121316
If the angle \(\theta\) between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane \(2 x-y+6 \sqrt{\lambda z}+4=0\) is such that \(\sin \theta=\frac{1}{3}\), then the value of \(\lambda\) is
1 \(\frac{-3}{5}\)
2 \(\frac{-5}{3}\)
3 \(\frac{5}{3}\)
4 \(\frac{3}{5}\)
Explanation:
C Given, \(\sin \theta=\frac{1}{3}\)
The direction ratios of the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) are 1,2,2 and the direction ratios of the plane \(2 x-y+\sqrt{\lambda} z+4=0\) are \(2,-1, \sqrt{\lambda}\).
\(\left \vert\frac{(1)(2)+2(-1)+2(\sqrt{\lambda})}{\sqrt{1^2+2^2+2^2} \sqrt{2^2+(-1)^2+(\sqrt{\lambda})^2}}\right \vert=\frac{1}{3}\)
\(\left \vert\frac{2-2+2 \sqrt{\lambda}}{\sqrt{9} \sqrt{4+1+\lambda}}\right \vert=\frac{1}{3}\)
\(\frac{2 \sqrt{\lambda}}{3 \sqrt{5+\lambda}}=\frac{1}{3}\)
\(2 \sqrt{\lambda}=\sqrt{5+\lambda}\)
On squaring both sides we get -
\(4 \lambda=5+\lambda \Rightarrow 3 \lambda=5 \Rightarrow \lambda=\frac{5}{3}\)
MHT CET-2019
Three Dimensional Geometry
121318
If the line passing through the origin makes angles \(\boldsymbol{\theta}_1, \boldsymbol{\theta}_2, \boldsymbol{\theta}_3\) with the planes XOY, YOZ, ZOX respectively, then
A Let \(P \equiv(x, y, z)\) be a point on the line passing through origin direction ratios of \(\mathrm{OP}\) are \(\mathrm{x}, \mathrm{y}\), \(\mathrm{z}\) and direction ratios of XOY plane are \(0,0,1\). \(\therefore \sin \theta_1=\frac{\mathrm{x}(0)+\mathrm{y}(0)+\mathrm{z}(1)}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2} \sqrt{0+0+1}}=\frac{\mathrm{z}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) Similarly, \(\sin \theta_2=\frac{x}{\sqrt{x^2+y^2+z^2}}\) and \(\sin \theta_3=\frac{\mathrm{y}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3\) \(=\frac{\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{x}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{y}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\) \(=\frac{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}=1\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3=1\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=1-\sin ^2 \theta_3\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=\cos ^2 \theta_3\)
MHT CET-2019
Three Dimensional Geometry
121319
If lines \(\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and \(\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}\) are perpendicular to each other, then \(\lambda=\)
1 6
2 \(-\frac{7}{6}\)
3 7
4 \(-\frac{6}{7}\)
Explanation:
D The given lines are Line \(1, \frac{2(x-2)}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and Line \(2 \frac{x-1}{1}=\frac{3\left(y-\frac{1}{3}\right)}{\lambda}=\frac{z-2}{1}\) \(\frac{x-2}{\left(\frac{\lambda}{2}\right)}=\frac{y-1}{2}=\frac{z-3}{1} \quad \text { and } x-1=\frac{\left(y-\frac{1}{3}\right)}{\left(\frac{\lambda}{3}\right)}=\frac{z-2}{1}\) The direction ratios of the lines are \(\frac{\lambda}{2}, 2,1\) and \(1, \frac{\lambda}{3}, 1\) As the given lines are perpendicular - \(\therefore\left(\frac{\lambda}{2}\right)(1)+(2)\left(\frac{\lambda}{3}\right)+(1)(1)=0\) \(\frac{\lambda}{2}+\frac{2 \lambda}{3}=-1 \Rightarrow 7 \lambda=-6 \Rightarrow \lambda=-\frac{6}{7}\)
MHT CET-2019
Three Dimensional Geometry
121322
The angle between the lines
\(\overline{\mathbf{r}}=3 \hat{\mathbf{i}}+\mathbf{2} \mathbf{j}-\mathbf{4} \hat{\mathbf{k}}+\boldsymbol{\lambda}(\hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}+\mathbf{2 \hat { k }})\)
1 \(\cos ^{-1}\left(\frac{18}{21}\right)\)
