B Given, \(P_1: x+2 y+2 z-5=0\)
\(P_2: 3 x+3 y+2 z-8=0\)
From equation (i) and (ii) we get-
\(\overrightarrow{\mathrm{n}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{n}}_2=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Angle between planes \(\mathrm{P}_1\) and \(\mathrm{P}_2\) is
\(\cos \theta =\left \vert\frac{\left(\vec{n}_1\right) \cdot\left(\vec{n}_2\right)}{\left \vert\vec{n}_1\right \vert \cdot\left \vert\vec{n}_2\right \vert}\right \vert\)
\(\cos \theta =\left \vert\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+3 \hat{j}+2 \hat{k})}{\sqrt{1+4+4} \cdot \sqrt{9+9+4}}\right \vert\)
\(\cos \theta =\left \vert\frac{3+6+4}{\sqrt{9} \cdot \sqrt{22}}\right \vert=\frac{13}{3 \sqrt{22}}\)
\(\theta \quad =\cos ^{-1}\left(\frac{13}{3 \sqrt{22}}\right)\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121336
The foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\). If \((2,2\), \(-1),(3,4,2),(3,4,2)(3,3,0)\) are three points on the plane \(\pi_2\), then the angle between the planes \(\pi_1\) and \(\pi_2\) is
1 \(\frac{\pi}{2}\)
2 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
3 \(\frac{\pi}{6}\)
4 \(\cos ^{-1}\left(\frac{2}{5}\right)\)
Explanation:
A We know that, foot of perpendicular from a point \(\left(x_1, y_1, z_1\right)\) on the plane \(a x+b y+c z+d=0\) is \((\mathrm{x}, \mathrm{y}, \mathrm{z})\) where \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-\left(a_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}\)
Since, the foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\)
\(\therefore \quad \frac{1-1}{\mathrm{a}}=\frac{3-1}{\mathrm{~b}}=\frac{5-1}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \frac{0}{\mathrm{a}}=\frac{2}{\mathrm{~b}}=\frac{4}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \mathrm{a}=0, \mathrm{~b}=2 \mathrm{k}, \mathrm{c}=4 \mathrm{k}\)
Equation of plane \(\pi_1\) through point \((1,3,5)\) and having directions of normal to plane are \(a, b, c\) is
\( 0(\mathrm{x}-1)+2 \mathrm{k}(\mathrm{y}-3)+4 \mathrm{k}(\mathrm{z}-5) =0\)
\(\Rightarrow (\mathrm{y}-3)+2(\mathrm{z}-5) =0\)
\(\Rightarrow \mathrm{y}+2 \mathrm{z}-13 =0\)
And equation of plane \(\pi_2\) is
\(\begin{aligned} &\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 3-2 & 4-2 & 2+1 \\ 3-2 & 3-2 & 0+1\end{array}\right \vert=0 \\ & \quad\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 1 & 2 & 3 \\ 1 & 1 & 1\end{array}\right \vert=0\end{aligned}\)
\(\Rightarrow (x-2)[2-3]-(y-2)[1-3]+(z+1)[1-2]=0\)
\(\Rightarrow -(x-2)+2(y-2)-1(z+1)=0\)
\(\Rightarrow (x-2)-2(y-2)+(z+1)=0\)
\(\Rightarrow x-2 y+z+3=0\)
\(\therefore\) Angle between plane \(\pi_1\) and \(\pi_2\)
\(\cos \theta =\left \vert\left(\frac{0 \times 1+1 \times-2+2 \times 1}{\sqrt{0+1+4} \sqrt{1+4+1}}\right)\right \vert\)
\(\cos \theta =0\)
\(\cos \theta =\cos \frac{\pi}{2}\)
\(\theta =\frac{\pi}{2}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121337
A tetrahedron has vertices \(O(0,0,0),(A(1,2\), 1) \(\mathrm{B}(2,1,3), \mathrm{C}(-1,1,2)\). If \(\theta\) is the angle between the faces \(O A B\) and \(A B C\), then \(\cos \theta=\)
1 \(\frac{1}{\sqrt{2}}\)
2 \(\frac{19}{35}\)
3 \(\frac{\sqrt{3}}{2}\)
4 \(\frac{17}{31}\)
Explanation:
B Equation of plane OAB is given by, \(\left \vert\begin{array}{ccc}\mathrm{x}-0 & \mathrm{y}-0 & \mathrm{z}-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0\end{array}\right \vert=0 \Rightarrow\left \vert\begin{array}{ccc}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right \vert=0\)
