NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121262
The lines \(\quad \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}\) intersect each other at point
1 \((-2,-4,5)\)
2 \((-2,-4,-5)\)
3 \((2,4,-5)\)
4 \((2,-4,-5)\)
Explanation:
B Given, Lines, \(L_1: \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}=\lambda\) Lines, \(L_2: \frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}=\mu\) From equation (i), we get - \(\mathrm{x}=2 \lambda+1, \mathrm{y}=2 \lambda-1, \mathrm{z}=4 \lambda+1\) And points, \(\mathrm{P}(2 \lambda+1,2 \lambda-1,4 \lambda+1)\) Now putting the value of \(\mathrm{P}\) in equation (ii) we get- \(\frac{2 \lambda+1-3}{1}=\frac{2 \lambda-1-6}{2}=\frac{4 \lambda+1}{1}\) Now, taking first two terms- \(4 \lambda-2=2 \lambda-7\) \(2 \lambda=-7+2\) \(\lambda=\frac{-3}{2}\) Put, \(\lambda=\frac{-3}{2}\) we get point of intersection, \(\mathrm{Q}=\left[2 \times \frac{-3}{2}+1,2 \times\left(\frac{-3}{2}\right)-1,4 \times\left(\frac{-3}{2}\right)+1\right]\) \(\mathrm{Q}=(-2,-4,-5)\)
MHT CET-2017
Three Dimensional Geometry
121263
The equation of line is \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}\) A point on the line at a distance of 3 units from the point \((1,-1,-1)\) is
1 \((7,-7,2)\)
2 \((3,-3,0)\)
3 \((6,7,-2)\)
4 \((-3,3,0)\)
Explanation:
B Given, Let, \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}=\lambda\) Coordinates of any point on the line are \(\mathrm{x}=1+2 \lambda, \mathrm{y}=-1-2 \lambda, \mathrm{z}=-1+\lambda\) Let, \(\mathrm{P} \equiv(1+2 \lambda,-1-2 \lambda,-1+\lambda)\) The point \(\mathrm{P}\) on the line is at a distance of 3 units from the point \(\mathrm{A}(1,-1,-1)\) \(\therefore \sqrt{(1+2 \lambda-1)^2+(-1-2 \lambda+1)^2+(-1+\lambda+1)^2}=3\) \(\therefore 4 \lambda^2+4 \lambda^2+\lambda^2=9 \Rightarrow 9 \lambda^2=9 \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1\) \(\therefore \mathrm{P}=(1+2,-1-2,-1+1)\) \(\mathrm{P} \equiv(1-2,-1+2,-1-1)\) \(\therefore \mathrm{P} \equiv(3,-3,0) \text { or } \mathrm{P} \equiv(-1,1,-2)\)
MHT CET-2010
Three Dimensional Geometry
121264
If a plane meets the coordinate axes at \(A, B\) and \(C\) in such a way that the centroid of triangle \(\mathrm{ABC}\) is at the point \((1,2,3)\) then the equation of the plane is
B Suppose, equation of plane \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) Its meets the coordinate axis on point \(\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0, \mathrm{~b}, 0)\) and \(\mathrm{C}(0,0, \mathrm{c})\), The centroid of \(\Delta \mathrm{ABC}\) will be \(\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{3}\right)\) \(\frac{a}{3}=1, \frac{b}{3}=2, \frac{c}{3}=3\) \(a=3, b=6, c=9\)Hence, required equation of plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
Karnataka CET-2021]
Three Dimensional Geometry
121265
A plane which is passing through the point (3, \(2,0)\) and the line \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) is
1 \(x-y+z=1\)
2 \(x+y+z=1\)
3 \(x-y+2 z=0\)