2 \(\cos ^{-1}\left(\frac{19}{21}\right)\)
3 \(\cos ^{-1}\left(\frac{20}{21}\right)\)
4 \(\cos ^{-1}\left(\frac{17}{21}\right)\)
Explanation:
B Let, \(\theta\) be the angle between the given lines. The given lines are parallel to the vectors \(\overline{\mathrm{b}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}_2=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) The angle \(\theta\) between them is given by, \(\cos \theta=\frac{\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2}{\left \vert\overline{\mathrm{b}}_1\right \vert\left \vert\overline{\mathrm{b}}_2\right \vert}\)
\(\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=(1)(3)+2(2)+2(6)\)
\(\left \vert\overline{\mathrm{b}}_1\right \vert=\sqrt{1+4+4}=\sqrt{9}=3 \text { and }\left \vert\overline{\mathrm{b}}_2\right \vert=\sqrt{9+4+36}=\sqrt{49}=7\)
\(\therefore \cos \theta=\frac{19}{3 \times 7}=\frac{19}{21} \Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)\)
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Three Dimensional Geometry
121316
If the angle \(\theta\) between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane \(2 x-y+6 \sqrt{\lambda z}+4=0\) is such that \(\sin \theta=\frac{1}{3}\), then the value of \(\lambda\) is
1 \(\frac{-3}{5}\)
2 \(\frac{-5}{3}\)
3 \(\frac{5}{3}\)
4 \(\frac{3}{5}\)
Explanation:
C Given, \(\sin \theta=\frac{1}{3}\)
The direction ratios of the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) are 1,2,2 and the direction ratios of the plane \(2 x-y+\sqrt{\lambda} z+4=0\) are \(2,-1, \sqrt{\lambda}\).
\(\left \vert\frac{(1)(2)+2(-1)+2(\sqrt{\lambda})}{\sqrt{1^2+2^2+2^2} \sqrt{2^2+(-1)^2+(\sqrt{\lambda})^2}}\right \vert=\frac{1}{3}\)
\(\left \vert\frac{2-2+2 \sqrt{\lambda}}{\sqrt{9} \sqrt{4+1+\lambda}}\right \vert=\frac{1}{3}\)
\(\frac{2 \sqrt{\lambda}}{3 \sqrt{5+\lambda}}=\frac{1}{3}\)
\(2 \sqrt{\lambda}=\sqrt{5+\lambda}\)
On squaring both sides we get -
\(4 \lambda=5+\lambda \Rightarrow 3 \lambda=5 \Rightarrow \lambda=\frac{5}{3}\)
MHT CET-2019
Three Dimensional Geometry
121318
If the line passing through the origin makes angles \(\boldsymbol{\theta}_1, \boldsymbol{\theta}_2, \boldsymbol{\theta}_3\) with the planes XOY, YOZ, ZOX respectively, then
A Let \(P \equiv(x, y, z)\) be a point on the line passing through origin direction ratios of \(\mathrm{OP}\) are \(\mathrm{x}, \mathrm{y}\), \(\mathrm{z}\) and direction ratios of XOY plane are \(0,0,1\). \(\therefore \sin \theta_1=\frac{\mathrm{x}(0)+\mathrm{y}(0)+\mathrm{z}(1)}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2} \sqrt{0+0+1}}=\frac{\mathrm{z}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) Similarly, \(\sin \theta_2=\frac{x}{\sqrt{x^2+y^2+z^2}}\) and \(\sin \theta_3=\frac{\mathrm{y}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3\) \(=\frac{\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{x}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{y}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\) \(=\frac{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}=1\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3=1\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=1-\sin ^2 \theta_3\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=\cos ^2 \theta_3\)
MHT CET-2019
Three Dimensional Geometry
121319
If lines \(\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and \(\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}\) are perpendicular to each other, then \(\lambda=\)
1 6
2 \(-\frac{7}{6}\)
