NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121100
The line segment joining the points \(A(2,3,4)\) and \(B(-3,5,-4)\) intersects \(y z\) plane at the point.
1 \(\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
2 \((0,4,5)\)
3 \(\left(9, \frac{14}{5}, 4\right)\)
4 \((0,0,0)\)
Explanation:
A Given, \(A=(2,3,4)\) \(B=(-3,5,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in ratio \(\lambda: 1\). Coordinates of \(\mathrm{P}\) will be - \(\mathrm{P}=\left(\frac{-3 \lambda+2}{\lambda+1}, \frac{5 \lambda+3}{\lambda+1}, \frac{-4 \lambda+4}{\lambda+1}\right)\) \(\because\) P lies on \(\mathrm{yz}=\) plane so \(\mathrm{x}\)-coordinate will be zero. \(\frac{-3 \lambda+2}{\lambda+1}=0\) \(\lambda=\frac{2}{3}\) Put \(t=2 / 3\) in equation (i) \(\mathrm{P}=\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
AP EAMCET-21.09.2020
Three Dimensional Geometry
121101
The line \(x-2 y+4 z+4=0, x+y+z-8=0\) intersect the plane \(x-y+2 z+1=0\) at the point
1 \((-2,5,1)\)
2 \((2,-5,1)\)
3 \((2,5,-1)\)
4 \((2,5,1)\)
Explanation:
D We have equation of line, \(x-2 y+4 z+4=0\) \(x+y+z-8=0\) Subtracting equation (i) and (ii) we get - \(y-z=4\) Given, equation of plane - \(x-y+2 z+1=0\) According to question on solving equation (iii) and (iv) we get \(\mathrm{x}+\mathrm{z}=3\) Hence, \(y=5, x=2\) and \(z=1\),
WB JEE-2022
Three Dimensional Geometry
121102
What will be the coordinates of foot of perpendicular from the point \((1,1.5,2)\) to the plane \(2 x-2 y+4 z+5=0\) ?
1 \((1,1 / 2,1)\)
2 \((0,5 / 2,0)\)
3 \((0,0,-5 / 4) /\)
4 \((6,1,-15 / 4)\)
Explanation:
B Let A \((1,1.5,2)\) be the given point and let \(\mathrm{B}\) be the foot of perpendicular drawn from \(A\) to plane \(2 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+5=0\) Direction ratio of normal to the plane (i) is \((2,-2,4)\) equation of line \(A B\) is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-1.5}{-2}=\frac{\mathrm{z}-2}{4}=\lambda\) \(\mathrm{B}(1,1.5,2)\) \(\mathrm{x}, \mathrm{y}, 2)\) Any point on line \(\mathrm{AB}\) is \(\mathrm{B}(2 \lambda+1,1.5-2 \lambda, 4 \lambda+2)\) This point lies in plane (i) \(2(2 \lambda+1)-2(1.5-2 \lambda)+4(4 \lambda+2)+5=0\) \(24 \lambda=-12\) \(\lambda=\frac{-1}{2}\) So, coordinate of \(\mathrm{B}=(0,2.5,0)\) \(=(0,5 / 2,0)\)
J and K CET-2018
Three Dimensional Geometry
121103
Two lines in three-dimensional system are given as \((x-1) / 2=(y-2) / 3=(z-3) / 4\) and \((x-\) 4) \(/ 5=(y-1) / 2=z\). Their point of intersection is
1 \((1,2,3)\)
2 \((-1,-1,-1)\)
3 \((4,1,0)\)
4 \((1,1,1)\)
Explanation:
C Given line are - \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda\) And \(\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}\) Let the point on the line (i) is \((2 \lambda+1,3 \lambda+2,4 \lambda+3)\) According to question, line (i) and (ii) intersect each other- So, the above point satisfies the line (ii), \(\frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3-0}{1}\) \(\Rightarrow \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}=\frac{4 \lambda+3}{1}\) \(\Rightarrow 4 \lambda-6=15 \lambda+5\) \(\Rightarrow 11 \lambda=-11\) \(\Rightarrow \lambda=-1\) So, point of intersection is, \((2(-1)+1),(3(-1)+2),(4(-1)+3)\) \(=(-1,-1,-1)\)
121100
The line segment joining the points \(A(2,3,4)\) and \(B(-3,5,-4)\) intersects \(y z\) plane at the point.
