NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121296
Two circles \(S_1=\mathbf{p x}^2+\mathbf{p y}^2+2 \mathrm{~g}^{\prime} \mathbf{x}+\mathbf{2} \mathrm{f}^{\prime} \mathbf{y}+\mathbf{d}=\mathbf{0}\) and \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) have a common chord \(P Q\). The equation of \(P Q\) is
1 \(\mathrm{S}_1-\mathrm{S}_2=0\)
2 \(\mathrm{S}_1+\mathrm{S}_2=0\)
3 \(\mathrm{S}_1-\mathrm{pS}_2=0\)
4 \(\mathrm{S}_1+\mathrm{pS}_2=0\)
Explanation:
C Given, \(S_1=p x^2+p y^2+2 g x+2 f^{\prime} y+d=0\) \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) \(\mathrm{S}_1\) can be written as - \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2+\frac{2 \mathrm{~g}^{\prime}}{\mathrm{p}} \mathrm{x}+\frac{2 \mathrm{f}^{\prime}}{\mathrm{p}} \mathrm{y}+\frac{\mathrm{d}}{\mathrm{p}}=0\) Equation of common chord is - \(\mathrm{S}_1-\mathrm{S}_2=0\) \(2 \mathrm{x}\left(\frac{\mathrm{g}^{\prime}}{\mathrm{p}}-\mathrm{g}\right)+2 \mathrm{y}\left(\frac{\mathrm{f}^{\prime}}{\mathrm{p}}-\mathrm{f}\right)+\left(\frac{\mathrm{d}}{\mathrm{p}}-\mathrm{d}^{\prime}\right)=0\) \(2 \mathrm{x}\left(\mathrm{g}^{\prime}-\mathrm{gp}\right)+2 \mathrm{y}\left(\mathrm{f}^{\prime}-\mathrm{fp}\right)+\left(\mathrm{d}-\mathrm{d}^{\prime} \mathrm{p}\right)=0\) \(\mathrm{~S}_1-\mathrm{pS}_2=0\)
WB JEE-2022
Three Dimensional Geometry
121297
The Cartesian equation of a line \(2 x-3=3 y+1\) \(=5-6 \mathrm{z}\). The vector equation of the line passing through the point \((7,-5,0)\) and parallel to the given line is
C Given, the cartesian equation \(l_1: 2 \mathrm{x}-3=3 \mathrm{y}+1=5-6 \mathrm{z}\) \(\text { or } l_1:=\frac{x-3 / 2}{1 / 2}=\frac{y+1 / 3}{1 / 3}=\frac{z-5 / 6}{-1 / 6}\) Equation of required line passing through \((7,-5,0)\) And, parallel to \(l_1\) is - \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) \(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda\left(\frac{\hat{\mathrm{i}}}{2}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{6}\right)\) Or\(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121299
Equation of the plane passing through the intersection of the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}\) and \(\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}\) and parallel to the \(x y-\) plane is
1 \(z=4\)
2 \(z=2\)
3 \(\mathrm{z}=5\)
4 \(z=-5\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-5}{-3}=\lambda\) \(l_2: \frac{\mathrm{x}+5}{3}=\frac{\mathrm{y}-4}{-1}=\frac{\mathrm{z}+3}{4}=\mu\) Any point on \(l_1\) is \(\mathrm{M}(\lambda+1,2 \lambda+2,-3 \lambda+5)\) Any point on \(l_2\) is also \(M(3 \mu-5,-\mu+4,4 \mu-3)\) Since, both points are same, \(\therefore \lambda+1=3 \mu-5,2 \lambda+2=-\mu+4\) and \(-3 \lambda+5=4 \mu-3\) From above equations. \(\lambda=0 \text { and } \mu=2\) So, \(\mathrm{M}=(1,2,5)\) Hence, equation of plane containing \(\mathrm{M}\) and parallel to \(\mathrm{xy}\)-plane is - \(0(\mathrm{x}-1)+0(\mathrm{y}-2)+1(\mathrm{z}-5)=0\) \(\mathrm{z}-5=0\) \(\mathrm{z}=5\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121301
Find equation of the plane passing through the point \((2,1,3)\) and perpendicular to the planes \(x\) \(-2 y+2 z+3=0\) and \(3 x-2 y+4 z-4=0\).