3 7
4 \(-\frac{6}{7}\)
Explanation:
D The given lines are Line \(1, \frac{2(x-2)}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and Line \(2 \frac{x-1}{1}=\frac{3\left(y-\frac{1}{3}\right)}{\lambda}=\frac{z-2}{1}\) \(\frac{x-2}{\left(\frac{\lambda}{2}\right)}=\frac{y-1}{2}=\frac{z-3}{1} \quad \text { and } x-1=\frac{\left(y-\frac{1}{3}\right)}{\left(\frac{\lambda}{3}\right)}=\frac{z-2}{1}\) The direction ratios of the lines are \(\frac{\lambda}{2}, 2,1\) and \(1, \frac{\lambda}{3}, 1\) As the given lines are perpendicular - \(\therefore\left(\frac{\lambda}{2}\right)(1)+(2)\left(\frac{\lambda}{3}\right)+(1)(1)=0\) \(\frac{\lambda}{2}+\frac{2 \lambda}{3}=-1 \Rightarrow 7 \lambda=-6 \Rightarrow \lambda=-\frac{6}{7}\)
MHT CET-2019
Three Dimensional Geometry
121322
The angle between the lines
\(\overline{\mathbf{r}}=3 \hat{\mathbf{i}}+\mathbf{2} \mathbf{j}-\mathbf{4} \hat{\mathbf{k}}+\boldsymbol{\lambda}(\hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}+\mathbf{2 \hat { k }})\)
1 \(\cos ^{-1}\left(\frac{18}{21}\right)\)
2 \(\cos ^{-1}\left(\frac{19}{21}\right)\)
3 \(\cos ^{-1}\left(\frac{20}{21}\right)\)
4 \(\cos ^{-1}\left(\frac{17}{21}\right)\)
Explanation:
B Let, \(\theta\) be the angle between the given lines. The given lines are parallel to the vectors \(\overline{\mathrm{b}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}_2=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) The angle \(\theta\) between them is given by, \(\cos \theta=\frac{\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2}{\left \vert\overline{\mathrm{b}}_1\right \vert\left \vert\overline{\mathrm{b}}_2\right \vert}\)
\(\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=(1)(3)+2(2)+2(6)\)
\(\left \vert\overline{\mathrm{b}}_1\right \vert=\sqrt{1+4+4}=\sqrt{9}=3 \text { and }\left \vert\overline{\mathrm{b}}_2\right \vert=\sqrt{9+4+36}=\sqrt{49}=7\)
\(\therefore \cos \theta=\frac{19}{3 \times 7}=\frac{19}{21} \Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)\)
121316
If the angle \(\theta\) between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane \(2 x-y+6 \sqrt{\lambda z}+4=0\) is such that \(\sin \theta=\frac{1}{3}\), then the value of \(\lambda\) is
1 \(\frac{-3}{5}\)
2 \(\frac{-5}{3}\)
3 \(\frac{5}{3}\)
4 \(\frac{3}{5}\)
Explanation:
C Given, \(\sin \theta=\frac{1}{3}\)
The direction ratios of the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) are 1,2,2 and the direction ratios of the plane \(2 x-y+\sqrt{\lambda} z+4=0\) are \(2,-1, \sqrt{\lambda}\).
\(\left \vert\frac{(1)(2)+2(-1)+2(\sqrt{\lambda})}{\sqrt{1^2+2^2+2^2} \sqrt{2^2+(-1)^2+(\sqrt{\lambda})^2}}\right \vert=\frac{1}{3}\)
\(\left \vert\frac{2-2+2 \sqrt{\lambda}}{\sqrt{9} \sqrt{4+1+\lambda}}\right \vert=\frac{1}{3}\)
\(\frac{2 \sqrt{\lambda}}{3 \sqrt{5+\lambda}}=\frac{1}{3}\)
\(2 \sqrt{\lambda}=\sqrt{5+\lambda}\)
On squaring both sides we get -
\(4 \lambda=5+\lambda \Rightarrow 3 \lambda=5 \Rightarrow \lambda=\frac{5}{3}\)
MHT CET-2019
Three Dimensional Geometry
121318
If the line passing through the origin makes angles \(\boldsymbol{\theta}_1, \boldsymbol{\theta}_2, \boldsymbol{\theta}_3\) with the planes XOY, YOZ, ZOX respectively, then