\(\mathrm{x}(6-1)-\mathrm{y}(3-2)+\mathrm{z}(1-4)=0\)
\(5 x-y-3 z=0\)
Equation of plane \(\mathrm{ABC}\) is given by,
\(\begin{aligned} & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1\end{array}\right \vert=0 \\ & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right \vert=0\end{aligned}\)
\((x-1)(-1+2)-(y-2)(1+4)+(z-1)(-1-2)=0\)
\(x-1-5 y+10-3 z+3=0\)
\(x-5 y-3 z+12=0\)
Then angle between planes represented by Eqs. (i) and (ii) is given by -
\(\cos \theta=\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9} \sqrt{1+25+9}}=\frac{19}{35}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121307
If a line makes angles of measure \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\) with \(X\) and \(Y\) axes respectively, then the angle made by the line with \(\mathrm{Z}\) axis is
B Given, \(P_1: x+2 y+2 z-5=0\)
\(P_2: 3 x+3 y+2 z-8=0\)
From equation (i) and (ii) we get-
\(\overrightarrow{\mathrm{n}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{n}}_2=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Angle between planes \(\mathrm{P}_1\) and \(\mathrm{P}_2\) is
\(\cos \theta =\left \vert\frac{\left(\vec{n}_1\right) \cdot\left(\vec{n}_2\right)}{\left \vert\vec{n}_1\right \vert \cdot\left \vert\vec{n}_2\right \vert}\right \vert\)
\(\cos \theta =\left \vert\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+3 \hat{j}+2 \hat{k})}{\sqrt{1+4+4} \cdot \sqrt{9+9+4}}\right \vert\)
\(\cos \theta =\left \vert\frac{3+6+4}{\sqrt{9} \cdot \sqrt{22}}\right \vert=\frac{13}{3 \sqrt{22}}\)
\(\theta \quad =\cos ^{-1}\left(\frac{13}{3 \sqrt{22}}\right)\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121336
The foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\). If \((2,2\), \(-1),(3,4,2),(3,4,2)(3,3,0)\) are three points on the plane \(\pi_2\), then the angle between the planes \(\pi_1\) and \(\pi_2\) is
1 \(\frac{\pi}{2}\)
2 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
3 \(\frac{\pi}{6}\)
4 \(\cos ^{-1}\left(\frac{2}{5}\right)\)
Explanation:
A We know that, foot of perpendicular from a point \(\left(x_1, y_1, z_1\right)\) on the plane \(a x+b y+c z+d=0\) is \((\mathrm{x}, \mathrm{y}, \mathrm{z})\) where \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-\left(a_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}\)
Since, the foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\)
\(\therefore \quad \frac{1-1}{\mathrm{a}}=\frac{3-1}{\mathrm{~b}}=\frac{5-1}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \frac{0}{\mathrm{a}}=\frac{2}{\mathrm{~b}}=\frac{4}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \mathrm{a}=0, \mathrm{~b}=2 \mathrm{k}, \mathrm{c}=4 \mathrm{k}\)
Equation of plane \(\pi_1\) through point \((1,3,5)\) and having directions of normal to plane are \(a, b, c\) is
\( 0(\mathrm{x}-1)+2 \mathrm{k}(\mathrm{y}-3)+4 \mathrm{k}(\mathrm{z}-5) =0\)
\(\Rightarrow (\mathrm{y}-3)+2(\mathrm{z}-5) =0\)
\(\Rightarrow \mathrm{y}+2 \mathrm{z}-13 =0\)
And equation of plane \(\pi_2\) is
\(\begin{aligned} &\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 3-2 & 4-2 & 2+1 \\ 3-2 & 3-2 & 0+1\end{array}\right \vert=0 \\ & \quad\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 1 & 2 & 3 \\ 1 & 1 & 1\end{array}\right \vert=0\end{aligned}\)
\(\Rightarrow (x-2)[2-3]-(y-2)[1-3]+(z+1)[1-2]=0\)
\(\Rightarrow -(x-2)+2(y-2)-1(z+1)=0\)
\(\Rightarrow (x-2)-2(y-2)+(z+1)=0\)
\(\Rightarrow x-2 y+z+3=0\)
\(\therefore\) Angle between plane \(\pi_1\) and \(\pi_2\)