1 \(\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
2 \((0,4,5)\)
3 \(\left(9, \frac{14}{5}, 4\right)\)
4 \((0,0,0)\)
Explanation:
A Given, \(A=(2,3,4)\) \(B=(-3,5,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in ratio \(\lambda: 1\). Coordinates of \(\mathrm{P}\) will be - \(\mathrm{P}=\left(\frac{-3 \lambda+2}{\lambda+1}, \frac{5 \lambda+3}{\lambda+1}, \frac{-4 \lambda+4}{\lambda+1}\right)\) \(\because\) P lies on \(\mathrm{yz}=\) plane so \(\mathrm{x}\)-coordinate will be zero. \(\frac{-3 \lambda+2}{\lambda+1}=0\) \(\lambda=\frac{2}{3}\) Put \(t=2 / 3\) in equation (i) \(\mathrm{P}=\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
AP EAMCET-21.09.2020
Three Dimensional Geometry
121101
The line \(x-2 y+4 z+4=0, x+y+z-8=0\) intersect the plane \(x-y+2 z+1=0\) at the point
1 \((-2,5,1)\)
2 \((2,-5,1)\)
3 \((2,5,-1)\)
4 \((2,5,1)\)
Explanation:
D We have equation of line, \(x-2 y+4 z+4=0\) \(x+y+z-8=0\) Subtracting equation (i) and (ii) we get - \(y-z=4\) Given, equation of plane - \(x-y+2 z+1=0\) According to question on solving equation (iii) and (iv) we get \(\mathrm{x}+\mathrm{z}=3\) Hence, \(y=5, x=2\) and \(z=1\),
WB JEE-2022
Three Dimensional Geometry
121102
What will be the coordinates of foot of perpendicular from the point \((1,1.5,2)\) to the plane \(2 x-2 y+4 z+5=0\) ?
1 \((1,1 / 2,1)\)
2 \((0,5 / 2,0)\)
3 \((0,0,-5 / 4) /\)
4 \((6,1,-15 / 4)\)
Explanation:
B Let A \((1,1.5,2)\) be the given point and let \(\mathrm{B}\) be the foot of perpendicular drawn from \(A\) to plane \(2 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+5=0\) Direction ratio of normal to the plane (i) is \((2,-2,4)\) equation of line \(A B\) is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-1.5}{-2}=\frac{\mathrm{z}-2}{4}=\lambda\) \(\mathrm{B}(1,1.5,2)\) \(\mathrm{x}, \mathrm{y}, 2)\) Any point on line \(\mathrm{AB}\) is \(\mathrm{B}(2 \lambda+1,1.5-2 \lambda, 4 \lambda+2)\) This point lies in plane (i) \(2(2 \lambda+1)-2(1.5-2 \lambda)+4(4 \lambda+2)+5=0\) \(24 \lambda=-12\) \(\lambda=\frac{-1}{2}\) So, coordinate of \(\mathrm{B}=(0,2.5,0)\) \(=(0,5 / 2,0)\)
J and K CET-2018
Three Dimensional Geometry
121103
Two lines in three-dimensional system are given as \((x-1) / 2=(y-2) / 3=(z-3) / 4\) and \((x-\) 4) \(/ 5=(y-1) / 2=z\). Their point of intersection is
1 \((1,2,3)\)
2 \((-1,-1,-1)\)
3 \((4,1,0)\)
4 \((1,1,1)\)
Explanation:
C Given line are - \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda\) And \(\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}\) Let the point on the line (i) is \((2 \lambda+1,3 \lambda+2,4 \lambda+3)\) According to question, line (i) and (ii) intersect each other- So, the above point satisfies the line (ii), \(\frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3-0}{1}\) \(\Rightarrow \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}=\frac{4 \lambda+3}{1}\) \(\Rightarrow 4 \lambda-6=15 \lambda+5\) \(\Rightarrow 11 \lambda=-11\) \(\Rightarrow \lambda=-1\) So, point of intersection is, \((2(-1)+1),(3(-1)+2),(4(-1)+3)\) \(=(-1,-1,-1)\)
121100
The line segment joining the points \(A(2,3,4)\) and \(B(-3,5,-4)\) intersects \(y z\) plane at the point.