4 \(x+y+2 z=0\)
Explanation:
A Given, point \((3,2,0)\) Line : \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) Let equation of required plane be - \(\mathrm{a}(\mathrm{x}-3)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) It must pass through points \((3,6,4)\) So, \(a(3-3)+b(6-2)+c(4-0)=0\) \(4 b+4 c=0\) \(b+c=0\) Also, \(1 \times a+5 \times b+4 \times c=0\) \(a+5 b+4 c=0\) By solving (ii) and (iii), we get - \(\frac{\mathrm{a}}{4-5}=\frac{\mathrm{b}}{1-0}=\frac{\mathrm{c}}{0-1}=\lambda \text { (say) }\) \(\mathrm{a}=-\lambda, \mathrm{b}=\lambda, \mathrm{c}=-\lambda \quad(\text { put in (i)) }\) \(-\lambda(\mathrm{x}-3)+\lambda(\mathrm{y}-2)-\lambda(\mathrm{z}-0)=0\) \(\mathrm{x}-\mathrm{y}+\mathrm{z}-1=0 \text { or } \mathrm{x}-\mathrm{y}+\mathrm{z}=1\)
121262
The lines \(\quad \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}\) intersect each other at point
1 \((-2,-4,5)\)
2 \((-2,-4,-5)\)
3 \((2,4,-5)\)
4 \((2,-4,-5)\)
Explanation:
B Given, Lines, \(L_1: \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}=\lambda\) Lines, \(L_2: \frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}=\mu\) From equation (i), we get - \(\mathrm{x}=2 \lambda+1, \mathrm{y}=2 \lambda-1, \mathrm{z}=4 \lambda+1\) And points, \(\mathrm{P}(2 \lambda+1,2 \lambda-1,4 \lambda+1)\) Now putting the value of \(\mathrm{P}\) in equation (ii) we get- \(\frac{2 \lambda+1-3}{1}=\frac{2 \lambda-1-6}{2}=\frac{4 \lambda+1}{1}\) Now, taking first two terms- \(4 \lambda-2=2 \lambda-7\) \(2 \lambda=-7+2\) \(\lambda=\frac{-3}{2}\) Put, \(\lambda=\frac{-3}{2}\) we get point of intersection, \(\mathrm{Q}=\left[2 \times \frac{-3}{2}+1,2 \times\left(\frac{-3}{2}\right)-1,4 \times\left(\frac{-3}{2}\right)+1\right]\) \(\mathrm{Q}=(-2,-4,-5)\)
MHT CET-2017
Three Dimensional Geometry
121263
The equation of line is \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}\) A point on the line at a distance of 3 units from the point \((1,-1,-1)\) is
1 \((7,-7,2)\)
2 \((3,-3,0)\)
3 \((6,7,-2)\)
4 \((-3,3,0)\)
Explanation:
B Given, Let, \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}=\lambda\) Coordinates of any point on the line are \(\mathrm{x}=1+2 \lambda, \mathrm{y}=-1-2 \lambda, \mathrm{z}=-1+\lambda\) Let, \(\mathrm{P} \equiv(1+2 \lambda,-1-2 \lambda,-1+\lambda)\) The point \(\mathrm{P}\) on the line is at a distance of 3 units from the point \(\mathrm{A}(1,-1,-1)\) \(\therefore \sqrt{(1+2 \lambda-1)^2+(-1-2 \lambda+1)^2+(-1+\lambda+1)^2}=3\) \(\therefore 4 \lambda^2+4 \lambda^2+\lambda^2=9 \Rightarrow 9 \lambda^2=9 \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1\) \(\therefore \mathrm{P}=(1+2,-1-2,-1+1)\) \(\mathrm{P} \equiv(1-2,-1+2,-1-1)\) \(\therefore \mathrm{P} \equiv(3,-3,0) \text { or } \mathrm{P} \equiv(-1,1,-2)\)
MHT CET-2010
Three Dimensional Geometry
121264
If a plane meets the coordinate axes at \(A, B\) and \(C\) in such a way that the centroid of triangle \(\mathrm{ABC}\) is at the point \((1,2,3)\) then the equation of the plane is
