121379
Let the image of the point \(P(1,2,6)\) in the plane passing through the points \(A(1,2,0), B(1,4,1)\) and \(\mathbf{C}(0,5,1)\) be \(\mathrm{Q}\left(\alpha^2+\beta^2+\gamma^2\right)\) is equal to :
1 65
2 70
3 76
4 62
Explanation:
A Given, \(\quad\)\(\quad\)\(\quad\)\(\mathrm{P}(1,2,6) \& \mathrm{~A}(1,2,0), \mathrm{B}(1,4,1), \mathrm{C}(0,5,1)\) Equation of required plane will be \(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\mathrm{a}(\mathrm{x}-1)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) Putting \((1,4,1) \Rightarrow 2 b+c=0\) putting \((0,5,1) \Rightarrow-a+3 b+c=0\) On solving (i) \& (ii) we get - \(\mathrm{a}=\mathrm{b} \text { and } \mathrm{c}=-2 \mathrm{~b}\) So, \(1(x-1)+1(y-2)-2(z-0)=0\) \(\Rightarrow x+y-2 z-3=0\) Image of \(P\) on this plane is - \(\frac{x-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2(1+2-12-3)}{6}\) \(\Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4\) \(\Rightarrow \alpha=5, \beta=6, \gamma=-2\) \(\text { So, } \alpha^2+\beta^2+\gamma^2=25+36+4\) \(\Rightarrow \quad \alpha^2+\beta^2+\gamma^2=65\)
JEE Main-10.04.2023
Three Dimensional Geometry
121381
The perpendicular distance from origin to the plane \(x+2 y-2 z+5=0\) equals \(\qquad\) units.
1 \(\frac{3}{5}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{9}\)
4 5
Explanation:
B Distance from \((0,0)\) to the plane \(x+2 y-2 z+5=0\) is
\(d =\left \vert\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right \vert\)
\(=\left \vert\frac{1(0)+2(0)-2(0)+5}{\sqrt{1^2+2^2+(-2)^2}}\right \vert\)
\(\Rightarrow \quad d =\frac{5}{3}\)
APEAPCET-20.08.2021
Three Dimensional Geometry
121382
If \((3,4,-7)\) is the foot of the perpendicular drawn from the point \((-2,3,6)\) to the plane \(\pi\) then the sum of the intercepts made by the plane \(\pi\) on the \(x\) and \(y\)-axes is
1 132
2 142
3 210
4 175
Explanation:
A Given, \(\mathrm{A}(3,4,-7), \mathrm{B}(-2,3,6)\) \(\Rightarrow \overrightarrow{\mathrm{AB}}=-5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+13 \hat{\mathrm{k}}\) Equation of plane \(\lambda\) containing point \(A\) having normal vector \(\overrightarrow{\mathrm{AB}}\) is \(\quad\)\(\lambda:-5(x-3)-1(y-4)+13(z+7)=0\) \(\Rightarrow-5 \mathrm{x}-\mathrm{y}+13 \mathrm{z}=-110\) Or \(5 x+y-13 z=110\) \(\quad\)\(\frac{\mathrm{x}}{22}+\frac{\mathrm{y}}{110}-\frac{\mathrm{z}}{110 / 13}=1\) Sum of intercept on \(\mathrm{x}\) and \(\mathrm{y}\) axes is \(22+110=132\)
AP EAMCET-06.07.2022
Three Dimensional Geometry
121383
If the point \((a, 8,-2)\) divides the line segment joining the points \((1,4,6)\) and \((5,2,10)\) in the ratio \(\mathrm{m}\) : \(\mathrm{n}\) then \(\frac{2 \mathrm{~m}}{\mathrm{n}}-\frac{\mathrm{a}}{3}=\)
121379
Let the image of the point \(P(1,2,6)\) in the plane passing through the points \(A(1,2,0), B(1,4,1)\) and \(\mathbf{C}(0,5,1)\) be \(\mathrm{Q}\left(\alpha^2+\beta^2+\gamma^2\right)\) is equal to :
1 65
2 70
3 76
4 62
Explanation:
A Given, \(\quad\)\(\quad\)\(\quad\)\(\mathrm{P}(1,2,6) \& \mathrm{~A}(1,2,0), \mathrm{B}(1,4,1), \mathrm{C}(0,5,1)\) Equation of required plane will be \(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\mathrm{a}(\mathrm{x}-1)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) Putting \((1,4,1) \Rightarrow 2 b+c=0\) putting \((0,5,1) \Rightarrow-a+3 b+c=0\) On solving (i) \& (ii) we get - \(\mathrm{a}=\mathrm{b} \text { and } \mathrm{c}=-2 \mathrm{~b}\) So, \(1(x-1)+1(y-2)-2(z-0)=0\) \(\Rightarrow x+y-2 z-3=0\) Image of \(P\) on this plane is - \(\frac{x-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2(1+2-12-3)}{6}\) \(\Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4\) \(\Rightarrow \alpha=5, \beta=6, \gamma=-2\) \(\text { So, } \alpha^2+\beta^2+\gamma^2=25+36+4\) \(\Rightarrow \quad \alpha^2+\beta^2+\gamma^2=65\)
JEE Main-10.04.2023
Three Dimensional Geometry
121381
The perpendicular distance from origin to the plane \(x+2 y-2 z+5=0\) equals \(\qquad\) units.
