121327
The angle between the planes \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\)
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given, \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\) On comparing with \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}\), we get - \(a_1=3, b_1=-4, c_1=5\) And, \(\mathrm{a}_2=2, \mathrm{~b}_2=-1, \mathrm{c}_2=2\) Angle between planes is \(\cos \theta=\frac{(3)(2)+(-4)(-1)+(5)(-2)}{\left(\sqrt{3^2+(-4)^2+(5)^2}\right)\left(\sqrt{(2)^2+(-1)^2+(-2)^2}\right)}\) \(\Rightarrow \quad \cos \theta=\frac{6+4-10}{(\sqrt{9+16+25})(\sqrt{4+1+4})}=0\) \(\Rightarrow \quad \cos \theta=0\) \(\Rightarrow \quad \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2}\)
CG PET- 2017
Three Dimensional Geometry
121329
The acute angle between the planes \(P_1\) and \(P_2\), when \(P_1\) and \(P_2\) are the planes passing through the intersection of the planes \(5 x+8 y+13 z-29\) \(=0\) and \(8 x-7 y+z-20=0\) and the points \((2,1\), 3 ) and \((0,1,2)\) respectively, is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{12}\)
Explanation:
A Equation of plane passing through the intersection of planes \(5 x+8 y+13 z-29=0\) and \(8 x-\) \(7 \mathrm{y}+\mathrm{z}-20=0\) is \(5 x+8 y+3 z-29+\lambda(8 x-7 y+z-20)=0\) and if it is passing through \((2,1,3)\) then \(\lambda=\frac{7}{2}\). \(P_1\) : Equation of plane through intersection of \(5 x+8 y+\) \(3 z-19+\frac{7}{2}(8 x-7 y+z-20)=0\) \(\Rightarrow \quad 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=6\) Similarly \(\mathrm{P}_2\) : Equation of plane through intersection of \(5 x+8 y+13 z-29=0\) and \(8 x-7 y+z-20=0\) and the point \((0,1,2)\) is \(\Rightarrow \quad \mathrm{x}+\mathrm{y}+2 \mathrm{z}=5\) Angle between planes \(=\theta=\cos ^{-1}\left(\frac{3}{\sqrt{6} \sqrt{6}}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)
JEE Main-28.06.2022
Three Dimensional Geometry
121331
A plane is making intercepts \(2,3,4\) on \(X, Y\) and Z-axes respectively. Another plane is passing through the point \((-1,6,2)\) and is perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,4)\). Then angle between the two planes is
1 \(90^{\circ}\)
2 \(\cos ^{-1} \sqrt{\frac{12}{61}}\)
3 \(\cos ^{-1} \sqrt{\frac{11}{61}}\)
4 \(\cos ^{-1} \sqrt{\frac{5}{6}}\)
Explanation:
C Equation of plane ' \(\mathrm{P}_1\) ' making intercept 2, 3, 4 on \(X, Y\) and \(Z\) axes is \(\mathrm{P}_1: \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{4}=1\)
or \(6 x+4 y+3 z=12\)
Line \(\overrightarrow{\mathrm{AB}}\) is \(=(-2-1) \hat{i}+(3-2) \hat{j}+(4-3) \hat{k}\)
(Given A \& B)
\(\overrightarrow{\mathrm{AB}} \text { or } \overrightarrow{\mathrm{n}}=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Now, equation of plane \(\left(\mathrm{P}_2\right)\) containing point \((-1,6,2) \mid\) and perpendicular to \(\overrightarrow{\mathrm{AB}}\) is-
\(-3(\mathrm{x}+1)+1(\mathrm{y}-6)+1(\mathrm{z}-2)=0\)
\(\Rightarrow -3 \mathrm{x}-3+\mathrm{y}-6+\mathrm{z}-2=0\)
\(\Rightarrow -3 \mathrm{x}+\mathrm{y}+\mathrm{z}=11\)
Now, angle between \(P_1\) and \(P_2\) is -
\(\cos \theta =\left \vert\frac{(6 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+\hat{k})}{\sqrt{36+16+9} \cdot \sqrt{9+1+1}}\right \vert\)
