121412
A straight line joining the points \((1,1,1)\) and \((0,0,0)\) intersects the place \(2 x+2 y+z=10\) at
1 \((1,2,5)\)
2 \((2,2,2)\)
3 \((2,1,5)\)
4 \((1,1,6)\)
Explanation:
B Given, Equation of line : \((\hat{i}+\hat{j}+\hat{k}),(1,1,1)\) Intersecting plane \(2 x+2 y+z=10\) Let at point \((\lambda, \lambda . \lambda)\) it intersects plane So, \(2 \lambda+2 \lambda+\lambda=10\) \( 5 \lambda =10\) \(\Rightarrow \lambda =2\)Hence, point is \((2,2,2)\)
WB JEE-2016
Three Dimensional Geometry
121413
The equation of the plane through \((1,2,-3)\) and \((2,-2,1)\) and parallel to \(\mathrm{X}\)-axis is
121414
A point \(P\) lies on a line through \(Q(1,-2,3)\) and is parallel to the line \(\frac{x}{1}=\frac{y}{4}=\frac{z}{5}\). If \(P\) lies on the plane \(2 x+3 y-4 z+22=0\), then segment \(P Q\) equals
1 \(\sqrt{42}\) units
2 \(\sqrt{32}\) units
3 4 units
4 5 units
Explanation:
A Given, \(\mathrm{Q} =(1,-2,3)\) \(\text { Line : } \quad \frac{\mathrm{x}}{1} =\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) \(\text { Plane }: \mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Line : \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) Plane : \(\mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Equation of line containing point \(\mathrm{Q}\) and parallel to ' \(\mathrm{l}\) ' is \(\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+2}{4}=\frac{\mathrm{z}-3}{5}=\mathrm{k} \text { (say) }\) Now, \(\mathrm{P}=(\mathrm{k}+1,4 \mathrm{k}-2,5 \mathrm{k}+3)\) Also, 'P' lies on plane \(\mathrm{P}_1\) \(2(\mathrm{k}+1)+3(4 \mathrm{k}-2)-4(5 \mathrm{k}+3)+22=0\) \(\Rightarrow 2 \mathrm{k}+2+12 \mathrm{k}-6-20 \mathrm{k}-12+22=0\) \(\Rightarrow -6 \mathrm{k}=-6\) \(\mathrm{k}=1\) Hence, \(\mathrm{P}=(2,2,8)\) Ling segment \(P Q=\sqrt{1+16+25}\) \(\mathrm{PQ}=\sqrt{42}\)
WB JEE-2018
Three Dimensional Geometry
121411
The value of \(\lambda\) for which the lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}\) \(=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular to each other is
1 -1
2 -2
3 1
4 2
Explanation:
D The two lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular \(-3 \times 3 \lambda+2 \lambda \times 1+7 \times 2=0\) \(-9 \lambda+2 \lambda+14=0 \Rightarrow-7 \lambda=-14 \Rightarrow \lambda=2\)
121412
A straight line joining the points \((1,1,1)\) and \((0,0,0)\) intersects the place \(2 x+2 y+z=10\) at
1 \((1,2,5)\)
2 \((2,2,2)\)
3 \((2,1,5)\)
4 \((1,1,6)\)
Explanation:
B Given, Equation of line : \((\hat{i}+\hat{j}+\hat{k}),(1,1,1)\) Intersecting plane \(2 x+2 y+z=10\) Let at point \((\lambda, \lambda . \lambda)\) it intersects plane So, \(2 \lambda+2 \lambda+\lambda=10\) \( 5 \lambda =10\) \(\Rightarrow \lambda =2\)Hence, point is \((2,2,2)\)
WB JEE-2016
Three Dimensional Geometry
121413
The equation of the plane through \((1,2,-3)\) and \((2,-2,1)\) and parallel to \(\mathrm{X}\)-axis is
121414
A point \(P\) lies on a line through \(Q(1,-2,3)\) and is parallel to the line \(\frac{x}{1}=\frac{y}{4}=\frac{z}{5}\). If \(P\) lies on the plane \(2 x+3 y-4 z+22=0\), then segment \(P Q\) equals