1 \(\frac{3}{5}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{9}\)
4 5
Explanation:
B Distance from \((0,0)\) to the plane \(x+2 y-2 z+5=0\) is
\(d =\left \vert\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right \vert\)
\(=\left \vert\frac{1(0)+2(0)-2(0)+5}{\sqrt{1^2+2^2+(-2)^2}}\right \vert\)
\(\Rightarrow \quad d =\frac{5}{3}\)
APEAPCET-20.08.2021
Three Dimensional Geometry
121382
If \((3,4,-7)\) is the foot of the perpendicular drawn from the point \((-2,3,6)\) to the plane \(\pi\) then the sum of the intercepts made by the plane \(\pi\) on the \(x\) and \(y\)-axes is
1 132
2 142
3 210
4 175
Explanation:
A Given, \(\mathrm{A}(3,4,-7), \mathrm{B}(-2,3,6)\) \(\Rightarrow \overrightarrow{\mathrm{AB}}=-5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+13 \hat{\mathrm{k}}\) Equation of plane \(\lambda\) containing point \(A\) having normal vector \(\overrightarrow{\mathrm{AB}}\) is \(\quad\)\(\lambda:-5(x-3)-1(y-4)+13(z+7)=0\) \(\Rightarrow-5 \mathrm{x}-\mathrm{y}+13 \mathrm{z}=-110\) Or \(5 x+y-13 z=110\) \(\quad\)\(\frac{\mathrm{x}}{22}+\frac{\mathrm{y}}{110}-\frac{\mathrm{z}}{110 / 13}=1\) Sum of intercept on \(\mathrm{x}\) and \(\mathrm{y}\) axes is \(22+110=132\)
AP EAMCET-06.07.2022
Three Dimensional Geometry
121383
If the point \((a, 8,-2)\) divides the line segment joining the points \((1,4,6)\) and \((5,2,10)\) in the ratio \(\mathrm{m}\) : \(\mathrm{n}\) then \(\frac{2 \mathrm{~m}}{\mathrm{n}}-\frac{\mathrm{a}}{3}=\)
121379
Let the image of the point \(P(1,2,6)\) in the plane passing through the points \(A(1,2,0), B(1,4,1)\) and \(\mathbf{C}(0,5,1)\) be \(\mathrm{Q}\left(\alpha^2+\beta^2+\gamma^2\right)\) is equal to :
1 65
2 70
3 76
4 62
Explanation:
A Given, \(\quad\)\(\quad\)\(\quad\)\(\mathrm{P}(1,2,6) \& \mathrm{~A}(1,2,0), \mathrm{B}(1,4,1), \mathrm{C}(0,5,1)\) Equation of required plane will be \(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\mathrm{a}(\mathrm{x}-1)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) Putting \((1,4,1) \Rightarrow 2 b+c=0\) putting \((0,5,1) \Rightarrow-a+3 b+c=0\) On solving (i) \& (ii) we get - \(\mathrm{a}=\mathrm{b} \text { and } \mathrm{c}=-2 \mathrm{~b}\) So, \(1(x-1)+1(y-2)-2(z-0)=0\) \(\Rightarrow x+y-2 z-3=0\) Image of \(P\) on this plane is - \(\frac{x-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2(1+2-12-3)}{6}\) \(\Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4\) \(\Rightarrow \alpha=5, \beta=6, \gamma=-2\) \(\text { So, } \alpha^2+\beta^2+\gamma^2=25+36+4\) \(\Rightarrow \quad \alpha^2+\beta^2+\gamma^2=65\)
JEE Main-10.04.2023
Three Dimensional Geometry
121381
The perpendicular distance from origin to the plane \(x+2 y-2 z+5=0\) equals \(\qquad\) units.