\(\cos \theta =\left \vert\left(\frac{0 \times 1+1 \times-2+2 \times 1}{\sqrt{0+1+4} \sqrt{1+4+1}}\right)\right \vert\)
\(\cos \theta =0\)
\(\cos \theta =\cos \frac{\pi}{2}\)
\(\theta =\frac{\pi}{2}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121337
A tetrahedron has vertices \(O(0,0,0),(A(1,2\), 1) \(\mathrm{B}(2,1,3), \mathrm{C}(-1,1,2)\). If \(\theta\) is the angle between the faces \(O A B\) and \(A B C\), then \(\cos \theta=\)
1 \(\frac{1}{\sqrt{2}}\)
2 \(\frac{19}{35}\)
3 \(\frac{\sqrt{3}}{2}\)
4 \(\frac{17}{31}\)
Explanation:
B Equation of plane OAB is given by, \(\left \vert\begin{array}{ccc}\mathrm{x}-0 & \mathrm{y}-0 & \mathrm{z}-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0\end{array}\right \vert=0 \Rightarrow\left \vert\begin{array}{ccc}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right \vert=0\)
\(\mathrm{x}(6-1)-\mathrm{y}(3-2)+\mathrm{z}(1-4)=0\)
\(5 x-y-3 z=0\)
Equation of plane \(\mathrm{ABC}\) is given by,
\(\begin{aligned} & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1\end{array}\right \vert=0 \\ & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right \vert=0\end{aligned}\)
\((x-1)(-1+2)-(y-2)(1+4)+(z-1)(-1-2)=0\)
\(x-1-5 y+10-3 z+3=0\)
\(x-5 y-3 z+12=0\)
Then angle between planes represented by Eqs. (i) and (ii) is given by -
\(\cos \theta=\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9} \sqrt{1+25+9}}=\frac{19}{35}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121307
If a line makes angles of measure \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\) with \(X\) and \(Y\) axes respectively, then the angle made by the line with \(\mathrm{Z}\) axis is
B Given, \(P_1: x+2 y+2 z-5=0\)
\(P_2: 3 x+3 y+2 z-8=0\)
From equation (i) and (ii) we get-
\(\overrightarrow{\mathrm{n}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{n}}_2=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Angle between planes \(\mathrm{P}_1\) and \(\mathrm{P}_2\) is
\(\cos \theta =\left \vert\frac{\left(\vec{n}_1\right) \cdot\left(\vec{n}_2\right)}{\left \vert\vec{n}_1\right \vert \cdot\left \vert\vec{n}_2\right \vert}\right \vert\)
\(\cos \theta =\left \vert\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+3 \hat{j}+2 \hat{k})}{\sqrt{1+4+4} \cdot \sqrt{9+9+4}}\right \vert\)
\(\cos \theta =\left \vert\frac{3+6+4}{\sqrt{9} \cdot \sqrt{22}}\right \vert=\frac{13}{3 \sqrt{22}}\)
\(\theta \quad =\cos ^{-1}\left(\frac{13}{3 \sqrt{22}}\right)\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121336
The foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\). If \((2,2\), \(-1),(3,4,2),(3,4,2)(3,3,0)\) are three points on the plane \(\pi_2\), then the angle between the planes \(\pi_1\) and \(\pi_2\) is
1 \(\frac{\pi}{2}\)
2 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
3 \(\frac{\pi}{6}\)
4 \(\cos ^{-1}\left(\frac{2}{5}\right)\)
Explanation:
A We know that, foot of perpendicular from a point \(\left(x_1, y_1, z_1\right)\) on the plane \(a x+b y+c z+d=0\) is \((\mathrm{x}, \mathrm{y}, \mathrm{z})\) where \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-\left(a_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}\)
Since, the foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\)
\(\therefore \quad \frac{1-1}{\mathrm{a}}=\frac{3-1}{\mathrm{~b}}=\frac{5-1}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \frac{0}{\mathrm{a}}=\frac{2}{\mathrm{~b}}=\frac{4}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \mathrm{a}=0, \mathrm{~b}=2 \mathrm{k}, \mathrm{c}=4 \mathrm{k}\)