1 \(\sqrt{42}\) units
2 \(\sqrt{32}\) units
3 4 units
4 5 units
Explanation:
A Given, \(\mathrm{Q} =(1,-2,3)\) \(\text { Line : } \quad \frac{\mathrm{x}}{1} =\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) \(\text { Plane }: \mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Line : \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) Plane : \(\mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Equation of line containing point \(\mathrm{Q}\) and parallel to ' \(\mathrm{l}\) ' is \(\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+2}{4}=\frac{\mathrm{z}-3}{5}=\mathrm{k} \text { (say) }\) Now, \(\mathrm{P}=(\mathrm{k}+1,4 \mathrm{k}-2,5 \mathrm{k}+3)\) Also, 'P' lies on plane \(\mathrm{P}_1\) \(2(\mathrm{k}+1)+3(4 \mathrm{k}-2)-4(5 \mathrm{k}+3)+22=0\) \(\Rightarrow 2 \mathrm{k}+2+12 \mathrm{k}-6-20 \mathrm{k}-12+22=0\) \(\Rightarrow -6 \mathrm{k}=-6\) \(\mathrm{k}=1\) Hence, \(\mathrm{P}=(2,2,8)\) Ling segment \(P Q=\sqrt{1+16+25}\) \(\mathrm{PQ}=\sqrt{42}\)
WB JEE-2018
Three Dimensional Geometry
121411
The value of \(\lambda\) for which the lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}\) \(=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular to each other is
1 -1
2 -2
3 1
4 2
Explanation:
D The two lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular \(-3 \times 3 \lambda+2 \lambda \times 1+7 \times 2=0\) \(-9 \lambda+2 \lambda+14=0 \Rightarrow-7 \lambda=-14 \Rightarrow \lambda=2\)
121412
A straight line joining the points \((1,1,1)\) and \((0,0,0)\) intersects the place \(2 x+2 y+z=10\) at
1 \((1,2,5)\)
2 \((2,2,2)\)
3 \((2,1,5)\)
4 \((1,1,6)\)
Explanation:
B Given, Equation of line : \((\hat{i}+\hat{j}+\hat{k}),(1,1,1)\) Intersecting plane \(2 x+2 y+z=10\) Let at point \((\lambda, \lambda . \lambda)\) it intersects plane So, \(2 \lambda+2 \lambda+\lambda=10\) \( 5 \lambda =10\) \(\Rightarrow \lambda =2\)Hence, point is \((2,2,2)\)
WB JEE-2016
Three Dimensional Geometry
121413
The equation of the plane through \((1,2,-3)\) and \((2,-2,1)\) and parallel to \(\mathrm{X}\)-axis is
121414
A point \(P\) lies on a line through \(Q(1,-2,3)\) and is parallel to the line \(\frac{x}{1}=\frac{y}{4}=\frac{z}{5}\). If \(P\) lies on the plane \(2 x+3 y-4 z+22=0\), then segment \(P Q\) equals
1 \(\sqrt{42}\) units
2 \(\sqrt{32}\) units
3 4 units
4 5 units
Explanation:
A Given, \(\mathrm{Q} =(1,-2,3)\) \(\text { Line : } \quad \frac{\mathrm{x}}{1} =\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) \(\text { Plane }: \mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Line : \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) Plane : \(\mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Equation of line containing point \(\mathrm{Q}\) and parallel to ' \(\mathrm{l}\) ' is \(\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+2}{4}=\frac{\mathrm{z}-3}{5}=\mathrm{k} \text { (say) }\) Now, \(\mathrm{P}=(\mathrm{k}+1,4 \mathrm{k}-2,5 \mathrm{k}+3)\) Also, 'P' lies on plane \(\mathrm{P}_1\) \(2(\mathrm{k}+1)+3(4 \mathrm{k}-2)-4(5 \mathrm{k}+3)+22=0\) \(\Rightarrow 2 \mathrm{k}+2+12 \mathrm{k}-6-20 \mathrm{k}-12+22=0\) \(\Rightarrow -6 \mathrm{k}=-6\) \(\mathrm{k}=1\) Hence, \(\mathrm{P}=(2,2,8)\) Ling segment \(P Q=\sqrt{1+16+25}\) \(\mathrm{PQ}=\sqrt{42}\)