1 \(\frac{3}{5}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{9}\)
4 5
Explanation:
B Distance from \((0,0)\) to the plane \(x+2 y-2 z+5=0\) is
\(d =\left \vert\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right \vert\)
\(=\left \vert\frac{1(0)+2(0)-2(0)+5}{\sqrt{1^2+2^2+(-2)^2}}\right \vert\)
\(\Rightarrow \quad d =\frac{5}{3}\)
APEAPCET-20.08.2021
Three Dimensional Geometry
121382
If \((3,4,-7)\) is the foot of the perpendicular drawn from the point \((-2,3,6)\) to the plane \(\pi\) then the sum of the intercepts made by the plane \(\pi\) on the \(x\) and \(y\)-axes is
1 132
2 142
3 210
4 175
Explanation:
A Given, \(\mathrm{A}(3,4,-7), \mathrm{B}(-2,3,6)\) \(\Rightarrow \overrightarrow{\mathrm{AB}}=-5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+13 \hat{\mathrm{k}}\) Equation of plane \(\lambda\) containing point \(A\) having normal vector \(\overrightarrow{\mathrm{AB}}\) is \(\quad\)\(\lambda:-5(x-3)-1(y-4)+13(z+7)=0\) \(\Rightarrow-5 \mathrm{x}-\mathrm{y}+13 \mathrm{z}=-110\) Or \(5 x+y-13 z=110\) \(\quad\)\(\frac{\mathrm{x}}{22}+\frac{\mathrm{y}}{110}-\frac{\mathrm{z}}{110 / 13}=1\) Sum of intercept on \(\mathrm{x}\) and \(\mathrm{y}\) axes is \(22+110=132\)
AP EAMCET-06.07.2022
Three Dimensional Geometry
121383
If the point \((a, 8,-2)\) divides the line segment joining the points \((1,4,6)\) and \((5,2,10)\) in the ratio \(\mathrm{m}\) : \(\mathrm{n}\) then \(\frac{2 \mathrm{~m}}{\mathrm{n}}-\frac{\mathrm{a}}{3}=\)
121379
Let the image of the point \(P(1,2,6)\) in the plane passing through the points \(A(1,2,0), B(1,4,1)\) and \(\mathbf{C}(0,5,1)\) be \(\mathrm{Q}\left(\alpha^2+\beta^2+\gamma^2\right)\) is equal to :
1 65
2 70
3 76
4 62
Explanation:
A Given, \(\quad\)\(\quad\)\(\quad\)\(\mathrm{P}(1,2,6) \& \mathrm{~A}(1,2,0), \mathrm{B}(1,4,1), \mathrm{C}(0,5,1)\) Equation of required plane will be \(\quad\)\(\quad\)\(\quad\)\(\quad\)\(\mathrm{a}(\mathrm{x}-1)+\mathrm{b}(\mathrm{y}-2)+\mathrm{c}(\mathrm{z}-0)=0\) Putting \((1,4,1) \Rightarrow 2 b+c=0\) putting \((0,5,1) \Rightarrow-a+3 b+c=0\) On solving (i) \& (ii) we get - \(\mathrm{a}=\mathrm{b} \text { and } \mathrm{c}=-2 \mathrm{~b}\) So, \(1(x-1)+1(y-2)-2(z-0)=0\) \(\Rightarrow x+y-2 z-3=0\) Image of \(P\) on this plane is - \(\frac{x-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2(1+2-12-3)}{6}\) \(\Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4\) \(\Rightarrow \alpha=5, \beta=6, \gamma=-2\) \(\text { So, } \alpha^2+\beta^2+\gamma^2=25+36+4\) \(\Rightarrow \quad \alpha^2+\beta^2+\gamma^2=65\)
JEE Main-10.04.2023
Three Dimensional Geometry
121381
The perpendicular distance from origin to the plane \(x+2 y-2 z+5=0\) equals \(\qquad\) units.
1 \(\frac{3}{5}\)
2 \(\frac{5}{3}\)
3 \(\frac{5}{9}\)
4 5
Explanation:
B Distance from \((0,0)\) to the plane \(x+2 y-2 z+5=0\) is
\(d =\left \vert\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right \vert\)
\(=\left \vert\frac{1(0)+2(0)-2(0)+5}{\sqrt{1^2+2^2+(-2)^2}}\right \vert\)
\(\Rightarrow \quad d =\frac{5}{3}\)
APEAPCET-20.08.2021
Three Dimensional Geometry
121382
If \((3,4,-7)\) is the foot of the perpendicular drawn from the point \((-2,3,6)\) to the plane \(\pi\) then the sum of the intercepts made by the plane \(\pi\) on the \(x\) and \(y\)-axes is
1 132
2 142
3 210
4 175
Explanation:
A Given, \(\mathrm{A}(3,4,-7), \mathrm{B}(-2,3,6)\) \(\Rightarrow \overrightarrow{\mathrm{AB}}=-5 \hat{\mathrm{i}}-\hat{\mathrm{j}}+13 \hat{\mathrm{k}}\) Equation of plane \(\lambda\) containing point \(A\) having normal vector \(\overrightarrow{\mathrm{AB}}\) is \(\quad\)\(\lambda:-5(x-3)-1(y-4)+13(z+7)=0\) \(\Rightarrow-5 \mathrm{x}-\mathrm{y}+13 \mathrm{z}=-110\) Or \(5 x+y-13 z=110\) \(\quad\)\(\frac{\mathrm{x}}{22}+\frac{\mathrm{y}}{110}-\frac{\mathrm{z}}{110 / 13}=1\) Sum of intercept on \(\mathrm{x}\) and \(\mathrm{y}\) axes is \(22+110=132\)
AP EAMCET-06.07.2022
Three Dimensional Geometry
121383
If the point \((a, 8,-2)\) divides the line segment joining the points \((1,4,6)\) and \((5,2,10)\) in the ratio \(\mathrm{m}\) : \(\mathrm{n}\) then \(\frac{2 \mathrm{~m}}{\mathrm{n}}-\frac{\mathrm{a}}{3}=\)
