 2x+y+z=7. \quad \text{ISC Specimen 2021}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Direction ratios of line joining } P(3,-4,-5) \text{ and } Q(2,-3,1) \text{ are } (2-3,-3+4,1+5)=(-1,1,6)$
$\displaystyle \text{Equation of line through } (3,-4,-5) \text{ is }$
$\displaystyle \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$
$\displaystyle \text{Let } \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda$
$\displaystyle \Rightarrow x=3-\lambda,\; y=\lambda-4,\; z=6\lambda-5$
$\displaystyle \text{General point is } (3-\lambda,\lambda-4,6\lambda-5)$
$\displaystyle \text{This lies on plane } 2x+y+z=7$
$\displaystyle \Rightarrow 2(3-\lambda)+(\lambda-4)+(6\lambda-5)=7$
$\displaystyle \Rightarrow 6-2\lambda+\lambda-4+6\lambda-5=7$
$\displaystyle \Rightarrow 5\lambda-3=7 \Rightarrow \lambda=2$
$\displaystyle \text{Point of intersection is } (1,-2,7)$
$\displaystyle \text{Distance between } (3,4,4) \text{ and } (1,-2,7)$
$\displaystyle =\sqrt{(3-1)^2+(4+2)^2+(4-7)^2}$
$\displaystyle =\sqrt{4+36+9}=\sqrt{49}=7$
$\displaystyle \text{Answer: } 7$
$\\$