\(\cos \theta =\left \vert\left(\frac{-18+4+3}{\sqrt{61} \cdot \sqrt{11}}\right)\right \vert=\left \vert\frac{-11}{\sqrt{61} \cdot \sqrt{11}}\right \vert\)
\(\Rightarrow \quad \theta =\cos ^{-1}\left(\sqrt{\frac{11}{61}}\right)\)
AP EAMCET-20.04.2019
Three Dimensional Geometry
121333
The angle between the line \(x-2 y+z=0=x+\) \(2 y-2 z\) and the plane \(5 x-2 y-z+17=0\) is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(0^0\)
Explanation:
D Given, Plane \(5 x-2 y-z+17=0\)
And also a line \(x-2 y+z=0=x+2 y-2 z\)
symmetric form of the line
\(\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 2 & -2\end{array}\right \vert=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
Thus direction ratio is of the line \((2,3,4)\)
Angle between line and plane is given by
\(\sin \theta=\left \vert\frac{\mathrm{b} \cdot \mathrm{n}}{ \vert\mathrm{b} \vert \mid \mathrm{n}}\right \vert\)
\(\Rightarrow \quad \sin \theta=\left \vert\frac{5 \times 2-2 \times 3-1 \times 4}{\sqrt{30} \sqrt{29}}\right \vert\)
\(\sin \theta=\sin 0^{\circ}\)
\(\theta=0^{\circ}\)
AMU-2018
Three Dimensional Geometry
121334
The plane \(x+m y=0\) is rotated about its line of intersection with the plane \(z=0\) through an angle \(\alpha\). The equation changes to
121327
The angle between the planes \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\)
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given, \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\) On comparing with \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}\), we get - \(a_1=3, b_1=-4, c_1=5\) And, \(\mathrm{a}_2=2, \mathrm{~b}_2=-1, \mathrm{c}_2=2\) Angle between planes is \(\cos \theta=\frac{(3)(2)+(-4)(-1)+(5)(-2)}{\left(\sqrt{3^2+(-4)^2+(5)^2}\right)\left(\sqrt{(2)^2+(-1)^2+(-2)^2}\right)}\) \(\Rightarrow \quad \cos \theta=\frac{6+4-10}{(\sqrt{9+16+25})(\sqrt{4+1+4})}=0\) \(\Rightarrow \quad \cos \theta=0\) \(\Rightarrow \quad \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2}\)
CG PET- 2017
Three Dimensional Geometry
121329
The acute angle between the planes \(P_1\) and \(P_2\), when \(P_1\) and \(P_2\) are the planes passing through the intersection of the planes \(5 x+8 y+13 z-29\) \(=0\) and \(8 x-7 y+z-20=0\) and the points \((2,1\), 3 ) and \((0,1,2)\) respectively, is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{12}\)
Explanation:
A Equation of plane passing through the intersection of planes \(5 x+8 y+13 z-29=0\) and \(8 x-\) \(7 \mathrm{y}+\mathrm{z}-20=0\) is \(5 x+8 y+3 z-29+\lambda(8 x-7 y+z-20)=0\) and if it is passing through \((2,1,3)\) then \(\lambda=\frac{7}{2}\). \(P_1\) : Equation of plane through intersection of \(5 x+8 y+\) \(3 z-19+\frac{7}{2}(8 x-7 y+z-20)=0\) \(\Rightarrow \quad 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=6\) Similarly \(\mathrm{P}_2\) : Equation of plane through intersection of \(5 x+8 y+13 z-29=0\) and \(8 x-7 y+z-20=0\) and the point \((0,1,2)\) is \(\Rightarrow \quad \mathrm{x}+\mathrm{y}+2 \mathrm{z}=5\) Angle between planes \(=\theta=\cos ^{-1}\left(\frac{3}{\sqrt{6} \sqrt{6}}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)
JEE Main-28.06.2022
Three Dimensional Geometry
121331
