C Given, Equation of line is \(3 x+4=4 y-1=1-4 z \Rightarrow \frac{3\left(x+\frac{4}{3}\right)}{1}=\frac{4\left(y-\frac{1}{4}\right)}{1}=\frac{-4\left(z-\frac{1}{4}\right)}{1}\) \(\frac{x+\frac{4}{3}}{\left(\frac{1}{3}\right)}=\frac{y-\frac{1}{4}}{\left(\frac{1}{4}\right)}=\frac{z-\frac{1}{4}}{\left(-\frac{1}{4}\right)}\) \(\therefore \frac{1}{3}, \frac{1}{4},-\frac{1}{4}\) are direction ratios of line i.e. \(4,3,-3\). Hence equation of required line is - \(\frac{\mathrm{x}-2}{4}=\frac{\mathrm{y}-4}{3}=\frac{\mathrm{z}-6}{-3}\)
MHT CET-2020
Three Dimensional Geometry
121252
The equation of the line passing through \((1,2,3)\) and perpendicular to the lines \(x-1=\frac{y+2}{2}=\frac{z+4}{4}\) and \(\frac{x-1}{2}=\frac{y-2}{2}=z+3\) is
1 \(\frac{x-1}{6}=\frac{y-2}{7}=\frac{z-3}{2}\)
2 \(\frac{x-1}{4}=\frac{2-y}{5}=\frac{z-3}{2}\)
3 \(x-1=\frac{y-2}{7}=\frac{z-3}{4}\)
4 \(\frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
Explanation:
D The vector perpendicular to both the given lines is given by - \(\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1\end{array}\right|=\hat{i}(-6)-\hat{j}(-7)+\hat{k}(-2)=-6 \hat{i}+7 \hat{j}-2 \hat{k}\)
Hence direction ratios of required line are \(6,-7,2\).
Thus equation of required line is
\(\frac{x-1}{6}=\frac{y-2}{-7}=\frac{z-3}{2} \text { i.e. } \frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
MHT CET-2020
Three Dimensional Geometry
121254
The parametric equations of the line passing through \(A(3,4,-7)\) and \(B(1,-1,6)\) are
A Given, Let, \(\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(3,4,-7)\) \(\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(1,-1,6)\) Required equation is - \(\mathrm{x}=\mathrm{x}_1+\lambda\left(\mathrm{x}_2-\mathrm{x}_1\right) \Rightarrow \mathrm{x}=3-2 \lambda\) \(\mathrm{y}=\mathrm{y}_1+\lambda\left(\mathrm{y}_2-\mathrm{y}_1\right) \Rightarrow \mathrm{y}=4-5 \lambda\) \(\mathrm{z}=\mathrm{z}_1+\lambda\left(\mathrm{z}_2-\mathrm{z}_1\right) \Rightarrow \mathrm{z}=-7+13 \lambda\)
MHT CET-2020
Three Dimensional Geometry
121255
Points on the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{6}\), which are at 7 unit distance from the origin are
1 \((0,0,7)\) and \((7,0,0)\)
2 \((2,3,6)\) and \((-2,-3,-6)\)
3 \((-7,0,0)\) and \((7,0,0)\)
4 \((-2,3,6)\) and \((2,-3,6)\)
Explanation:
B Let, \(\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) be any point on the line \(\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{6}\). \(\therefore\) The coordinates of \(\mathrm{P} \equiv(2 \lambda, 3 \lambda, 6 \lambda)\). The given line is passing through origin \(\mathrm{O}(0,0,0)\) and \(\mathrm{OP}=7\) \(\sqrt{(2 \lambda)^2+(3 \lambda)^2+(6 \lambda)^2}=7\) \(4 \lambda^2+9 \lambda^2+36 \lambda^2=49\) \(49 \lambda^2=49\) \(\lambda= \pm 1\) \(\therefore\) The coordinates of \(\mathrm{P}\) are \((2,3,6)\) or \((-2,-3,-6)\)