A Let \(P \equiv(x, y, z)\) be a point on the line passing through origin direction ratios of \(\mathrm{OP}\) are \(\mathrm{x}, \mathrm{y}\), \(\mathrm{z}\) and direction ratios of XOY plane are \(0,0,1\). \(\therefore \sin \theta_1=\frac{\mathrm{x}(0)+\mathrm{y}(0)+\mathrm{z}(1)}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2} \sqrt{0+0+1}}=\frac{\mathrm{z}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) Similarly, \(\sin \theta_2=\frac{x}{\sqrt{x^2+y^2+z^2}}\) and \(\sin \theta_3=\frac{\mathrm{y}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3\) \(=\frac{\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{x}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{y}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\) \(=\frac{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}=1\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3=1\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=1-\sin ^2 \theta_3\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=\cos ^2 \theta_3\)
MHT CET-2019
Three Dimensional Geometry
121319
If lines \(\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and \(\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}\) are perpendicular to each other, then \(\lambda=\)
1 6
2 \(-\frac{7}{6}\)
3 7
4 \(-\frac{6}{7}\)
Explanation:
D The given lines are Line \(1, \frac{2(x-2)}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and Line \(2 \frac{x-1}{1}=\frac{3\left(y-\frac{1}{3}\right)}{\lambda}=\frac{z-2}{1}\) \(\frac{x-2}{\left(\frac{\lambda}{2}\right)}=\frac{y-1}{2}=\frac{z-3}{1} \quad \text { and } x-1=\frac{\left(y-\frac{1}{3}\right)}{\left(\frac{\lambda}{3}\right)}=\frac{z-2}{1}\) The direction ratios of the lines are \(\frac{\lambda}{2}, 2,1\) and \(1, \frac{\lambda}{3}, 1\) As the given lines are perpendicular - \(\therefore\left(\frac{\lambda}{2}\right)(1)+(2)\left(\frac{\lambda}{3}\right)+(1)(1)=0\) \(\frac{\lambda}{2}+\frac{2 \lambda}{3}=-1 \Rightarrow 7 \lambda=-6 \Rightarrow \lambda=-\frac{6}{7}\)
MHT CET-2019
Three Dimensional Geometry
121322
The angle between the lines
\(\overline{\mathbf{r}}=3 \hat{\mathbf{i}}+\mathbf{2} \mathbf{j}-\mathbf{4} \hat{\mathbf{k}}+\boldsymbol{\lambda}(\hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}+\mathbf{2 \hat { k }})\)
1 \(\cos ^{-1}\left(\frac{18}{21}\right)\)
2 \(\cos ^{-1}\left(\frac{19}{21}\right)\)
3 \(\cos ^{-1}\left(\frac{20}{21}\right)\)
4 \(\cos ^{-1}\left(\frac{17}{21}\right)\)
Explanation:
B Let, \(\theta\) be the angle between the given lines. The given lines are parallel to the vectors \(\overline{\mathrm{b}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}_2=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) The angle \(\theta\) between them is given by, \(\cos \theta=\frac{\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2}{\left \vert\overline{\mathrm{b}}_1\right \vert\left \vert\overline{\mathrm{b}}_2\right \vert}\)
\(\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=(1)(3)+2(2)+2(6)\)
\(\left \vert\overline{\mathrm{b}}_1\right \vert=\sqrt{1+4+4}=\sqrt{9}=3 \text { and }\left \vert\overline{\mathrm{b}}_2\right \vert=\sqrt{9+4+36}=\sqrt{49}=7\)
\(\therefore \cos \theta=\frac{19}{3 \times 7}=\frac{19}{21} \Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)\)
121316
If the angle \(\theta\) between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane \(2 x-y+6 \sqrt{\lambda z}+4=0\) is such that \(\sin \theta=\frac{1}{3}\), then the value of \(\lambda\) is
1 \(\frac{-3}{5}\)
2 \(\frac{-5}{3}\)
3 \(\frac{5}{3}\)
4 \(\frac{3}{5}\)
Explanation:
C Given, \(\sin \theta=\frac{1}{3}\)
The direction ratios of the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) are 1,2,2 and the direction ratios of the plane \(2 x-y+\sqrt{\lambda} z+4=0\) are \(2,-1, \sqrt{\lambda}\).