A plane is making intercepts \(2,3,4\) on \(X, Y\) and Z-axes respectively. Another plane is passing through the point \((-1,6,2)\) and is perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,4)\). Then angle between the two planes is
1 \(90^{\circ}\)
2 \(\cos ^{-1} \sqrt{\frac{12}{61}}\)
3 \(\cos ^{-1} \sqrt{\frac{11}{61}}\)
4 \(\cos ^{-1} \sqrt{\frac{5}{6}}\)
Explanation:
C Equation of plane ' \(\mathrm{P}_1\) ' making intercept 2, 3, 4 on \(X, Y\) and \(Z\) axes is \(\mathrm{P}_1: \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{4}=1\)
or \(6 x+4 y+3 z=12\)
Line \(\overrightarrow{\mathrm{AB}}\) is \(=(-2-1) \hat{i}+(3-2) \hat{j}+(4-3) \hat{k}\)
(Given A \& B)
\(\overrightarrow{\mathrm{AB}} \text { or } \overrightarrow{\mathrm{n}}=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Now, equation of plane \(\left(\mathrm{P}_2\right)\) containing point \((-1,6,2) \mid\) and perpendicular to \(\overrightarrow{\mathrm{AB}}\) is-
\(-3(\mathrm{x}+1)+1(\mathrm{y}-6)+1(\mathrm{z}-2)=0\)
\(\Rightarrow -3 \mathrm{x}-3+\mathrm{y}-6+\mathrm{z}-2=0\)
\(\Rightarrow -3 \mathrm{x}+\mathrm{y}+\mathrm{z}=11\)
Now, angle between \(P_1\) and \(P_2\) is -
\(\cos \theta =\left \vert\frac{(6 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+\hat{k})}{\sqrt{36+16+9} \cdot \sqrt{9+1+1}}\right \vert\)
\(\cos \theta =\left \vert\left(\frac{-18+4+3}{\sqrt{61} \cdot \sqrt{11}}\right)\right \vert=\left \vert\frac{-11}{\sqrt{61} \cdot \sqrt{11}}\right \vert\)
\(\Rightarrow \quad \theta =\cos ^{-1}\left(\sqrt{\frac{11}{61}}\right)\)
AP EAMCET-20.04.2019
Three Dimensional Geometry
121333
The angle between the line \(x-2 y+z=0=x+\) \(2 y-2 z\) and the plane \(5 x-2 y-z+17=0\) is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(0^0\)
Explanation:
D Given, Plane \(5 x-2 y-z+17=0\)
And also a line \(x-2 y+z=0=x+2 y-2 z\)
symmetric form of the line
\(\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 2 & -2\end{array}\right \vert=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
Thus direction ratio is of the line \((2,3,4)\)
Angle between line and plane is given by
\(\sin \theta=\left \vert\frac{\mathrm{b} \cdot \mathrm{n}}{ \vert\mathrm{b} \vert \mid \mathrm{n}}\right \vert\)
\(\Rightarrow \quad \sin \theta=\left \vert\frac{5 \times 2-2 \times 3-1 \times 4}{\sqrt{30} \sqrt{29}}\right \vert\)
\(\sin \theta=\sin 0^{\circ}\)
\(\theta=0^{\circ}\)
AMU-2018
Three Dimensional Geometry
121334
The plane \(x+m y=0\) is rotated about its line of intersection with the plane \(z=0\) through an angle \(\alpha\). The equation changes to
121327
The angle between the planes \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\)
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given, \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\) On comparing with \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}\), we get - \(a_1=3, b_1=-4, c_1=5\) And, \(\mathrm{a}_2=2, \mathrm{~b}_2=-1, \mathrm{c}_2=2\) Angle between planes is \(\cos \theta=\frac{(3)(2)+(-4)(-1)+(5)(-2)}{\left(\sqrt{3^2+(-4)^2+(5)^2}\right)\left(\sqrt{(2)^2+(-1)^2+(-2)^2}\right)}\) \(\Rightarrow \quad \cos \theta=\frac{6+4-10}{(\sqrt{9+16+25})(\sqrt{4+1+4})}=0\) \(\Rightarrow \quad \cos \theta=0\) \(\Rightarrow \quad \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2}\)
