C We know that, \(\ell, \mathrm{m}, \mathrm{n}\) are the direction cosines of a line only if, \(\ell^2+m^2+n^2=1\) According to options- \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2} \neq 1\) \(\therefore \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\) are not direction cosine ratio of line.
MHT CET-2019
Three Dimensional Geometry
121128
The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes \(x+2 y+3 z=4\) and \(4 x+3 y+2 z=1\) are
1 \(3,2,1\)
2 \(2,3,1\)
3 \(1,2,3\)
4 \(3,1,2\)
Explanation:
A The equation of the plane passing through the line of intersection of the planes \(x+2 y+3 z-4=0 \quad\) and \(4 x+3 y+2 z-1=0\) is \((x+2 y+3 z-4)+\lambda(4 x+3 y+2 z-1)=0\) \(\Rightarrow \quad(1+4 \lambda) \mathrm{x}+(2+3 \lambda) \mathrm{y}+(3+2 \lambda) \mathrm{z}+(-4-\lambda)=0\) ..(i) But the plane passes through the origin \((0,0,0)\) \(\therefore-4-\lambda=0 \Rightarrow \lambda=-4\) Substituting the value of \(\lambda\) in eq. (i), we get- \(-15 x-10 y-5 z=0 \Rightarrow 3 x+2 y+z=0\) Hence, the direction ratio of the normal to the plane are \(3,2,1\).
MHT CET-2019
Three Dimensional Geometry
121129
The equation of line equally inclined to coordinate axes and passing through \((-3,2,-5)\) is
1 \(\frac{x+3}{1}=\frac{y-2}{1}=\frac{z+5}{1}\)
2 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{5+z}{-1}\)
3 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{1}\)
4 \(\frac{x+3}{-1}=\frac{2-y}{1}=\frac{z+5}{-1}\)
Explanation:
B Let equation of the line passing through \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) and direction cosines be \(l, \mathrm{~m}, \mathrm{n}\) is- \(\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{l}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~m}}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{n}}\) Line is equally inclined to coordinates axes. \(\therefore \quad l=-1, \mathrm{~m}=1, \mathrm{n}=-1\) Equation of line, \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{-1}\)
C We know that, \(\ell, \mathrm{m}, \mathrm{n}\) are the direction cosines of a line only if, \(\ell^2+m^2+n^2=1\) According to options- \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2} \neq 1\) \(\therefore \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\) are not direction cosine ratio of line.
MHT CET-2019
Three Dimensional Geometry
121128
The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes \(x+2 y+3 z=4\) and \(4 x+3 y+2 z=1\) are
1 \(3,2,1\)
2 \(2,3,1\)
3 \(1,2,3\)
4 \(3,1,2\)
Explanation:
A The equation of the plane passing through the line of intersection of the planes \(x+2 y+3 z-4=0 \quad\) and \(4 x+3 y+2 z-1=0\) is \((x+2 y+3 z-4)+\lambda(4 x+3 y+2 z-1)=0\) \(\Rightarrow \quad(1+4 \lambda) \mathrm{x}+(2+3 \lambda) \mathrm{y}+(3+2 \lambda) \mathrm{z}+(-4-\lambda)=0\) ..(i) But the plane passes through the origin \((0,0,0)\) \(\therefore-4-\lambda=0 \Rightarrow \lambda=-4\) Substituting the value of \(\lambda\) in eq. (i), we get- \(-15 x-10 y-5 z=0 \Rightarrow 3 x+2 y+z=0\) Hence, the direction ratio of the normal to the plane are \(3,2,1\).
MHT CET-2019
Three Dimensional Geometry
121129
The equation of line equally inclined to coordinate axes and passing through \((-3,2,-5)\) is
1 \(\frac{x+3}{1}=\frac{y-2}{1}=\frac{z+5}{1}\)
2 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{5+z}{-1}\)
3 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{1}\)
4 \(\frac{x+3}{-1}=\frac{2-y}{1}=\frac{z+5}{-1}\)
Explanation:
B Let equation of the line passing through \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) and direction cosines be \(l, \mathrm{~m}, \mathrm{n}\) is- \(\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{l}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~m}}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{n}}\) Line is equally inclined to coordinates axes. \(\therefore \quad l=-1, \mathrm{~m}=1, \mathrm{n}=-1\) Equation of line, \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{-1}\)
C We know that, \(\ell, \mathrm{m}, \mathrm{n}\) are the direction cosines of a line only if, \(\ell^2+m^2+n^2=1\) According to options- \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2} \neq 1\) \(\therefore \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\) are not direction cosine ratio of line.