Equation of plane \(\pi_1\) through point \((1,3,5)\) and having directions of normal to plane are \(a, b, c\) is
\( 0(\mathrm{x}-1)+2 \mathrm{k}(\mathrm{y}-3)+4 \mathrm{k}(\mathrm{z}-5) =0\)
\(\Rightarrow (\mathrm{y}-3)+2(\mathrm{z}-5) =0\)
\(\Rightarrow \mathrm{y}+2 \mathrm{z}-13 =0\)
And equation of plane \(\pi_2\) is
\(\begin{aligned} &\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 3-2 & 4-2 & 2+1 \\ 3-2 & 3-2 & 0+1\end{array}\right \vert=0 \\ & \quad\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 1 & 2 & 3 \\ 1 & 1 & 1\end{array}\right \vert=0\end{aligned}\)
\(\Rightarrow (x-2)[2-3]-(y-2)[1-3]+(z+1)[1-2]=0\)
\(\Rightarrow -(x-2)+2(y-2)-1(z+1)=0\)
\(\Rightarrow (x-2)-2(y-2)+(z+1)=0\)
\(\Rightarrow x-2 y+z+3=0\)
\(\therefore\) Angle between plane \(\pi_1\) and \(\pi_2\)
\(\cos \theta =\left \vert\left(\frac{0 \times 1+1 \times-2+2 \times 1}{\sqrt{0+1+4} \sqrt{1+4+1}}\right)\right \vert\)
\(\cos \theta =0\)
\(\cos \theta =\cos \frac{\pi}{2}\)
\(\theta =\frac{\pi}{2}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121337
A tetrahedron has vertices \(O(0,0,0),(A(1,2\), 1) \(\mathrm{B}(2,1,3), \mathrm{C}(-1,1,2)\). If \(\theta\) is the angle between the faces \(O A B\) and \(A B C\), then \(\cos \theta=\)
1 \(\frac{1}{\sqrt{2}}\)
2 \(\frac{19}{35}\)
3 \(\frac{\sqrt{3}}{2}\)
4 \(\frac{17}{31}\)
Explanation:
B Equation of plane OAB is given by, \(\left \vert\begin{array}{ccc}\mathrm{x}-0 & \mathrm{y}-0 & \mathrm{z}-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0\end{array}\right \vert=0 \Rightarrow\left \vert\begin{array}{ccc}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right \vert=0\)
\(\mathrm{x}(6-1)-\mathrm{y}(3-2)+\mathrm{z}(1-4)=0\)
\(5 x-y-3 z=0\)
Equation of plane \(\mathrm{ABC}\) is given by,
\(\begin{aligned} & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1\end{array}\right \vert=0 \\ & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right \vert=0\end{aligned}\)
\((x-1)(-1+2)-(y-2)(1+4)+(z-1)(-1-2)=0\)
\(x-1-5 y+10-3 z+3=0\)
\(x-5 y-3 z+12=0\)
Then angle between planes represented by Eqs. (i) and (ii) is given by -
\(\cos \theta=\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9} \sqrt{1+25+9}}=\frac{19}{35}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121307
If a line makes angles of measure \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\) with \(X\) and \(Y\) axes respectively, then the angle made by the line with \(\mathrm{Z}\) axis is
B Given, \(P_1: x+2 y+2 z-5=0\)
\(P_2: 3 x+3 y+2 z-8=0\)
From equation (i) and (ii) we get-
\(\overrightarrow{\mathrm{n}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
\(\overrightarrow{\mathrm{n}}_2=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
Angle between planes \(\mathrm{P}_1\) and \(\mathrm{P}_2\) is
\(\cos \theta =\left \vert\frac{\left(\vec{n}_1\right) \cdot\left(\vec{n}_2\right)}{\left \vert\vec{n}_1\right \vert \cdot\left \vert\vec{n}_2\right \vert}\right \vert\)
\(\cos \theta =\left \vert\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+3 \hat{j}+2 \hat{k})}{\sqrt{1+4+4} \cdot \sqrt{9+9+4}}\right \vert\)
\(\cos \theta =\left \vert\frac{3+6+4}{\sqrt{9} \cdot \sqrt{22}}\right \vert=\frac{13}{3 \sqrt{22}}\)
\(\theta \quad =\cos ^{-1}\left(\frac{13}{3 \sqrt{22}}\right)\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121336
The foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\). If \((2,2\), \(-1),(3,4,2),(3,4,2)(3,3,0)\) are three points on the plane \(\pi_2\), then the angle between the planes \(\pi_1\) and \(\pi_2\) is