CG PET- 2017
Three Dimensional Geometry
121329
The acute angle between the planes \(P_1\) and \(P_2\), when \(P_1\) and \(P_2\) are the planes passing through the intersection of the planes \(5 x+8 y+13 z-29\) \(=0\) and \(8 x-7 y+z-20=0\) and the points \((2,1\), 3 ) and \((0,1,2)\) respectively, is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{12}\)
Explanation:
A Equation of plane passing through the intersection of planes \(5 x+8 y+13 z-29=0\) and \(8 x-\) \(7 \mathrm{y}+\mathrm{z}-20=0\) is \(5 x+8 y+3 z-29+\lambda(8 x-7 y+z-20)=0\) and if it is passing through \((2,1,3)\) then \(\lambda=\frac{7}{2}\). \(P_1\) : Equation of plane through intersection of \(5 x+8 y+\) \(3 z-19+\frac{7}{2}(8 x-7 y+z-20)=0\) \(\Rightarrow \quad 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=6\) Similarly \(\mathrm{P}_2\) : Equation of plane through intersection of \(5 x+8 y+13 z-29=0\) and \(8 x-7 y+z-20=0\) and the point \((0,1,2)\) is \(\Rightarrow \quad \mathrm{x}+\mathrm{y}+2 \mathrm{z}=5\) Angle between planes \(=\theta=\cos ^{-1}\left(\frac{3}{\sqrt{6} \sqrt{6}}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)
JEE Main-28.06.2022
Three Dimensional Geometry
121331
A plane is making intercepts \(2,3,4\) on \(X, Y\) and Z-axes respectively. Another plane is passing through the point \((-1,6,2)\) and is perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,4)\). Then angle between the two planes is
1 \(90^{\circ}\)
2 \(\cos ^{-1} \sqrt{\frac{12}{61}}\)
3 \(\cos ^{-1} \sqrt{\frac{11}{61}}\)
4 \(\cos ^{-1} \sqrt{\frac{5}{6}}\)
Explanation:
C Equation of plane ' \(\mathrm{P}_1\) ' making intercept 2, 3, 4 on \(X, Y\) and \(Z\) axes is \(\mathrm{P}_1: \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{4}=1\)
or \(6 x+4 y+3 z=12\)
Line \(\overrightarrow{\mathrm{AB}}\) is \(=(-2-1) \hat{i}+(3-2) \hat{j}+(4-3) \hat{k}\)
(Given A \& B)
\(\overrightarrow{\mathrm{AB}} \text { or } \overrightarrow{\mathrm{n}}=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Now, equation of plane \(\left(\mathrm{P}_2\right)\) containing point \((-1,6,2) \mid\) and perpendicular to \(\overrightarrow{\mathrm{AB}}\) is-
\(-3(\mathrm{x}+1)+1(\mathrm{y}-6)+1(\mathrm{z}-2)=0\)
\(\Rightarrow -3 \mathrm{x}-3+\mathrm{y}-6+\mathrm{z}-2=0\)
\(\Rightarrow -3 \mathrm{x}+\mathrm{y}+\mathrm{z}=11\)
Now, angle between \(P_1\) and \(P_2\) is -
\(\cos \theta =\left \vert\frac{(6 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+\hat{k})}{\sqrt{36+16+9} \cdot \sqrt{9+1+1}}\right \vert\)
\(\cos \theta =\left \vert\left(\frac{-18+4+3}{\sqrt{61} \cdot \sqrt{11}}\right)\right \vert=\left \vert\frac{-11}{\sqrt{61} \cdot \sqrt{11}}\right \vert\)
\(\Rightarrow \quad \theta =\cos ^{-1}\left(\sqrt{\frac{11}{61}}\right)\)
AP EAMCET-20.04.2019
Three Dimensional Geometry
121333
The angle between the line \(x-2 y+z=0=x+\) \(2 y-2 z\) and the plane \(5 x-2 y-z+17=0\) is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(0^0\)
Explanation:
D Given, Plane \(5 x-2 y-z+17=0\)
And also a line \(x-2 y+z=0=x+2 y-2 z\)
symmetric form of the line
\(\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 2 & -2\end{array}\right \vert=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