C Given, Equation of line is \(3 x+4=4 y-1=1-4 z \Rightarrow \frac{3\left(x+\frac{4}{3}\right)}{1}=\frac{4\left(y-\frac{1}{4}\right)}{1}=\frac{-4\left(z-\frac{1}{4}\right)}{1}\) \(\frac{x+\frac{4}{3}}{\left(\frac{1}{3}\right)}=\frac{y-\frac{1}{4}}{\left(\frac{1}{4}\right)}=\frac{z-\frac{1}{4}}{\left(-\frac{1}{4}\right)}\) \(\therefore \frac{1}{3}, \frac{1}{4},-\frac{1}{4}\) are direction ratios of line i.e. \(4,3,-3\). Hence equation of required line is - \(\frac{\mathrm{x}-2}{4}=\frac{\mathrm{y}-4}{3}=\frac{\mathrm{z}-6}{-3}\)
MHT CET-2020
Three Dimensional Geometry
121252
The equation of the line passing through \((1,2,3)\) and perpendicular to the lines \(x-1=\frac{y+2}{2}=\frac{z+4}{4}\) and \(\frac{x-1}{2}=\frac{y-2}{2}=z+3\) is
1 \(\frac{x-1}{6}=\frac{y-2}{7}=\frac{z-3}{2}\)
2 \(\frac{x-1}{4}=\frac{2-y}{5}=\frac{z-3}{2}\)
3 \(x-1=\frac{y-2}{7}=\frac{z-3}{4}\)
4 \(\frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
Explanation:
D The vector perpendicular to both the given lines is given by - \(\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1\end{array}\right|=\hat{i}(-6)-\hat{j}(-7)+\hat{k}(-2)=-6 \hat{i}+7 \hat{j}-2 \hat{k}\)
Hence direction ratios of required line are \(6,-7,2\).
Thus equation of required line is
\(\frac{x-1}{6}=\frac{y-2}{-7}=\frac{z-3}{2} \text { i.e. } \frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
MHT CET-2020
Three Dimensional Geometry
121254
The parametric equations of the line passing through \(A(3,4,-7)\) and \(B(1,-1,6)\) are
A Given, Let, \(\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(3,4,-7)\) \(\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(1,-1,6)\) Required equation is - \(\mathrm{x}=\mathrm{x}_1+\lambda\left(\mathrm{x}_2-\mathrm{x}_1\right) \Rightarrow \mathrm{x}=3-2 \lambda\) \(\mathrm{y}=\mathrm{y}_1+\lambda\left(\mathrm{y}_2-\mathrm{y}_1\right) \Rightarrow \mathrm{y}=4-5 \lambda\) \(\mathrm{z}=\mathrm{z}_1+\lambda\left(\mathrm{z}_2-\mathrm{z}_1\right) \Rightarrow \mathrm{z}=-7+13 \lambda\)
MHT CET-2020
Three Dimensional Geometry
121255
Points on the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{6}\), which are at 7 unit distance from the origin are
1 \((0,0,7)\) and \((7,0,0)\)
2 \((2,3,6)\) and \((-2,-3,-6)\)
3 \((-7,0,0)\) and \((7,0,0)\)
4 \((-2,3,6)\) and \((2,-3,6)\)
Explanation:
B Let, \(\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) be any point on the line \(\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{6}\). \(\therefore\) The coordinates of \(\mathrm{P} \equiv(2 \lambda, 3 \lambda, 6 \lambda)\). The given line is passing through origin \(\mathrm{O}(0,0,0)\) and \(\mathrm{OP}=7\) \(\sqrt{(2 \lambda)^2+(3 \lambda)^2+(6 \lambda)^2}=7\) \(4 \lambda^2+9 \lambda^2+36 \lambda^2=49\) \(49 \lambda^2=49\) \(\lambda= \pm 1\) \(\therefore\) The coordinates of \(\mathrm{P}\) are \((2,3,6)\) or \((-2,-3,-6)\)