1 \(2 x-y-2 z+3=0\)
2 \(x-2 y+2 z-3=0\)
3 \(2 x-y+2 z-3=0\)
4 \(2 x+y-2 z-3=0\)
Explanation:
A Given, the plane \(\mathrm{P}_1: \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+3=0\) \(\mathrm{P}_2: 3 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}-4=0\) Equation of required plane passing through \((2,1,3)\) is \(a(x-2)+b(y-1)+c(z-3)=0\) Also, it is perpendicular to \(P_1\) and \(P_2\) then, \(\mathrm{a}-2 \mathrm{~b}+2 \mathrm{c}=0\) and \(\quad 3 a-2 b+c=0\) From (ii) \& (iii) we get - \(-2 a-2 c=0\) \(a=-c\) \(a-2 b-2 a=0\) \(a=-2 b\) (put in (ii)) Putting, \(\mathrm{b}=\frac{-\mathrm{a}}{2}\) and \(\mathrm{c}=-\mathrm{a}\) (in equation (i)) \(a(x-2)-\frac{a}{2}(y-1)-a(z-3)=0\) \(x-2-\frac{y}{2}+\frac{1}{2}-z+3=0\) \(2 x-y-2 z+3=0\)
121296
Two circles \(S_1=\mathbf{p x}^2+\mathbf{p y}^2+2 \mathrm{~g}^{\prime} \mathbf{x}+\mathbf{2} \mathrm{f}^{\prime} \mathbf{y}+\mathbf{d}=\mathbf{0}\) and \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) have a common chord \(P Q\). The equation of \(P Q\) is
1 \(\mathrm{S}_1-\mathrm{S}_2=0\)
2 \(\mathrm{S}_1+\mathrm{S}_2=0\)
3 \(\mathrm{S}_1-\mathrm{pS}_2=0\)
4 \(\mathrm{S}_1+\mathrm{pS}_2=0\)
Explanation:
C Given, \(S_1=p x^2+p y^2+2 g x+2 f^{\prime} y+d=0\) \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) \(\mathrm{S}_1\) can be written as - \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2+\frac{2 \mathrm{~g}^{\prime}}{\mathrm{p}} \mathrm{x}+\frac{2 \mathrm{f}^{\prime}}{\mathrm{p}} \mathrm{y}+\frac{\mathrm{d}}{\mathrm{p}}=0\) Equation of common chord is - \(\mathrm{S}_1-\mathrm{S}_2=0\) \(2 \mathrm{x}\left(\frac{\mathrm{g}^{\prime}}{\mathrm{p}}-\mathrm{g}\right)+2 \mathrm{y}\left(\frac{\mathrm{f}^{\prime}}{\mathrm{p}}-\mathrm{f}\right)+\left(\frac{\mathrm{d}}{\mathrm{p}}-\mathrm{d}^{\prime}\right)=0\) \(2 \mathrm{x}\left(\mathrm{g}^{\prime}-\mathrm{gp}\right)+2 \mathrm{y}\left(\mathrm{f}^{\prime}-\mathrm{fp}\right)+\left(\mathrm{d}-\mathrm{d}^{\prime} \mathrm{p}\right)=0\) \(\mathrm{~S}_1-\mathrm{pS}_2=0\)
WB JEE-2022
Three Dimensional Geometry
121297
The Cartesian equation of a line \(2 x-3=3 y+1\) \(=5-6 \mathrm{z}\). The vector equation of the line passing through the point \((7,-5,0)\) and parallel to the given line is
C Given, the cartesian equation \(l_1: 2 \mathrm{x}-3=3 \mathrm{y}+1=5-6 \mathrm{z}\) \(\text { or } l_1:=\frac{x-3 / 2}{1 / 2}=\frac{y+1 / 3}{1 / 3}=\frac{z-5 / 6}{-1 / 6}\) Equation of required line passing through \((7,-5,0)\) And, parallel to \(l_1\) is - \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) \(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda\left(\frac{\hat{\mathrm{i}}}{2}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{6}\right)\) Or\(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121299
Equation of the plane passing through the intersection of the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}\) and \(\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}\) and parallel to the \(x y-\) plane is
1 \(z=4\)
2 \(z=2\)
3 \(\mathrm{z}=5\)
4 \(z=-5\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-5}{-3}=\lambda\) \(l_2: \frac{\mathrm{x}+5}{3}=\frac{\mathrm{y}-4}{-1}=\frac{\mathrm{z}+3}{4}=\mu\) Any point on \(l_1\) is \(\mathrm{M}(\lambda+1,2 \lambda+2,-3 \lambda+5)\) Any point on \(l_2\) is also \(M(3 \mu-5,-\mu+4,4 \mu-3)\) Since, both points are same, \(\therefore \lambda+1=3 \mu-5,2 \lambda+2=-\mu+4\) and \(-3 \lambda+5=4 \mu-3\) From above equations. \(\lambda=0 \text { and } \mu=2\) So, \(\mathrm{M}=(1,2,5)\) Hence, equation of plane containing \(\mathrm{M}\) and parallel to \(\mathrm{xy}\)-plane is - \(0(\mathrm{x}-1)+0(\mathrm{y}-2)+1(\mathrm{z}-5)=0\) \(\mathrm{z}-5=0\) \(\mathrm{z}=5\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121301
Find equation of the plane passing through the point \((2,1,3)\) and perpendicular to the planes \(x\) \(-2 y+2 z+3=0\) and \(3 x-2 y+4 z-4=0\).