MHT CET-2019
Three Dimensional Geometry
121128
The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes \(x+2 y+3 z=4\) and \(4 x+3 y+2 z=1\) are
1 \(3,2,1\)
2 \(2,3,1\)
3 \(1,2,3\)
4 \(3,1,2\)
Explanation:
A The equation of the plane passing through the line of intersection of the planes \(x+2 y+3 z-4=0 \quad\) and \(4 x+3 y+2 z-1=0\) is \((x+2 y+3 z-4)+\lambda(4 x+3 y+2 z-1)=0\) \(\Rightarrow \quad(1+4 \lambda) \mathrm{x}+(2+3 \lambda) \mathrm{y}+(3+2 \lambda) \mathrm{z}+(-4-\lambda)=0\) ..(i) But the plane passes through the origin \((0,0,0)\) \(\therefore-4-\lambda=0 \Rightarrow \lambda=-4\) Substituting the value of \(\lambda\) in eq. (i), we get- \(-15 x-10 y-5 z=0 \Rightarrow 3 x+2 y+z=0\) Hence, the direction ratio of the normal to the plane are \(3,2,1\).
MHT CET-2019
Three Dimensional Geometry
121129
The equation of line equally inclined to coordinate axes and passing through \((-3,2,-5)\) is
1 \(\frac{x+3}{1}=\frac{y-2}{1}=\frac{z+5}{1}\)
2 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{5+z}{-1}\)
3 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{1}\)
4 \(\frac{x+3}{-1}=\frac{2-y}{1}=\frac{z+5}{-1}\)
Explanation:
B Let equation of the line passing through \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) and direction cosines be \(l, \mathrm{~m}, \mathrm{n}\) is- \(\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{l}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~m}}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{n}}\) Line is equally inclined to coordinates axes. \(\therefore \quad l=-1, \mathrm{~m}=1, \mathrm{n}=-1\) Equation of line, \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{-1}\)
C We know that, \(\ell, \mathrm{m}, \mathrm{n}\) are the direction cosines of a line only if, \(\ell^2+m^2+n^2=1\) According to options- \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2+\left(\frac{-1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2} \neq 1\) \(\therefore \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}\) are not direction cosine ratio of line.
MHT CET-2019
Three Dimensional Geometry
121128
The direction ratios of the normal to the plane passing through origin and the line of intersection of the planes \(x+2 y+3 z=4\) and \(4 x+3 y+2 z=1\) are
1 \(3,2,1\)
2 \(2,3,1\)
3 \(1,2,3\)
4 \(3,1,2\)
Explanation:
A The equation of the plane passing through the line of intersection of the planes \(x+2 y+3 z-4=0 \quad\) and \(4 x+3 y+2 z-1=0\) is \((x+2 y+3 z-4)+\lambda(4 x+3 y+2 z-1)=0\) \(\Rightarrow \quad(1+4 \lambda) \mathrm{x}+(2+3 \lambda) \mathrm{y}+(3+2 \lambda) \mathrm{z}+(-4-\lambda)=0\) ..(i) But the plane passes through the origin \((0,0,0)\) \(\therefore-4-\lambda=0 \Rightarrow \lambda=-4\) Substituting the value of \(\lambda\) in eq. (i), we get- \(-15 x-10 y-5 z=0 \Rightarrow 3 x+2 y+z=0\) Hence, the direction ratio of the normal to the plane are \(3,2,1\).
MHT CET-2019
Three Dimensional Geometry
121129
The equation of line equally inclined to coordinate axes and passing through \((-3,2,-5)\) is
1 \(\frac{x+3}{1}=\frac{y-2}{1}=\frac{z+5}{1}\)
2 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{5+z}{-1}\)
3 \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{1}\)
4 \(\frac{x+3}{-1}=\frac{2-y}{1}=\frac{z+5}{-1}\)
Explanation:
B Let equation of the line passing through \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) and direction cosines be \(l, \mathrm{~m}, \mathrm{n}\) is- \(\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{l}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~m}}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{n}}\) Line is equally inclined to coordinates axes. \(\therefore \quad l=-1, \mathrm{~m}=1, \mathrm{n}=-1\) Equation of line, \(\frac{x+3}{-1}=\frac{y-2}{1}=\frac{z+5}{-1}\)