1 \(\frac{\pi}{2}\)
2 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
3 \(\frac{\pi}{6}\)
4 \(\cos ^{-1}\left(\frac{2}{5}\right)\)
Explanation:
A We know that, foot of perpendicular from a point \(\left(x_1, y_1, z_1\right)\) on the plane \(a x+b y+c z+d=0\) is \((\mathrm{x}, \mathrm{y}, \mathrm{z})\) where \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-\left(a_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}\)
Since, the foot of the perpendicular drawn from the point \((1,1,1)\) to the plane \(\pi_1\) is \((1,3,5)\)
\(\therefore \quad \frac{1-1}{\mathrm{a}}=\frac{3-1}{\mathrm{~b}}=\frac{5-1}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \frac{0}{\mathrm{a}}=\frac{2}{\mathrm{~b}}=\frac{4}{\mathrm{c}}=\frac{-(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\)
\(\Rightarrow \quad \mathrm{a}=0, \mathrm{~b}=2 \mathrm{k}, \mathrm{c}=4 \mathrm{k}\)
Equation of plane \(\pi_1\) through point \((1,3,5)\) and having directions of normal to plane are \(a, b, c\) is
\( 0(\mathrm{x}-1)+2 \mathrm{k}(\mathrm{y}-3)+4 \mathrm{k}(\mathrm{z}-5) =0\)
\(\Rightarrow (\mathrm{y}-3)+2(\mathrm{z}-5) =0\)
\(\Rightarrow \mathrm{y}+2 \mathrm{z}-13 =0\)
And equation of plane \(\pi_2\) is
\(\begin{aligned} &\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 3-2 & 4-2 & 2+1 \\ 3-2 & 3-2 & 0+1\end{array}\right \vert=0 \\ & \quad\left \vert\begin{array}{ccc}x-2 & y-2 & z+1 \\ 1 & 2 & 3 \\ 1 & 1 & 1\end{array}\right \vert=0\end{aligned}\)
\(\Rightarrow (x-2)[2-3]-(y-2)[1-3]+(z+1)[1-2]=0\)
\(\Rightarrow -(x-2)+2(y-2)-1(z+1)=0\)
\(\Rightarrow (x-2)-2(y-2)+(z+1)=0\)
\(\Rightarrow x-2 y+z+3=0\)
\(\therefore\) Angle between plane \(\pi_1\) and \(\pi_2\)
\(\cos \theta =\left \vert\left(\frac{0 \times 1+1 \times-2+2 \times 1}{\sqrt{0+1+4} \sqrt{1+4+1}}\right)\right \vert\)
\(\cos \theta =0\)
\(\cos \theta =\cos \frac{\pi}{2}\)
\(\theta =\frac{\pi}{2}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121337
A tetrahedron has vertices \(O(0,0,0),(A(1,2\), 1) \(\mathrm{B}(2,1,3), \mathrm{C}(-1,1,2)\). If \(\theta\) is the angle between the faces \(O A B\) and \(A B C\), then \(\cos \theta=\)
1 \(\frac{1}{\sqrt{2}}\)
2 \(\frac{19}{35}\)
3 \(\frac{\sqrt{3}}{2}\)
4 \(\frac{17}{31}\)
Explanation:
B Equation of plane OAB is given by, \(\left \vert\begin{array}{ccc}\mathrm{x}-0 & \mathrm{y}-0 & \mathrm{z}-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0\end{array}\right \vert=0 \Rightarrow\left \vert\begin{array}{ccc}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 1 & 2 & 1 \\ 2 & 1 & 3\end{array}\right \vert=0\)
\(\mathrm{x}(6-1)-\mathrm{y}(3-2)+\mathrm{z}(1-4)=0\)
\(5 x-y-3 z=0\)
Equation of plane \(\mathrm{ABC}\) is given by,
\(\begin{aligned} & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1\end{array}\right \vert=0 \\ & \left \vert\begin{array}{ccc}x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1\end{array}\right \vert=0\end{aligned}\)
\((x-1)(-1+2)-(y-2)(1+4)+(z-1)(-1-2)=0\)
\(x-1-5 y+10-3 z+3=0\)
\(x-5 y-3 z+12=0\)
Then angle between planes represented by Eqs. (i) and (ii) is given by -
\(\cos \theta=\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9} \sqrt{1+25+9}}=\frac{19}{35}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121307
If a line makes angles of measure \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\) with \(X\) and \(Y\) axes respectively, then the angle made by the line with \(\mathrm{Z}\) axis is