B Suppose, equation of plane \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) Its meets the coordinate axis on point \(\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0, \mathrm{~b}, 0)\) and \(\mathrm{C}(0,0, \mathrm{c})\), The centroid of \(\Delta \mathrm{ABC}\) will be \(\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{3}\right)\) \(\frac{a}{3}=1, \frac{b}{3}=2, \frac{c}{3}=3\) \(a=3, b=6, c=9\)Hence, required equation of plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
Karnataka CET-2021]
Three Dimensional Geometry
121265
A plane which is passing through the point (3, \(2,0)\) and the line \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) is
1 \(x-y+z=1\)
2 \(x+y+z=1\)
3 \(x-y+2 z=0\)
4 \(x+y+2 z=0\)
Explanation:
A Given, point \((3,2,0)\) Line : \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) Let equation of required plane be - \(\mathrm{a}(\mathrm{x}-3)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) It must pass through points \((3,6,4)\) So, \(a(3-3)+b(6-2)+c(4-0)=0\) \(4 b+4 c=0\) \(b+c=0\) Also, \(1 \times a+5 \times b+4 \times c=0\) \(a+5 b+4 c=0\) By solving (ii) and (iii), we get - \(\frac{\mathrm{a}}{4-5}=\frac{\mathrm{b}}{1-0}=\frac{\mathrm{c}}{0-1}=\lambda \text { (say) }\) \(\mathrm{a}=-\lambda, \mathrm{b}=\lambda, \mathrm{c}=-\lambda \quad(\text { put in (i)) }\) \(-\lambda(\mathrm{x}-3)+\lambda(\mathrm{y}-2)-\lambda(\mathrm{z}-0)=0\) \(\mathrm{x}-\mathrm{y}+\mathrm{z}-1=0 \text { or } \mathrm{x}-\mathrm{y}+\mathrm{z}=1\)
121262
The lines \(\quad \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}\) intersect each other at point
1 \((-2,-4,5)\)
2 \((-2,-4,-5)\)
3 \((2,4,-5)\)
4 \((2,-4,-5)\)
Explanation:
B Given, Lines, \(L_1: \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}=\lambda\) Lines, \(L_2: \frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}=\mu\) From equation (i), we get - \(\mathrm{x}=2 \lambda+1, \mathrm{y}=2 \lambda-1, \mathrm{z}=4 \lambda+1\) And points, \(\mathrm{P}(2 \lambda+1,2 \lambda-1,4 \lambda+1)\) Now putting the value of \(\mathrm{P}\) in equation (ii) we get- \(\frac{2 \lambda+1-3}{1}=\frac{2 \lambda-1-6}{2}=\frac{4 \lambda+1}{1}\) Now, taking first two terms- \(4 \lambda-2=2 \lambda-7\) \(2 \lambda=-7+2\) \(\lambda=\frac{-3}{2}\) Put, \(\lambda=\frac{-3}{2}\) we get point of intersection, \(\mathrm{Q}=\left[2 \times \frac{-3}{2}+1,2 \times\left(\frac{-3}{2}\right)-1,4 \times\left(\frac{-3}{2}\right)+1\right]\) \(\mathrm{Q}=(-2,-4,-5)\)
MHT CET-2017
Three Dimensional Geometry
121263
The equation of line is \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}\) A point on the line at a distance of 3 units from the point \((1,-1,-1)\) is
1 \((7,-7,2)\)
2 \((3,-3,0)\)
3 \((6,7,-2)\)
4 \((-3,3,0)\)
Explanation:
B Given, Let, \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}=\lambda\) Coordinates of any point on the line are \(\mathrm{x}=1+2 \lambda, \mathrm{y}=-1-2 \lambda, \mathrm{z}=-1+\lambda\) Let, \(\mathrm{P} \equiv(1+2 \lambda,-1-2 \lambda,-1+\lambda)\) The point \(\mathrm{P}\) on the line is at a distance of 3 units from the point \(\mathrm{A}(1,-1,-1)\) \(\therefore \sqrt{(1+2 \lambda-1)^2+(-1-2 \lambda+1)^2+(-1+\lambda+1)^2}=3\) \(\therefore 4 \lambda^2+4 \lambda^2+\lambda^2=9 \Rightarrow 9 \lambda^2=9 \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1\) \(\therefore \mathrm{P}=(1+2,-1-2,-1+1)\) \(\mathrm{P} \equiv(1-2,-1+2,-1-1)\) \(\therefore \mathrm{P} \equiv(3,-3,0) \text { or } \mathrm{P} \equiv(-1,1,-2)\)
MHT CET-2010
Three Dimensional Geometry
121264
If a plane meets the coordinate axes at \(A, B\) and \(C\) in such a way that the centroid of triangle \(\mathrm{ABC}\) is at the point \((1,2,3)\) then the equation of the plane is
B Suppose, equation of plane \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) Its meets the coordinate axis on point \(\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0, \mathrm{~b}, 0)\) and \(\mathrm{C}(0,0, \mathrm{c})\), The centroid of \(\Delta \mathrm{ABC}\) will be \(\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{3}\right)\) \(\frac{a}{3}=1, \frac{b}{3}=2, \frac{c}{3}=3\) \(a=3, b=6, c=9\)Hence, required equation of plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
Karnataka CET-2021]
Three Dimensional Geometry
121265
A plane which is passing through the point (3, \(2,0)\) and the line \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) is
1 \(x-y+z=1\)
2 \(x+y+z=1\)
3 \(x-y+2 z=0\)
4 \(x+y+2 z=0\)
Explanation:
A Given, point \((3,2,0)\) Line : \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) Let equation of required plane be - \(\mathrm{a}(\mathrm{x}-3)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) It must pass through points \((3,6,4)\) So, \(a(3-3)+b(6-2)+c(4-0)=0\) \(4 b+4 c=0\) \(b+c=0\) Also, \(1 \times a+5 \times b+4 \times c=0\) \(a+5 b+4 c=0\) By solving (ii) and (iii), we get - \(\frac{\mathrm{a}}{4-5}=\frac{\mathrm{b}}{1-0}=\frac{\mathrm{c}}{0-1}=\lambda \text { (say) }\) \(\mathrm{a}=-\lambda, \mathrm{b}=\lambda, \mathrm{c}=-\lambda \quad(\text { put in (i)) }\) \(-\lambda(\mathrm{x}-3)+\lambda(\mathrm{y}-2)-\lambda(\mathrm{z}-0)=0\) \(\mathrm{x}-\mathrm{y}+\mathrm{z}-1=0 \text { or } \mathrm{x}-\mathrm{y}+\mathrm{z}=1\)
121262
The lines \(\quad \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}\) \(\frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}\) intersect each other at point
1 \((-2,-4,5)\)
2 \((-2,-4,-5)\)
3 \((2,4,-5)\)
4 \((2,-4,-5)\)
Explanation:
B Given, Lines, \(L_1: \frac{x-1}{2}=\frac{y+1}{2}=\frac{z-1}{4}=\lambda\) Lines, \(L_2: \frac{x-3}{1}=\frac{y-6}{2}=\frac{z}{1}=\mu\) From equation (i), we get - \(\mathrm{x}=2 \lambda+1, \mathrm{y}=2 \lambda-1, \mathrm{z}=4 \lambda+1\) And points, \(\mathrm{P}(2 \lambda+1,2 \lambda-1,4 \lambda+1)\) Now putting the value of \(\mathrm{P}\) in equation (ii) we get- \(\frac{2 \lambda+1-3}{1}=\frac{2 \lambda-1-6}{2}=\frac{4 \lambda+1}{1}\) Now, taking first two terms- \(4 \lambda-2=2 \lambda-7\) \(2 \lambda=-7+2\) \(\lambda=\frac{-3}{2}\) Put, \(\lambda=\frac{-3}{2}\) we get point of intersection, \(\mathrm{Q}=\left[2 \times \frac{-3}{2}+1,2 \times\left(\frac{-3}{2}\right)-1,4 \times\left(\frac{-3}{2}\right)+1\right]\) \(\mathrm{Q}=(-2,-4,-5)\)
