C Let \(\alpha, \beta, \gamma\) be the angle made by the line with the positive \(\mathrm{Z}-\mathrm{Y}\) axes. \(\therefore\) Line bisect the angle between positive directions of \(\mathrm{Y}\) and \(\mathrm{Z}\) axes. \(\because\) Line is perpendicular to \(\mathrm{X}\)-axes \(\alpha=90^{\circ}, \beta=45^{\circ} \text {, and } \gamma=45^{\circ}\) We know that, If \(l, \mathrm{~m}, \mathrm{n}\) are the direction cosines of line, \(l=\cos \alpha=\cos 90^{\circ}=0\) \(\mathrm{~m}=\cos \beta=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\) \(\mathrm{n}=\cos \gamma=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)Direction cosines of the line are \(\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
MHT CET-2020
Three Dimensional Geometry
121121
If \(A \equiv(3,2,-1)\) and \(B \equiv(1,4,3)\), then equation of the plane which bisects the segment \(A B\) perpendicularly is
1 \(x-y+2 z-3=0\)
2 \(x+y-2 z-3=0\)
3 \(x-y-2 z+3=0\)
4 \(x+y+2 z+3=0\)
Explanation:
C Since plane bisects segment AB, the plane meets the line \(A B\) at the mid point i.e. \(\left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right)=(2,3,1)\) Now the line \(A B\) is perpendicular to the plane. So, direction ratios of the plane \(\mathrm{AB}\) is \((1-3,4-2,3+\) 1 ) i.e. \((-2,2,4)\) or \((-1,1,2)\) Thus the equation of the plane passing through \((2,3,1)\) and having the direction ratios \((-1,1,2)\) is \((\mathrm{x}-2)+(\mathrm{y}-3)+2(\mathrm{z}-1)=0\) \(\Rightarrow -\mathrm{x}+2+\mathrm{y}-3+2 \mathrm{z}-2=0\) \(\therefore \quad \mathrm{x}-\mathrm{y}-2 \mathrm{z}+3=0\)
MHT CET-2020
Three Dimensional Geometry
121123
If the lines \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) and \(\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}\) are perpendicular, then \(\lambda=\)
1 \(\frac{8}{3}\)
2 4
3 \(\frac{-8}{3}\)
4 -4
Explanation:
A Given, Line, \(\mathrm{L}_1: \frac{1-\mathrm{x}}{2}=\frac{\mathrm{y}-8}{\lambda}=\frac{\mathrm{z}-5}{2}\) and Line, \(\mathrm{L}_2: \frac{\mathrm{x}-11}{5}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{1}\) We have \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2} \Rightarrow \frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) Direction ratio of given lines are \(-2, \lambda, 2\) and \(5,3,1\). \(\because\) lines are perpendicular- \(\therefore-2(5)+\lambda(3)+2(1)=0\) \(-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}\)
MHT CET-2020
Three Dimensional Geometry
121124
A line makes an angle of \(45^{\circ}\) with \(x\)-axis and congruent angles with \(y\) and \(z\)-axis, then the direction cosines of the line are
1 \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)
Explanation:
C Let angle made by line with each of \(y\) and \(z\) axis be \(\theta\) \(\therefore \cos ^2 45^{\circ}+\cos ^2 \theta+\cos ^2 \theta=1\) \(\Rightarrow 2 \cos ^2 \theta=1-\frac{1}{2}=\frac{1}{2} \Rightarrow \cos ^2 \theta=\frac{1}{4} \Rightarrow \cos \theta= \pm \frac{1}{2}\) Hence, direction ratio is, \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} \text { or } \frac{1}{\sqrt{2}}, \frac{-1}{2}, \frac{-1}{2}\)
C Let \(\alpha, \beta, \gamma\) be the angle made by the line with the positive \(\mathrm{Z}-\mathrm{Y}\) axes. \(\therefore\) Line bisect the angle between positive directions of \(\mathrm{Y}\) and \(\mathrm{Z}\) axes. \(\because\) Line is perpendicular to \(\mathrm{X}\)-axes \(\alpha=90^{\circ}, \beta=45^{\circ} \text {, and } \gamma=45^{\circ}\) We know that, If \(l, \mathrm{~m}, \mathrm{n}\) are the direction cosines of line, \(l=\cos \alpha=\cos 90^{\circ}=0\) \(\mathrm{~m}=\cos \beta=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\) \(\mathrm{n}=\cos \gamma=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)Direction cosines of the line are \(\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
MHT CET-2020
Three Dimensional Geometry
121121
If \(A \equiv(3,2,-1)\) and \(B \equiv(1,4,3)\), then equation of the plane which bisects the segment \(A B\) perpendicularly is
1 \(x-y+2 z-3=0\)
2 \(x+y-2 z-3=0\)
3 \(x-y-2 z+3=0\)
4 \(x+y+2 z+3=0\)
Explanation:
C Since plane bisects segment AB, the plane meets the line \(A B\) at the mid point i.e. \(\left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right)=(2,3,1)\) Now the line \(A B\) is perpendicular to the plane. So, direction ratios of the plane \(\mathrm{AB}\) is \((1-3,4-2,3+\) 1 ) i.e. \((-2,2,4)\) or \((-1,1,2)\) Thus the equation of the plane passing through \((2,3,1)\) and having the direction ratios \((-1,1,2)\) is \((\mathrm{x}-2)+(\mathrm{y}-3)+2(\mathrm{z}-1)=0\) \(\Rightarrow -\mathrm{x}+2+\mathrm{y}-3+2 \mathrm{z}-2=0\) \(\therefore \quad \mathrm{x}-\mathrm{y}-2 \mathrm{z}+3=0\)
MHT CET-2020
Three Dimensional Geometry
121123
If the lines \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) and \(\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}\) are perpendicular, then \(\lambda=\)
1 \(\frac{8}{3}\)
2 4
3 \(\frac{-8}{3}\)
4 -4
Explanation:
A Given, Line, \(\mathrm{L}_1: \frac{1-\mathrm{x}}{2}=\frac{\mathrm{y}-8}{\lambda}=\frac{\mathrm{z}-5}{2}\) and Line, \(\mathrm{L}_2: \frac{\mathrm{x}-11}{5}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{1}\) We have \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2} \Rightarrow \frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) Direction ratio of given lines are \(-2, \lambda, 2\) and \(5,3,1\). \(\because\) lines are perpendicular- \(\therefore-2(5)+\lambda(3)+2(1)=0\) \(-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}\)
MHT CET-2020
Three Dimensional Geometry
121124
A line makes an angle of \(45^{\circ}\) with \(x\)-axis and congruent angles with \(y\) and \(z\)-axis, then the direction cosines of the line are
1 \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)
Explanation:
C Let angle made by line with each of \(y\) and \(z\) axis be \(\theta\) \(\therefore \cos ^2 45^{\circ}+\cos ^2 \theta+\cos ^2 \theta=1\) \(\Rightarrow 2 \cos ^2 \theta=1-\frac{1}{2}=\frac{1}{2} \Rightarrow \cos ^2 \theta=\frac{1}{4} \Rightarrow \cos \theta= \pm \frac{1}{2}\) Hence, direction ratio is, \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} \text { or } \frac{1}{\sqrt{2}}, \frac{-1}{2}, \frac{-1}{2}\)
C Let \(\alpha, \beta, \gamma\) be the angle made by the line with the positive \(\mathrm{Z}-\mathrm{Y}\) axes. \(\therefore\) Line bisect the angle between positive directions of \(\mathrm{Y}\) and \(\mathrm{Z}\) axes. \(\because\) Line is perpendicular to \(\mathrm{X}\)-axes \(\alpha=90^{\circ}, \beta=45^{\circ} \text {, and } \gamma=45^{\circ}\) We know that, If \(l, \mathrm{~m}, \mathrm{n}\) are the direction cosines of line, \(l=\cos \alpha=\cos 90^{\circ}=0\) \(\mathrm{~m}=\cos \beta=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\) \(\mathrm{n}=\cos \gamma=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)Direction cosines of the line are \(\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
MHT CET-2020
Three Dimensional Geometry
121121
If \(A \equiv(3,2,-1)\) and \(B \equiv(1,4,3)\), then equation of the plane which bisects the segment \(A B\) perpendicularly is
1 \(x-y+2 z-3=0\)
2 \(x+y-2 z-3=0\)
3 \(x-y-2 z+3=0\)
4 \(x+y+2 z+3=0\)
Explanation:
C Since plane bisects segment AB, the plane meets the line \(A B\) at the mid point i.e. \(\left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right)=(2,3,1)\) Now the line \(A B\) is perpendicular to the plane. So, direction ratios of the plane \(\mathrm{AB}\) is \((1-3,4-2,3+\) 1 ) i.e. \((-2,2,4)\) or \((-1,1,2)\) Thus the equation of the plane passing through \((2,3,1)\) and having the direction ratios \((-1,1,2)\) is \((\mathrm{x}-2)+(\mathrm{y}-3)+2(\mathrm{z}-1)=0\) \(\Rightarrow -\mathrm{x}+2+\mathrm{y}-3+2 \mathrm{z}-2=0\) \(\therefore \quad \mathrm{x}-\mathrm{y}-2 \mathrm{z}+3=0\)
MHT CET-2020
Three Dimensional Geometry
121123
If the lines \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) and \(\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}\) are perpendicular, then \(\lambda=\)
1 \(\frac{8}{3}\)
2 4
3 \(\frac{-8}{3}\)
4 -4
Explanation:
A Given, Line, \(\mathrm{L}_1: \frac{1-\mathrm{x}}{2}=\frac{\mathrm{y}-8}{\lambda}=\frac{\mathrm{z}-5}{2}\) and Line, \(\mathrm{L}_2: \frac{\mathrm{x}-11}{5}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{1}\) We have \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2} \Rightarrow \frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) Direction ratio of given lines are \(-2, \lambda, 2\) and \(5,3,1\). \(\because\) lines are perpendicular- \(\therefore-2(5)+\lambda(3)+2(1)=0\) \(-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}\)