C Given, Equation of line is \(3 x+4=4 y-1=1-4 z \Rightarrow \frac{3\left(x+\frac{4}{3}\right)}{1}=\frac{4\left(y-\frac{1}{4}\right)}{1}=\frac{-4\left(z-\frac{1}{4}\right)}{1}\) \(\frac{x+\frac{4}{3}}{\left(\frac{1}{3}\right)}=\frac{y-\frac{1}{4}}{\left(\frac{1}{4}\right)}=\frac{z-\frac{1}{4}}{\left(-\frac{1}{4}\right)}\) \(\therefore \frac{1}{3}, \frac{1}{4},-\frac{1}{4}\) are direction ratios of line i.e. \(4,3,-3\). Hence equation of required line is - \(\frac{\mathrm{x}-2}{4}=\frac{\mathrm{y}-4}{3}=\frac{\mathrm{z}-6}{-3}\)
MHT CET-2020
Three Dimensional Geometry
121252
The equation of the line passing through \((1,2,3)\) and perpendicular to the lines \(x-1=\frac{y+2}{2}=\frac{z+4}{4}\) and \(\frac{x-1}{2}=\frac{y-2}{2}=z+3\) is
1 \(\frac{x-1}{6}=\frac{y-2}{7}=\frac{z-3}{2}\)
2 \(\frac{x-1}{4}=\frac{2-y}{5}=\frac{z-3}{2}\)
3 \(x-1=\frac{y-2}{7}=\frac{z-3}{4}\)
4 \(\frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
Explanation:
D The vector perpendicular to both the given lines is given by - \(\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1\end{array}\right|=\hat{i}(-6)-\hat{j}(-7)+\hat{k}(-2)=-6 \hat{i}+7 \hat{j}-2 \hat{k}\)
Hence direction ratios of required line are \(6,-7,2\).
Thus equation of required line is
\(\frac{x-1}{6}=\frac{y-2}{-7}=\frac{z-3}{2} \text { i.e. } \frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
MHT CET-2020
Three Dimensional Geometry
121254
The parametric equations of the line passing through \(A(3,4,-7)\) and \(B(1,-1,6)\) are
A Given, Let, \(\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(3,4,-7)\) \(\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(1,-1,6)\) Required equation is - \(\mathrm{x}=\mathrm{x}_1+\lambda\left(\mathrm{x}_2-\mathrm{x}_1\right) \Rightarrow \mathrm{x}=3-2 \lambda\) \(\mathrm{y}=\mathrm{y}_1+\lambda\left(\mathrm{y}_2-\mathrm{y}_1\right) \Rightarrow \mathrm{y}=4-5 \lambda\) \(\mathrm{z}=\mathrm{z}_1+\lambda\left(\mathrm{z}_2-\mathrm{z}_1\right) \Rightarrow \mathrm{z}=-7+13 \lambda\)
MHT CET-2020
Three Dimensional Geometry
121255
Points on the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{6}\), which are at 7 unit distance from the origin are
1 \((0,0,7)\) and \((7,0,0)\)
2 \((2,3,6)\) and \((-2,-3,-6)\)
3 \((-7,0,0)\) and \((7,0,0)\)
4 \((-2,3,6)\) and \((2,-3,6)\)
Explanation:
B Let, \(\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) be any point on the line \(\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{6}\). \(\therefore\) The coordinates of \(\mathrm{P} \equiv(2 \lambda, 3 \lambda, 6 \lambda)\). The given line is passing through origin \(\mathrm{O}(0,0,0)\) and \(\mathrm{OP}=7\) \(\sqrt{(2 \lambda)^2+(3 \lambda)^2+(6 \lambda)^2}=7\) \(4 \lambda^2+9 \lambda^2+36 \lambda^2=49\) \(49 \lambda^2=49\) \(\lambda= \pm 1\) \(\therefore\) The coordinates of \(\mathrm{P}\) are \((2,3,6)\) or \((-2,-3,-6)\)