MHT CET-2017
Three Dimensional Geometry
121263
The equation of line is \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}\) A point on the line at a distance of 3 units from the point \((1,-1,-1)\) is
1 \((7,-7,2)\)
2 \((3,-3,0)\)
3 \((6,7,-2)\)
4 \((-3,3,0)\)
Explanation:
B Given, Let, \(\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z+1}{1}=\lambda\) Coordinates of any point on the line are \(\mathrm{x}=1+2 \lambda, \mathrm{y}=-1-2 \lambda, \mathrm{z}=-1+\lambda\) Let, \(\mathrm{P} \equiv(1+2 \lambda,-1-2 \lambda,-1+\lambda)\) The point \(\mathrm{P}\) on the line is at a distance of 3 units from the point \(\mathrm{A}(1,-1,-1)\) \(\therefore \sqrt{(1+2 \lambda-1)^2+(-1-2 \lambda+1)^2+(-1+\lambda+1)^2}=3\) \(\therefore 4 \lambda^2+4 \lambda^2+\lambda^2=9 \Rightarrow 9 \lambda^2=9 \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1\) \(\therefore \mathrm{P}=(1+2,-1-2,-1+1)\) \(\mathrm{P} \equiv(1-2,-1+2,-1-1)\) \(\therefore \mathrm{P} \equiv(3,-3,0) \text { or } \mathrm{P} \equiv(-1,1,-2)\)
MHT CET-2010
Three Dimensional Geometry
121264
If a plane meets the coordinate axes at \(A, B\) and \(C\) in such a way that the centroid of triangle \(\mathrm{ABC}\) is at the point \((1,2,3)\) then the equation of the plane is
B Suppose, equation of plane \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1\) Its meets the coordinate axis on point \(\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0, \mathrm{~b}, 0)\) and \(\mathrm{C}(0,0, \mathrm{c})\), The centroid of \(\Delta \mathrm{ABC}\) will be \(\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}, \frac{\mathrm{c}}{3}\right)\) \(\frac{a}{3}=1, \frac{b}{3}=2, \frac{c}{3}=3\) \(a=3, b=6, c=9\)Hence, required equation of plane is \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\)
Karnataka CET-2021]
Three Dimensional Geometry
121265
A plane which is passing through the point (3, \(2,0)\) and the line \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) is
1 \(x-y+z=1\)
2 \(x+y+z=1\)
3 \(x-y+2 z=0\)
4 \(x+y+2 z=0\)
Explanation:
A Given, point \((3,2,0)\) Line : \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) Let equation of required plane be - \(\mathrm{a}(\mathrm{x}-3)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) It must pass through points \((3,6,4)\) So, \(a(3-3)+b(6-2)+c(4-0)=0\) \(4 b+4 c=0\) \(b+c=0\) Also, \(1 \times a+5 \times b+4 \times c=0\) \(a+5 b+4 c=0\) By solving (ii) and (iii), we get - \(\frac{\mathrm{a}}{4-5}=\frac{\mathrm{b}}{1-0}=\frac{\mathrm{c}}{0-1}=\lambda \text { (say) }\) \(\mathrm{a}=-\lambda, \mathrm{b}=\lambda, \mathrm{c}=-\lambda \quad(\text { put in (i)) }\) \(-\lambda(\mathrm{x}-3)+\lambda(\mathrm{y}-2)-\lambda(\mathrm{z}-0)=0\) \(\mathrm{x}-\mathrm{y}+\mathrm{z}-1=0 \text { or } \mathrm{x}-\mathrm{y}+\mathrm{z}=1\)