1 \(\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
2 \((0,4,5)\)
3 \(\left(9, \frac{14}{5}, 4\right)\)
4 \((0,0,0)\)
Explanation:
A Given, \(A=(2,3,4)\) \(B=(-3,5,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in ratio \(\lambda: 1\). Coordinates of \(\mathrm{P}\) will be - \(\mathrm{P}=\left(\frac{-3 \lambda+2}{\lambda+1}, \frac{5 \lambda+3}{\lambda+1}, \frac{-4 \lambda+4}{\lambda+1}\right)\) \(\because\) P lies on \(\mathrm{yz}=\) plane so \(\mathrm{x}\)-coordinate will be zero. \(\frac{-3 \lambda+2}{\lambda+1}=0\) \(\lambda=\frac{2}{3}\) Put \(t=2 / 3\) in equation (i) \(\mathrm{P}=\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
AP EAMCET-21.09.2020
Three Dimensional Geometry
121101
The line \(x-2 y+4 z+4=0, x+y+z-8=0\) intersect the plane \(x-y+2 z+1=0\) at the point
1 \((-2,5,1)\)
2 \((2,-5,1)\)
3 \((2,5,-1)\)
4 \((2,5,1)\)
Explanation:
D We have equation of line, \(x-2 y+4 z+4=0\) \(x+y+z-8=0\) Subtracting equation (i) and (ii) we get - \(y-z=4\) Given, equation of plane - \(x-y+2 z+1=0\) According to question on solving equation (iii) and (iv) we get \(\mathrm{x}+\mathrm{z}=3\) Hence, \(y=5, x=2\) and \(z=1\),
WB JEE-2022
Three Dimensional Geometry
121102
What will be the coordinates of foot of perpendicular from the point \((1,1.5,2)\) to the plane \(2 x-2 y+4 z+5=0\) ?
1 \((1,1 / 2,1)\)
2 \((0,5 / 2,0)\)
3 \((0,0,-5 / 4) /\)
4 \((6,1,-15 / 4)\)
Explanation:
B Let A \((1,1.5,2)\) be the given point and let \(\mathrm{B}\) be the foot of perpendicular drawn from \(A\) to plane \(2 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+5=0\) Direction ratio of normal to the plane (i) is \((2,-2,4)\) equation of line \(A B\) is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-1.5}{-2}=\frac{\mathrm{z}-2}{4}=\lambda\) \(\mathrm{B}(1,1.5,2)\) \(\mathrm{x}, \mathrm{y}, 2)\) Any point on line \(\mathrm{AB}\) is \(\mathrm{B}(2 \lambda+1,1.5-2 \lambda, 4 \lambda+2)\) This point lies in plane (i) \(2(2 \lambda+1)-2(1.5-2 \lambda)+4(4 \lambda+2)+5=0\) \(24 \lambda=-12\) \(\lambda=\frac{-1}{2}\) So, coordinate of \(\mathrm{B}=(0,2.5,0)\) \(=(0,5 / 2,0)\)
J and K CET-2018
Three Dimensional Geometry
121103
Two lines in three-dimensional system are given as \((x-1) / 2=(y-2) / 3=(z-3) / 4\) and \((x-\) 4) \(/ 5=(y-1) / 2=z\). Their point of intersection is
1 \((1,2,3)\)
2 \((-1,-1,-1)\)
3 \((4,1,0)\)
4 \((1,1,1)\)
Explanation:
C Given line are - \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda\) And \(\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}\) Let the point on the line (i) is \((2 \lambda+1,3 \lambda+2,4 \lambda+3)\) According to question, line (i) and (ii) intersect each other- So, the above point satisfies the line (ii), \(\frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3-0}{1}\) \(\Rightarrow \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}=\frac{4 \lambda+3}{1}\) \(\Rightarrow 4 \lambda-6=15 \lambda+5\) \(\Rightarrow 11 \lambda=-11\) \(\Rightarrow \lambda=-1\) So, point of intersection is, \((2(-1)+1),(3(-1)+2),(4(-1)+3)\) \(=(-1,-1,-1)\)
121100
The line segment joining the points \(A(2,3,4)\) and \(B(-3,5,-4)\) intersects \(y z\) plane at the point.