Thus direction ratio is of the line \((2,3,4)\)
Angle between line and plane is given by
\(\sin \theta=\left \vert\frac{\mathrm{b} \cdot \mathrm{n}}{ \vert\mathrm{b} \vert \mid \mathrm{n}}\right \vert\)
\(\Rightarrow \quad \sin \theta=\left \vert\frac{5 \times 2-2 \times 3-1 \times 4}{\sqrt{30} \sqrt{29}}\right \vert\)
\(\sin \theta=\sin 0^{\circ}\)
\(\theta=0^{\circ}\)
AMU-2018
Three Dimensional Geometry
121334
The plane \(x+m y=0\) is rotated about its line of intersection with the plane \(z=0\) through an angle \(\alpha\). The equation changes to
121327
The angle between the planes \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\)
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given, \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\) On comparing with \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}\), we get - \(a_1=3, b_1=-4, c_1=5\) And, \(\mathrm{a}_2=2, \mathrm{~b}_2=-1, \mathrm{c}_2=2\) Angle between planes is \(\cos \theta=\frac{(3)(2)+(-4)(-1)+(5)(-2)}{\left(\sqrt{3^2+(-4)^2+(5)^2}\right)\left(\sqrt{(2)^2+(-1)^2+(-2)^2}\right)}\) \(\Rightarrow \quad \cos \theta=\frac{6+4-10}{(\sqrt{9+16+25})(\sqrt{4+1+4})}=0\) \(\Rightarrow \quad \cos \theta=0\) \(\Rightarrow \quad \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2}\)
CG PET- 2017
Three Dimensional Geometry
121329
The acute angle between the planes \(P_1\) and \(P_2\), when \(P_1\) and \(P_2\) are the planes passing through the intersection of the planes \(5 x+8 y+13 z-29\) \(=0\) and \(8 x-7 y+z-20=0\) and the points \((2,1\), 3 ) and \((0,1,2)\) respectively, is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{12}\)
Explanation:
A Equation of plane passing through the intersection of planes \(5 x+8 y+13 z-29=0\) and \(8 x-\) \(7 \mathrm{y}+\mathrm{z}-20=0\) is \(5 x+8 y+3 z-29+\lambda(8 x-7 y+z-20)=0\) and if it is passing through \((2,1,3)\) then \(\lambda=\frac{7}{2}\). \(P_1\) : Equation of plane through intersection of \(5 x+8 y+\) \(3 z-19+\frac{7}{2}(8 x-7 y+z-20)=0\) \(\Rightarrow \quad 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=6\) Similarly \(\mathrm{P}_2\) : Equation of plane through intersection of \(5 x+8 y+13 z-29=0\) and \(8 x-7 y+z-20=0\) and the point \((0,1,2)\) is \(\Rightarrow \quad \mathrm{x}+\mathrm{y}+2 \mathrm{z}=5\) Angle between planes \(=\theta=\cos ^{-1}\left(\frac{3}{\sqrt{6} \sqrt{6}}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)
JEE Main-28.06.2022
Three Dimensional Geometry
121331
A plane is making intercepts \(2,3,4\) on \(X, Y\) and Z-axes respectively. Another plane is passing through the point \((-1,6,2)\) and is perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,4)\). Then angle between the two planes is
1 \(90^{\circ}\)
2 \(\cos ^{-1} \sqrt{\frac{12}{61}}\)
3 \(\cos ^{-1} \sqrt{\frac{11}{61}}\)
4 \(\cos ^{-1} \sqrt{\frac{5}{6}}\)
Explanation:
C Equation of plane ' \(\mathrm{P}_1\) ' making intercept 2, 3, 4 on \(X, Y\) and \(Z\) axes is \(\mathrm{P}_1: \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{4}=1\)
or \(6 x+4 y+3 z=12\)
Line \(\overrightarrow{\mathrm{AB}}\) is \(=(-2-1) \hat{i}+(3-2) \hat{j}+(4-3) \hat{k}\)
(Given A \& B)