1 \(2 x-y-2 z+3=0\)
2 \(x-2 y+2 z-3=0\)
3 \(2 x-y+2 z-3=0\)
4 \(2 x+y-2 z-3=0\)
Explanation:
A Given, the plane \(\mathrm{P}_1: \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+3=0\) \(\mathrm{P}_2: 3 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}-4=0\) Equation of required plane passing through \((2,1,3)\) is \(a(x-2)+b(y-1)+c(z-3)=0\) Also, it is perpendicular to \(P_1\) and \(P_2\) then, \(\mathrm{a}-2 \mathrm{~b}+2 \mathrm{c}=0\) and \(\quad 3 a-2 b+c=0\) From (ii) \& (iii) we get - \(-2 a-2 c=0\) \(a=-c\) \(a-2 b-2 a=0\) \(a=-2 b\) (put in (ii)) Putting, \(\mathrm{b}=\frac{-\mathrm{a}}{2}\) and \(\mathrm{c}=-\mathrm{a}\) (in equation (i)) \(a(x-2)-\frac{a}{2}(y-1)-a(z-3)=0\) \(x-2-\frac{y}{2}+\frac{1}{2}-z+3=0\) \(2 x-y-2 z+3=0\)
121296
Two circles \(S_1=\mathbf{p x}^2+\mathbf{p y}^2+2 \mathrm{~g}^{\prime} \mathbf{x}+\mathbf{2} \mathrm{f}^{\prime} \mathbf{y}+\mathbf{d}=\mathbf{0}\) and \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) have a common chord \(P Q\). The equation of \(P Q\) is
1 \(\mathrm{S}_1-\mathrm{S}_2=0\)
2 \(\mathrm{S}_1+\mathrm{S}_2=0\)
3 \(\mathrm{S}_1-\mathrm{pS}_2=0\)
4 \(\mathrm{S}_1+\mathrm{pS}_2=0\)
Explanation:
C Given, \(S_1=p x^2+p y^2+2 g x+2 f^{\prime} y+d=0\) \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) \(\mathrm{S}_1\) can be written as - \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2+\frac{2 \mathrm{~g}^{\prime}}{\mathrm{p}} \mathrm{x}+\frac{2 \mathrm{f}^{\prime}}{\mathrm{p}} \mathrm{y}+\frac{\mathrm{d}}{\mathrm{p}}=0\) Equation of common chord is - \(\mathrm{S}_1-\mathrm{S}_2=0\) \(2 \mathrm{x}\left(\frac{\mathrm{g}^{\prime}}{\mathrm{p}}-\mathrm{g}\right)+2 \mathrm{y}\left(\frac{\mathrm{f}^{\prime}}{\mathrm{p}}-\mathrm{f}\right)+\left(\frac{\mathrm{d}}{\mathrm{p}}-\mathrm{d}^{\prime}\right)=0\) \(2 \mathrm{x}\left(\mathrm{g}^{\prime}-\mathrm{gp}\right)+2 \mathrm{y}\left(\mathrm{f}^{\prime}-\mathrm{fp}\right)+\left(\mathrm{d}-\mathrm{d}^{\prime} \mathrm{p}\right)=0\) \(\mathrm{~S}_1-\mathrm{pS}_2=0\)
WB JEE-2022
Three Dimensional Geometry
121297
The Cartesian equation of a line \(2 x-3=3 y+1\) \(=5-6 \mathrm{z}\). The vector equation of the line passing through the point \((7,-5,0)\) and parallel to the given line is
C Given, the cartesian equation \(l_1: 2 \mathrm{x}-3=3 \mathrm{y}+1=5-6 \mathrm{z}\) \(\text { or } l_1:=\frac{x-3 / 2}{1 / 2}=\frac{y+1 / 3}{1 / 3}=\frac{z-5 / 6}{-1 / 6}\) Equation of required line passing through \((7,-5,0)\) And, parallel to \(l_1\) is - \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) \(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda\left(\frac{\hat{\mathrm{i}}}{2}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{6}\right)\) Or\(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121299
Equation of the plane passing through the intersection of the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}\) and \(\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}\) and parallel to the \(x y-\) plane is