WB JEE-2018
Three Dimensional Geometry
121411
The value of \(\lambda\) for which the lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}\) \(=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular to each other is
1 -1
2 -2
3 1
4 2
Explanation:
D The two lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular \(-3 \times 3 \lambda+2 \lambda \times 1+7 \times 2=0\) \(-9 \lambda+2 \lambda+14=0 \Rightarrow-7 \lambda=-14 \Rightarrow \lambda=2\)
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Three Dimensional Geometry
121412
A straight line joining the points \((1,1,1)\) and \((0,0,0)\) intersects the place \(2 x+2 y+z=10\) at
1 \((1,2,5)\)
2 \((2,2,2)\)
3 \((2,1,5)\)
4 \((1,1,6)\)
Explanation:
B Given, Equation of line : \((\hat{i}+\hat{j}+\hat{k}),(1,1,1)\) Intersecting plane \(2 x+2 y+z=10\) Let at point \((\lambda, \lambda . \lambda)\) it intersects plane So, \(2 \lambda+2 \lambda+\lambda=10\) \( 5 \lambda =10\) \(\Rightarrow \lambda =2\)Hence, point is \((2,2,2)\)
WB JEE-2016
Three Dimensional Geometry
121413
The equation of the plane through \((1,2,-3)\) and \((2,-2,1)\) and parallel to \(\mathrm{X}\)-axis is
121414
A point \(P\) lies on a line through \(Q(1,-2,3)\) and is parallel to the line \(\frac{x}{1}=\frac{y}{4}=\frac{z}{5}\). If \(P\) lies on the plane \(2 x+3 y-4 z+22=0\), then segment \(P Q\) equals
1 \(\sqrt{42}\) units
2 \(\sqrt{32}\) units
3 4 units
4 5 units
Explanation:
A Given, \(\mathrm{Q} =(1,-2,3)\) \(\text { Line : } \quad \frac{\mathrm{x}}{1} =\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) \(\text { Plane }: \mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Line : \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}\) Plane : \(\mathrm{P}_1: 2 \mathrm{x}+3 \mathrm{y}-4 \mathrm{z}+22=0\) Equation of line containing point \(\mathrm{Q}\) and parallel to ' \(\mathrm{l}\) ' is \(\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+2}{4}=\frac{\mathrm{z}-3}{5}=\mathrm{k} \text { (say) }\) Now, \(\mathrm{P}=(\mathrm{k}+1,4 \mathrm{k}-2,5 \mathrm{k}+3)\) Also, 'P' lies on plane \(\mathrm{P}_1\) \(2(\mathrm{k}+1)+3(4 \mathrm{k}-2)-4(5 \mathrm{k}+3)+22=0\) \(\Rightarrow 2 \mathrm{k}+2+12 \mathrm{k}-6-20 \mathrm{k}-12+22=0\) \(\Rightarrow -6 \mathrm{k}=-6\) \(\mathrm{k}=1\) Hence, \(\mathrm{P}=(2,2,8)\) Ling segment \(P Q=\sqrt{1+16+25}\) \(\mathrm{PQ}=\sqrt{42}\)
WB JEE-2018
Three Dimensional Geometry
121411
The value of \(\lambda\) for which the lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}\) \(=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular to each other is
1 -1
2 -2
3 1
4 2
Explanation:
D The two lines \(\frac{1-x}{3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{6-z}{7}\) are perpendicular \(-3 \times 3 \lambda+2 \lambda \times 1+7 \times 2=0\) \(-9 \lambda+2 \lambda+14=0 \Rightarrow-7 \lambda=-14 \Rightarrow \lambda=2\)