\(\overrightarrow{\mathrm{AB}} \text { or } \overrightarrow{\mathrm{n}}=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Now, equation of plane \(\left(\mathrm{P}_2\right)\) containing point \((-1,6,2) \mid\) and perpendicular to \(\overrightarrow{\mathrm{AB}}\) is-
\(-3(\mathrm{x}+1)+1(\mathrm{y}-6)+1(\mathrm{z}-2)=0\)
\(\Rightarrow -3 \mathrm{x}-3+\mathrm{y}-6+\mathrm{z}-2=0\)
\(\Rightarrow -3 \mathrm{x}+\mathrm{y}+\mathrm{z}=11\)
Now, angle between \(P_1\) and \(P_2\) is -
\(\cos \theta =\left \vert\frac{(6 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+\hat{k})}{\sqrt{36+16+9} \cdot \sqrt{9+1+1}}\right \vert\)
\(\cos \theta =\left \vert\left(\frac{-18+4+3}{\sqrt{61} \cdot \sqrt{11}}\right)\right \vert=\left \vert\frac{-11}{\sqrt{61} \cdot \sqrt{11}}\right \vert\)
\(\Rightarrow \quad \theta =\cos ^{-1}\left(\sqrt{\frac{11}{61}}\right)\)
AP EAMCET-20.04.2019
Three Dimensional Geometry
121333
The angle between the line \(x-2 y+z=0=x+\) \(2 y-2 z\) and the plane \(5 x-2 y-z+17=0\) is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(0^0\)
Explanation:
D Given, Plane \(5 x-2 y-z+17=0\)
And also a line \(x-2 y+z=0=x+2 y-2 z\)
symmetric form of the line
\(\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 2 & -2\end{array}\right \vert=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
Thus direction ratio is of the line \((2,3,4)\)
Angle between line and plane is given by
\(\sin \theta=\left \vert\frac{\mathrm{b} \cdot \mathrm{n}}{ \vert\mathrm{b} \vert \mid \mathrm{n}}\right \vert\)
\(\Rightarrow \quad \sin \theta=\left \vert\frac{5 \times 2-2 \times 3-1 \times 4}{\sqrt{30} \sqrt{29}}\right \vert\)
\(\sin \theta=\sin 0^{\circ}\)
\(\theta=0^{\circ}\)
AMU-2018
Three Dimensional Geometry
121334
The plane \(x+m y=0\) is rotated about its line of intersection with the plane \(z=0\) through an angle \(\alpha\). The equation changes to
121327
The angle between the planes \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\)
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{2}\)
4 None of these
Explanation:
C Given, \(3 x-4 y+5 z=0\) \(2 x-y-2 z=5\) On comparing with \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}\), we get - \(a_1=3, b_1=-4, c_1=5\) And, \(\mathrm{a}_2=2, \mathrm{~b}_2=-1, \mathrm{c}_2=2\) Angle between planes is \(\cos \theta=\frac{(3)(2)+(-4)(-1)+(5)(-2)}{\left(\sqrt{3^2+(-4)^2+(5)^2}\right)\left(\sqrt{(2)^2+(-1)^2+(-2)^2}\right)}\) \(\Rightarrow \quad \cos \theta=\frac{6+4-10}{(\sqrt{9+16+25})(\sqrt{4+1+4})}=0\) \(\Rightarrow \quad \cos \theta=0\) \(\Rightarrow \quad \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2}\)
CG PET- 2017
Three Dimensional Geometry
121329
The acute angle between the planes \(P_1\) and \(P_2\), when \(P_1\) and \(P_2\) are the planes passing through the intersection of the planes \(5 x+8 y+13 z-29\) \(=0\) and \(8 x-7 y+z-20=0\) and the points \((2,1\), 3 ) and \((0,1,2)\) respectively, is
1 \(\frac{\pi}{3}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{6}\)
4 \(\frac{\pi}{12}\)
Explanation:
A Equation of plane passing through the intersection of planes \(5 x+8 y+13 z-29=0\) and \(8 x-\) \(7 \mathrm{y}+\mathrm{z}-20=0\) is \(5 x+8 y+3 z-29+\lambda(8 x-7 y+z-20)=0\) and if it is passing through \((2,1,3)\) then \(\lambda=\frac{7}{2}\). \(P_1\) : Equation of plane through intersection of \(5 x+8 y+\) \(3 z-19+\frac{7}{2}(8 x-7 y+z-20)=0\) \(\Rightarrow \quad 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=6\) Similarly \(\mathrm{P}_2\) : Equation of plane through intersection of \(5 x+8 y+13 z-29=0\) and \(8 x-7 y+z-20=0\) and the point \((0,1,2)\) is \(\Rightarrow \quad \mathrm{x}+\mathrm{y}+2 \mathrm{z}=5\) Angle between planes \(=\theta=\cos ^{-1}\left(\frac{3}{\sqrt{6} \sqrt{6}}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)
JEE Main-28.06.2022
Three Dimensional Geometry
121331
A plane is making intercepts \(2,3,4\) on \(X, Y\) and Z-axes respectively. Another plane is passing through the point \((-1,6,2)\) and is perpendicular to the line joining the points \((1,2,3)\) and \((-2,3,4)\). Then angle between the two planes is
1 \(90^{\circ}\)
2 \(\cos ^{-1} \sqrt{\frac{12}{61}}\)
3 \(\cos ^{-1} \sqrt{\frac{11}{61}}\)
4 \(\cos ^{-1} \sqrt{\frac{5}{6}}\)
Explanation:
C Equation of plane ' \(\mathrm{P}_1\) ' making intercept 2, 3, 4 on \(X, Y\) and \(Z\) axes is \(\mathrm{P}_1: \frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{4}=1\)
or \(6 x+4 y+3 z=12\)
Line \(\overrightarrow{\mathrm{AB}}\) is \(=(-2-1) \hat{i}+(3-2) \hat{j}+(4-3) \hat{k}\)
(Given A \& B)
\(\overrightarrow{\mathrm{AB}} \text { or } \overrightarrow{\mathrm{n}}=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Now, equation of plane \(\left(\mathrm{P}_2\right)\) containing point \((-1,6,2) \mid\) and perpendicular to \(\overrightarrow{\mathrm{AB}}\) is-
\(-3(\mathrm{x}+1)+1(\mathrm{y}-6)+1(\mathrm{z}-2)=0\)
\(\Rightarrow -3 \mathrm{x}-3+\mathrm{y}-6+\mathrm{z}-2=0\)
\(\Rightarrow -3 \mathrm{x}+\mathrm{y}+\mathrm{z}=11\)
Now, angle between \(P_1\) and \(P_2\) is -
\(\cos \theta =\left \vert\frac{(6 \hat{i}+4 \hat{j}+3 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+\hat{k})}{\sqrt{36+16+9} \cdot \sqrt{9+1+1}}\right \vert\)
\(\cos \theta =\left \vert\left(\frac{-18+4+3}{\sqrt{61} \cdot \sqrt{11}}\right)\right \vert=\left \vert\frac{-11}{\sqrt{61} \cdot \sqrt{11}}\right \vert\)
\(\Rightarrow \quad \theta =\cos ^{-1}\left(\sqrt{\frac{11}{61}}\right)\)
AP EAMCET-20.04.2019
Three Dimensional Geometry
121333
The angle between the line \(x-2 y+z=0=x+\) \(2 y-2 z\) and the plane \(5 x-2 y-z+17=0\) is
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(90^{\circ}\)
4 \(0^0\)
Explanation:
D Given, Plane \(5 x-2 y-z+17=0\)
And also a line \(x-2 y+z=0=x+2 y-2 z\)
symmetric form of the line
\(\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 1 & 2 & -2\end{array}\right \vert=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
Thus direction ratio is of the line \((2,3,4)\)
Angle between line and plane is given by
\(\sin \theta=\left \vert\frac{\mathrm{b} \cdot \mathrm{n}}{ \vert\mathrm{b} \vert \mid \mathrm{n}}\right \vert\)
\(\Rightarrow \quad \sin \theta=\left \vert\frac{5 \times 2-2 \times 3-1 \times 4}{\sqrt{30} \sqrt{29}}\right \vert\)
\(\sin \theta=\sin 0^{\circ}\)
\(\theta=0^{\circ}\)
AMU-2018
Three Dimensional Geometry
121334
The plane \(x+m y=0\) is rotated about its line of intersection with the plane \(z=0\) through an angle \(\alpha\). The equation changes to