1 \(\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
2 \((0,4,5)\)
3 \(\left(9, \frac{14}{5}, 4\right)\)
4 \((0,0,0)\)
Explanation:
A Given, \(A=(2,3,4)\) \(B=(-3,5,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in ratio \(\lambda: 1\). Coordinates of \(\mathrm{P}\) will be - \(\mathrm{P}=\left(\frac{-3 \lambda+2}{\lambda+1}, \frac{5 \lambda+3}{\lambda+1}, \frac{-4 \lambda+4}{\lambda+1}\right)\) \(\because\) P lies on \(\mathrm{yz}=\) plane so \(\mathrm{x}\)-coordinate will be zero. \(\frac{-3 \lambda+2}{\lambda+1}=0\) \(\lambda=\frac{2}{3}\) Put \(t=2 / 3\) in equation (i) \(\mathrm{P}=\left(0, \frac{19}{5}, \frac{4}{5}\right)\)
AP EAMCET-21.09.2020
Three Dimensional Geometry
121101
The line \(x-2 y+4 z+4=0, x+y+z-8=0\) intersect the plane \(x-y+2 z+1=0\) at the point
1 \((-2,5,1)\)
2 \((2,-5,1)\)
3 \((2,5,-1)\)
4 \((2,5,1)\)
Explanation:
D We have equation of line, \(x-2 y+4 z+4=0\) \(x+y+z-8=0\) Subtracting equation (i) and (ii) we get - \(y-z=4\) Given, equation of plane - \(x-y+2 z+1=0\) According to question on solving equation (iii) and (iv) we get \(\mathrm{x}+\mathrm{z}=3\) Hence, \(y=5, x=2\) and \(z=1\),
WB JEE-2022
Three Dimensional Geometry
121102
What will be the coordinates of foot of perpendicular from the point \((1,1.5,2)\) to the plane \(2 x-2 y+4 z+5=0\) ?
1 \((1,1 / 2,1)\)
2 \((0,5 / 2,0)\)
3 \((0,0,-5 / 4) /\)
4 \((6,1,-15 / 4)\)
Explanation:
B Let A \((1,1.5,2)\) be the given point and let \(\mathrm{B}\) be the foot of perpendicular drawn from \(A\) to plane \(2 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+5=0\) Direction ratio of normal to the plane (i) is \((2,-2,4)\) equation of line \(A B\) is \(\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-1.5}{-2}=\frac{\mathrm{z}-2}{4}=\lambda\) \(\mathrm{B}(1,1.5,2)\) \(\mathrm{x}, \mathrm{y}, 2)\) Any point on line \(\mathrm{AB}\) is \(\mathrm{B}(2 \lambda+1,1.5-2 \lambda, 4 \lambda+2)\) This point lies in plane (i) \(2(2 \lambda+1)-2(1.5-2 \lambda)+4(4 \lambda+2)+5=0\) \(24 \lambda=-12\) \(\lambda=\frac{-1}{2}\) So, coordinate of \(\mathrm{B}=(0,2.5,0)\) \(=(0,5 / 2,0)\)
J and K CET-2018
Three Dimensional Geometry
121103
Two lines in three-dimensional system are given as \((x-1) / 2=(y-2) / 3=(z-3) / 4\) and \((x-\) 4) \(/ 5=(y-1) / 2=z\). Their point of intersection is
1 \((1,2,3)\)
2 \((-1,-1,-1)\)
3 \((4,1,0)\)
4 \((1,1,1)\)
Explanation:
C Given line are - \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda\) And \(\frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1}\) Let the point on the line (i) is \((2 \lambda+1,3 \lambda+2,4 \lambda+3)\) According to question, line (i) and (ii) intersect each other- So, the above point satisfies the line (ii), \(\frac{2 \lambda+1-4}{5}=\frac{3 \lambda+2-1}{2}=\frac{4 \lambda+3-0}{1}\) \(\Rightarrow \frac{2 \lambda-3}{5}=\frac{3 \lambda+1}{2}=\frac{4 \lambda+3}{1}\) \(\Rightarrow 4 \lambda-6=15 \lambda+5\) \(\Rightarrow 11 \lambda=-11\) \(\Rightarrow \lambda=-1\) So, point of intersection is, \((2(-1)+1),(3(-1)+2),(4(-1)+3)\) \(=(-1,-1,-1)\)