\(\left \vert\frac{(1)(2)+2(-1)+2(\sqrt{\lambda})}{\sqrt{1^2+2^2+2^2} \sqrt{2^2+(-1)^2+(\sqrt{\lambda})^2}}\right \vert=\frac{1}{3}\)
\(\left \vert\frac{2-2+2 \sqrt{\lambda}}{\sqrt{9} \sqrt{4+1+\lambda}}\right \vert=\frac{1}{3}\)
\(\frac{2 \sqrt{\lambda}}{3 \sqrt{5+\lambda}}=\frac{1}{3}\)
\(2 \sqrt{\lambda}=\sqrt{5+\lambda}\)
On squaring both sides we get -
\(4 \lambda=5+\lambda \Rightarrow 3 \lambda=5 \Rightarrow \lambda=\frac{5}{3}\)
MHT CET-2019
Three Dimensional Geometry
121318
If the line passing through the origin makes angles \(\boldsymbol{\theta}_1, \boldsymbol{\theta}_2, \boldsymbol{\theta}_3\) with the planes XOY, YOZ, ZOX respectively, then
A Let \(P \equiv(x, y, z)\) be a point on the line passing through origin direction ratios of \(\mathrm{OP}\) are \(\mathrm{x}, \mathrm{y}\), \(\mathrm{z}\) and direction ratios of XOY plane are \(0,0,1\). \(\therefore \sin \theta_1=\frac{\mathrm{x}(0)+\mathrm{y}(0)+\mathrm{z}(1)}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2} \sqrt{0+0+1}}=\frac{\mathrm{z}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) Similarly, \(\sin \theta_2=\frac{x}{\sqrt{x^2+y^2+z^2}}\) and \(\sin \theta_3=\frac{\mathrm{y}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}}\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3\) \(=\frac{\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{x}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}+\frac{\mathrm{y}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\) \(=\frac{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}=1\) \(\therefore \sin ^2 \theta_1+\sin ^2 \theta_2+\sin ^2 \theta_3=1\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=1-\sin ^2 \theta_3\) \(\sin ^2 \theta_1+\sin ^2 \theta_2=\cos ^2 \theta_3\)
MHT CET-2019
Three Dimensional Geometry
121319
If lines \(\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and \(\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}\) are perpendicular to each other, then \(\lambda=\)
1 6
2 \(-\frac{7}{6}\)
3 7
4 \(-\frac{6}{7}\)
Explanation:
D The given lines are Line \(1, \frac{2(x-2)}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}\) and Line \(2 \frac{x-1}{1}=\frac{3\left(y-\frac{1}{3}\right)}{\lambda}=\frac{z-2}{1}\) \(\frac{x-2}{\left(\frac{\lambda}{2}\right)}=\frac{y-1}{2}=\frac{z-3}{1} \quad \text { and } x-1=\frac{\left(y-\frac{1}{3}\right)}{\left(\frac{\lambda}{3}\right)}=\frac{z-2}{1}\) The direction ratios of the lines are \(\frac{\lambda}{2}, 2,1\) and \(1, \frac{\lambda}{3}, 1\) As the given lines are perpendicular - \(\therefore\left(\frac{\lambda}{2}\right)(1)+(2)\left(\frac{\lambda}{3}\right)+(1)(1)=0\) \(\frac{\lambda}{2}+\frac{2 \lambda}{3}=-1 \Rightarrow 7 \lambda=-6 \Rightarrow \lambda=-\frac{6}{7}\)
MHT CET-2019
Three Dimensional Geometry
121322
The angle between the lines
\(\overline{\mathbf{r}}=3 \hat{\mathbf{i}}+\mathbf{2} \mathbf{j}-\mathbf{4} \hat{\mathbf{k}}+\boldsymbol{\lambda}(\hat{\mathbf{i}}+\mathbf{2} \hat{\mathbf{j}}+\mathbf{2 \hat { k }})\)
1 \(\cos ^{-1}\left(\frac{18}{21}\right)\)
2 \(\cos ^{-1}\left(\frac{19}{21}\right)\)
3 \(\cos ^{-1}\left(\frac{20}{21}\right)\)
4 \(\cos ^{-1}\left(\frac{17}{21}\right)\)
Explanation:
B Let, \(\theta\) be the angle between the given lines. The given lines are parallel to the vectors \(\overline{\mathrm{b}}_1=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}_2=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\) The angle \(\theta\) between them is given by, \(\cos \theta=\frac{\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2}{\left \vert\overline{\mathrm{b}}_1\right \vert\left \vert\overline{\mathrm{b}}_2\right \vert}\)
\(\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})=(1)(3)+2(2)+2(6)\)
\(\left \vert\overline{\mathrm{b}}_1\right \vert=\sqrt{1+4+4}=\sqrt{9}=3 \text { and }\left \vert\overline{\mathrm{b}}_2\right \vert=\sqrt{9+4+36}=\sqrt{49}=7\)
\(\therefore \cos \theta=\frac{19}{3 \times 7}=\frac{19}{21} \Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)\)