1 \(z=4\)
2 \(z=2\)
3 \(\mathrm{z}=5\)
4 \(z=-5\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-5}{-3}=\lambda\) \(l_2: \frac{\mathrm{x}+5}{3}=\frac{\mathrm{y}-4}{-1}=\frac{\mathrm{z}+3}{4}=\mu\) Any point on \(l_1\) is \(\mathrm{M}(\lambda+1,2 \lambda+2,-3 \lambda+5)\) Any point on \(l_2\) is also \(M(3 \mu-5,-\mu+4,4 \mu-3)\) Since, both points are same, \(\therefore \lambda+1=3 \mu-5,2 \lambda+2=-\mu+4\) and \(-3 \lambda+5=4 \mu-3\) From above equations. \(\lambda=0 \text { and } \mu=2\) So, \(\mathrm{M}=(1,2,5)\) Hence, equation of plane containing \(\mathrm{M}\) and parallel to \(\mathrm{xy}\)-plane is - \(0(\mathrm{x}-1)+0(\mathrm{y}-2)+1(\mathrm{z}-5)=0\) \(\mathrm{z}-5=0\) \(\mathrm{z}=5\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121301
Find equation of the plane passing through the point \((2,1,3)\) and perpendicular to the planes \(x\) \(-2 y+2 z+3=0\) and \(3 x-2 y+4 z-4=0\).
1 \(2 x-y-2 z+3=0\)
2 \(x-2 y+2 z-3=0\)
3 \(2 x-y+2 z-3=0\)
4 \(2 x+y-2 z-3=0\)
Explanation:
A Given, the plane \(\mathrm{P}_1: \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+3=0\) \(\mathrm{P}_2: 3 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}-4=0\) Equation of required plane passing through \((2,1,3)\) is \(a(x-2)+b(y-1)+c(z-3)=0\) Also, it is perpendicular to \(P_1\) and \(P_2\) then, \(\mathrm{a}-2 \mathrm{~b}+2 \mathrm{c}=0\) and \(\quad 3 a-2 b+c=0\) From (ii) \& (iii) we get - \(-2 a-2 c=0\) \(a=-c\) \(a-2 b-2 a=0\) \(a=-2 b\) (put in (ii)) Putting, \(\mathrm{b}=\frac{-\mathrm{a}}{2}\) and \(\mathrm{c}=-\mathrm{a}\) (in equation (i)) \(a(x-2)-\frac{a}{2}(y-1)-a(z-3)=0\) \(x-2-\frac{y}{2}+\frac{1}{2}-z+3=0\) \(2 x-y-2 z+3=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Three Dimensional Geometry
121296
Two circles \(S_1=\mathbf{p x}^2+\mathbf{p y}^2+2 \mathrm{~g}^{\prime} \mathbf{x}+\mathbf{2} \mathrm{f}^{\prime} \mathbf{y}+\mathbf{d}=\mathbf{0}\) and \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) have a common chord \(P Q\). The equation of \(P Q\) is
1 \(\mathrm{S}_1-\mathrm{S}_2=0\)
2 \(\mathrm{S}_1+\mathrm{S}_2=0\)
3 \(\mathrm{S}_1-\mathrm{pS}_2=0\)
4 \(\mathrm{S}_1+\mathrm{pS}_2=0\)
Explanation:
C Given, \(S_1=p x^2+p y^2+2 g x+2 f^{\prime} y+d=0\) \(S_2=x^2+y^2+2 g x+2 f y+d^{\prime}=0\) \(\mathrm{S}_1\) can be written as - \(\mathrm{S}_1=\mathrm{x}^2+\mathrm{y}^2+\frac{2 \mathrm{~g}^{\prime}}{\mathrm{p}} \mathrm{x}+\frac{2 \mathrm{f}^{\prime}}{\mathrm{p}} \mathrm{y}+\frac{\mathrm{d}}{\mathrm{p}}=0\) Equation of common chord is - \(\mathrm{S}_1-\mathrm{S}_2=0\) \(2 \mathrm{x}\left(\frac{\mathrm{g}^{\prime}}{\mathrm{p}}-\mathrm{g}\right)+2 \mathrm{y}\left(\frac{\mathrm{f}^{\prime}}{\mathrm{p}}-\mathrm{f}\right)+\left(\frac{\mathrm{d}}{\mathrm{p}}-\mathrm{d}^{\prime}\right)=0\) \(2 \mathrm{x}\left(\mathrm{g}^{\prime}-\mathrm{gp}\right)+2 \mathrm{y}\left(\mathrm{f}^{\prime}-\mathrm{fp}\right)+\left(\mathrm{d}-\mathrm{d}^{\prime} \mathrm{p}\right)=0\) \(\mathrm{~S}_1-\mathrm{pS}_2=0\)