C Given, Equation of line is \(3 x+4=4 y-1=1-4 z \Rightarrow \frac{3\left(x+\frac{4}{3}\right)}{1}=\frac{4\left(y-\frac{1}{4}\right)}{1}=\frac{-4\left(z-\frac{1}{4}\right)}{1}\) \(\frac{x+\frac{4}{3}}{\left(\frac{1}{3}\right)}=\frac{y-\frac{1}{4}}{\left(\frac{1}{4}\right)}=\frac{z-\frac{1}{4}}{\left(-\frac{1}{4}\right)}\) \(\therefore \frac{1}{3}, \frac{1}{4},-\frac{1}{4}\) are direction ratios of line i.e. \(4,3,-3\). Hence equation of required line is - \(\frac{\mathrm{x}-2}{4}=\frac{\mathrm{y}-4}{3}=\frac{\mathrm{z}-6}{-3}\)
MHT CET-2020
Three Dimensional Geometry
121252
The equation of the line passing through \((1,2,3)\) and perpendicular to the lines \(x-1=\frac{y+2}{2}=\frac{z+4}{4}\) and \(\frac{x-1}{2}=\frac{y-2}{2}=z+3\) is
1 \(\frac{x-1}{6}=\frac{y-2}{7}=\frac{z-3}{2}\)
2 \(\frac{x-1}{4}=\frac{2-y}{5}=\frac{z-3}{2}\)
3 \(x-1=\frac{y-2}{7}=\frac{z-3}{4}\)
4 \(\frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
Explanation:
D The vector perpendicular to both the given lines is given by - \(\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1\end{array}\right|=\hat{i}(-6)-\hat{j}(-7)+\hat{k}(-2)=-6 \hat{i}+7 \hat{j}-2 \hat{k}\)
Hence direction ratios of required line are \(6,-7,2\).
Thus equation of required line is
\(\frac{x-1}{6}=\frac{y-2}{-7}=\frac{z-3}{2} \text { i.e. } \frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}\)
MHT CET-2020
Three Dimensional Geometry
121254
The parametric equations of the line passing through \(A(3,4,-7)\) and \(B(1,-1,6)\) are
A Given, Let, \(\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)=(3,4,-7)\) \(\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2\right)=(1,-1,6)\) Required equation is - \(\mathrm{x}=\mathrm{x}_1+\lambda\left(\mathrm{x}_2-\mathrm{x}_1\right) \Rightarrow \mathrm{x}=3-2 \lambda\) \(\mathrm{y}=\mathrm{y}_1+\lambda\left(\mathrm{y}_2-\mathrm{y}_1\right) \Rightarrow \mathrm{y}=4-5 \lambda\) \(\mathrm{z}=\mathrm{z}_1+\lambda\left(\mathrm{z}_2-\mathrm{z}_1\right) \Rightarrow \mathrm{z}=-7+13 \lambda\)
MHT CET-2020
Three Dimensional Geometry
121255
Points on the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{6}\), which are at 7 unit distance from the origin are
1 \((0,0,7)\) and \((7,0,0)\)
2 \((2,3,6)\) and \((-2,-3,-6)\)
3 \((-7,0,0)\) and \((7,0,0)\)
4 \((-2,3,6)\) and \((2,-3,6)\)
Explanation:
B Let, \(\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) be any point on the line \(\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{3}=\frac{\mathrm{z}}{6}\). \(\therefore\) The coordinates of \(\mathrm{P} \equiv(2 \lambda, 3 \lambda, 6 \lambda)\). The given line is passing through origin \(\mathrm{O}(0,0,0)\) and \(\mathrm{OP}=7\) \(\sqrt{(2 \lambda)^2+(3 \lambda)^2+(6 \lambda)^2}=7\) \(4 \lambda^2+9 \lambda^2+36 \lambda^2=49\) \(49 \lambda^2=49\) \(\lambda= \pm 1\) \(\therefore\) The coordinates of \(\mathrm{P}\) are \((2,3,6)\) or \((-2,-3,-6)\)