MHT CET-2020
Three Dimensional Geometry
121124
A line makes an angle of \(45^{\circ}\) with \(x\)-axis and congruent angles with \(y\) and \(z\)-axis, then the direction cosines of the line are
1 \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)
Explanation:
C Let angle made by line with each of \(y\) and \(z\) axis be \(\theta\) \(\therefore \cos ^2 45^{\circ}+\cos ^2 \theta+\cos ^2 \theta=1\) \(\Rightarrow 2 \cos ^2 \theta=1-\frac{1}{2}=\frac{1}{2} \Rightarrow \cos ^2 \theta=\frac{1}{4} \Rightarrow \cos \theta= \pm \frac{1}{2}\) Hence, direction ratio is, \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} \text { or } \frac{1}{\sqrt{2}}, \frac{-1}{2}, \frac{-1}{2}\)
C Let \(\alpha, \beta, \gamma\) be the angle made by the line with the positive \(\mathrm{Z}-\mathrm{Y}\) axes. \(\therefore\) Line bisect the angle between positive directions of \(\mathrm{Y}\) and \(\mathrm{Z}\) axes. \(\because\) Line is perpendicular to \(\mathrm{X}\)-axes \(\alpha=90^{\circ}, \beta=45^{\circ} \text {, and } \gamma=45^{\circ}\) We know that, If \(l, \mathrm{~m}, \mathrm{n}\) are the direction cosines of line, \(l=\cos \alpha=\cos 90^{\circ}=0\) \(\mathrm{~m}=\cos \beta=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\) \(\mathrm{n}=\cos \gamma=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)Direction cosines of the line are \(\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
MHT CET-2020
Three Dimensional Geometry
121121
If \(A \equiv(3,2,-1)\) and \(B \equiv(1,4,3)\), then equation of the plane which bisects the segment \(A B\) perpendicularly is
1 \(x-y+2 z-3=0\)
2 \(x+y-2 z-3=0\)
3 \(x-y-2 z+3=0\)
4 \(x+y+2 z+3=0\)
Explanation:
C Since plane bisects segment AB, the plane meets the line \(A B\) at the mid point i.e. \(\left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right)=(2,3,1)\) Now the line \(A B\) is perpendicular to the plane. So, direction ratios of the plane \(\mathrm{AB}\) is \((1-3,4-2,3+\) 1 ) i.e. \((-2,2,4)\) or \((-1,1,2)\) Thus the equation of the plane passing through \((2,3,1)\) and having the direction ratios \((-1,1,2)\) is \((\mathrm{x}-2)+(\mathrm{y}-3)+2(\mathrm{z}-1)=0\) \(\Rightarrow -\mathrm{x}+2+\mathrm{y}-3+2 \mathrm{z}-2=0\) \(\therefore \quad \mathrm{x}-\mathrm{y}-2 \mathrm{z}+3=0\)
MHT CET-2020
Three Dimensional Geometry
121123
If the lines \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) and \(\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}\) are perpendicular, then \(\lambda=\)
1 \(\frac{8}{3}\)
2 4
3 \(\frac{-8}{3}\)
4 -4
Explanation:
A Given, Line, \(\mathrm{L}_1: \frac{1-\mathrm{x}}{2}=\frac{\mathrm{y}-8}{\lambda}=\frac{\mathrm{z}-5}{2}\) and Line, \(\mathrm{L}_2: \frac{\mathrm{x}-11}{5}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{1}\) We have \(\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2} \Rightarrow \frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}\) Direction ratio of given lines are \(-2, \lambda, 2\) and \(5,3,1\). \(\because\) lines are perpendicular- \(\therefore-2(5)+\lambda(3)+2(1)=0\) \(-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}\)
MHT CET-2020
Three Dimensional Geometry
121124
A line makes an angle of \(45^{\circ}\) with \(x\)-axis and congruent angles with \(y\) and \(z\)-axis, then the direction cosines of the line are
1 \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)
2 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\) and \(\frac{1}{\sqrt{2}},-\frac{1}{2},-\frac{1}{2}\)
4 \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\) and \(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)
Explanation:
C Let angle made by line with each of \(y\) and \(z\) axis be \(\theta\) \(\therefore \cos ^2 45^{\circ}+\cos ^2 \theta+\cos ^2 \theta=1\) \(\Rightarrow 2 \cos ^2 \theta=1-\frac{1}{2}=\frac{1}{2} \Rightarrow \cos ^2 \theta=\frac{1}{4} \Rightarrow \cos \theta= \pm \frac{1}{2}\) Hence, direction ratio is, \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} \text { or } \frac{1}{\sqrt{2}}, \frac{-1}{2}, \frac{-1}{2}\)