WB JEE-2022
Three Dimensional Geometry
121297
The Cartesian equation of a line \(2 x-3=3 y+1\) \(=5-6 \mathrm{z}\). The vector equation of the line passing through the point \((7,-5,0)\) and parallel to the given line is
C Given, the cartesian equation \(l_1: 2 \mathrm{x}-3=3 \mathrm{y}+1=5-6 \mathrm{z}\) \(\text { or } l_1:=\frac{x-3 / 2}{1 / 2}=\frac{y+1 / 3}{1 / 3}=\frac{z-5 / 6}{-1 / 6}\) Equation of required line passing through \((7,-5,0)\) And, parallel to \(l_1\) is - \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}\) \(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda\left(\frac{\hat{\mathrm{i}}}{2}+\frac{\hat{\mathrm{j}}}{3}-\frac{\hat{\mathrm{k}}}{6}\right)\) Or\(\overrightarrow{\mathrm{r}}=(7 \hat{\mathrm{i}}-5 \hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121299
Equation of the plane passing through the intersection of the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}\) and \(\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}\) and parallel to the \(x y-\) plane is
1 \(z=4\)
2 \(z=2\)
3 \(\mathrm{z}=5\)
4 \(z=-5\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{2}=\frac{\mathrm{z}-5}{-3}=\lambda\) \(l_2: \frac{\mathrm{x}+5}{3}=\frac{\mathrm{y}-4}{-1}=\frac{\mathrm{z}+3}{4}=\mu\) Any point on \(l_1\) is \(\mathrm{M}(\lambda+1,2 \lambda+2,-3 \lambda+5)\) Any point on \(l_2\) is also \(M(3 \mu-5,-\mu+4,4 \mu-3)\) Since, both points are same, \(\therefore \lambda+1=3 \mu-5,2 \lambda+2=-\mu+4\) and \(-3 \lambda+5=4 \mu-3\) From above equations. \(\lambda=0 \text { and } \mu=2\) So, \(\mathrm{M}=(1,2,5)\) Hence, equation of plane containing \(\mathrm{M}\) and parallel to \(\mathrm{xy}\)-plane is - \(0(\mathrm{x}-1)+0(\mathrm{y}-2)+1(\mathrm{z}-5)=0\) \(\mathrm{z}-5=0\) \(\mathrm{z}=5\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121301
Find equation of the plane passing through the point \((2,1,3)\) and perpendicular to the planes \(x\) \(-2 y+2 z+3=0\) and \(3 x-2 y+4 z-4=0\).
1 \(2 x-y-2 z+3=0\)
2 \(x-2 y+2 z-3=0\)
3 \(2 x-y+2 z-3=0\)
4 \(2 x+y-2 z-3=0\)
Explanation:
A Given, the plane \(\mathrm{P}_1: \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+3=0\) \(\mathrm{P}_2: 3 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}-4=0\) Equation of required plane passing through \((2,1,3)\) is \(a(x-2)+b(y-1)+c(z-3)=0\) Also, it is perpendicular to \(P_1\) and \(P_2\) then, \(\mathrm{a}-2 \mathrm{~b}+2 \mathrm{c}=0\) and \(\quad 3 a-2 b+c=0\) From (ii) \& (iii) we get - \(-2 a-2 c=0\) \(a=-c\) \(a-2 b-2 a=0\) \(a=-2 b\) (put in (ii)) Putting, \(\mathrm{b}=\frac{-\mathrm{a}}{2}\) and \(\mathrm{c}=-\mathrm{a}\) (in equation (i)) \(a(x-2)-\frac{a}{2}(y-1)-a(z-3)=0\) \(x-2-\frac{y}{2}+\frac{1}{2}-z+3=0\) \(2 x-y-2 z+3=0\)