121110
The point on the plane \(2 x-2 y+4 z+5=0\) that is nearer to \(\left(1, \frac{3}{2}, 2\right)\) is
1 \(\left(0, \frac{5}{2}, 0\right)\)
2 \(\left(-5, \frac{-5}{2}, 0\right)\)
3 \(\left(0,0, \frac{-5}{2}\right)\)
4 \(\left(-, \frac{1}{2}, 0,-1\right)\)
Explanation:
A The direction ratios of the normal to the plane are \(2,-2,4\). So, equation of line passing through \(\left(1, \frac{3}{2}, 2\right)\) and direction ratios are equal to the direction ratio of the normal to the plane i.e. 2, -2 , 4 is, \(\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda\) So, \((x, y, z)=(2 \lambda+1),\left(-2 \lambda+\frac{3}{2}\right),(4 \lambda+2)\) Above point also line on the given plane, \(2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5=0\) \(24 \lambda+12=0\) \(\lambda=-\frac{1}{2}\) So, the co-ordinates of the point in the plane are \(=\left(0, \frac{5}{2}, 0\right)\)
TS EAMCET-04.08.2021
Three Dimensional Geometry
121111
If \(M\) is the foot of the perpendicular drawn from the origin \(O\) on to the variable line \(L\), passing through a fixed point \((a, b)\), then the locus of the mid-point of \(O M\) is
B \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is mid-point of \(\mathrm{OM}\). \(\therefore\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}\right)=(\mathrm{h}, \mathrm{k}) \Rightarrow \alpha=2 \mathrm{~h}, \beta=2 \mathrm{k}\) Coordinates of \(\mathrm{M}\) are ( \(2 \mathrm{~h}, 2 \mathrm{k}\) ). Now, slope of \(\mathrm{OM}=\frac{2 \mathrm{k}-0}{2 \mathrm{~h}-0}=\frac{\mathrm{k}}{\mathrm{h}}\) and slope of MQ \(=\frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}\) According to question \(\mathrm{OM} \perp \mathrm{MQ}\) \(\therefore \frac{\mathrm{k}}{\mathrm{h}} \times \frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}=-1 \Rightarrow 2 \mathrm{k}^2-\mathrm{bk}=-2 \mathrm{~h}^2+\mathrm{ah}\) \(2 \mathrm{~h}^2+2 \mathrm{k}^2-\mathrm{ah}-\mathrm{bk}=0\) So, locus of \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is, \(2 x^2+2 y^2-a x-b y=0\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121087
The points \((1,2,3),(-1,-1,-1)\) and \((3,5,7)\) are the vertices of
1 an equilateral triangle
2 an isosceles triangle
3 a right triangle
4 None of these
Explanation:
D Here, the points are \(A(1,2,3), B(-1,-1,-1)\) and \(C(3,5,7)\).
\[
\begin{aligned}
& \vertA B \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29} \\
& \vertB C \vert=\sqrt{4^2+6^2+8^2}=\sqrt{116}=2 \sqrt{29} \\
& \text { and } \vertC A \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29}
\end{aligned}
\]
So, \( \vertB C \vert= \vertA B \vert+ \vertC A \vert\)
\(\Rightarrow\) The points \(A, B, C\) are collinear.
121110
The point on the plane \(2 x-2 y+4 z+5=0\) that is nearer to \(\left(1, \frac{3}{2}, 2\right)\) is
1 \(\left(0, \frac{5}{2}, 0\right)\)
2 \(\left(-5, \frac{-5}{2}, 0\right)\)
3 \(\left(0,0, \frac{-5}{2}\right)\)
4 \(\left(-, \frac{1}{2}, 0,-1\right)\)
Explanation:
A The direction ratios of the normal to the plane are \(2,-2,4\). So, equation of line passing through \(\left(1, \frac{3}{2}, 2\right)\) and direction ratios are equal to the direction ratio of the normal to the plane i.e. 2, -2 , 4 is, \(\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda\) So, \((x, y, z)=(2 \lambda+1),\left(-2 \lambda+\frac{3}{2}\right),(4 \lambda+2)\) Above point also line on the given plane, \(2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5=0\) \(24 \lambda+12=0\) \(\lambda=-\frac{1}{2}\) So, the co-ordinates of the point in the plane are \(=\left(0, \frac{5}{2}, 0\right)\)
TS EAMCET-04.08.2021
Three Dimensional Geometry
121111
If \(M\) is the foot of the perpendicular drawn from the origin \(O\) on to the variable line \(L\), passing through a fixed point \((a, b)\), then the locus of the mid-point of \(O M\) is
B \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is mid-point of \(\mathrm{OM}\). \(\therefore\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}\right)=(\mathrm{h}, \mathrm{k}) \Rightarrow \alpha=2 \mathrm{~h}, \beta=2 \mathrm{k}\) Coordinates of \(\mathrm{M}\) are ( \(2 \mathrm{~h}, 2 \mathrm{k}\) ). Now, slope of \(\mathrm{OM}=\frac{2 \mathrm{k}-0}{2 \mathrm{~h}-0}=\frac{\mathrm{k}}{\mathrm{h}}\) and slope of MQ \(=\frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}\) According to question \(\mathrm{OM} \perp \mathrm{MQ}\) \(\therefore \frac{\mathrm{k}}{\mathrm{h}} \times \frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}=-1 \Rightarrow 2 \mathrm{k}^2-\mathrm{bk}=-2 \mathrm{~h}^2+\mathrm{ah}\) \(2 \mathrm{~h}^2+2 \mathrm{k}^2-\mathrm{ah}-\mathrm{bk}=0\) So, locus of \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is, \(2 x^2+2 y^2-a x-b y=0\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121087
The points \((1,2,3),(-1,-1,-1)\) and \((3,5,7)\) are the vertices of
1 an equilateral triangle
2 an isosceles triangle
3 a right triangle
4 None of these
Explanation:
D Here, the points are \(A(1,2,3), B(-1,-1,-1)\) and \(C(3,5,7)\).
\[
\begin{aligned}
& \vertA B \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29} \\
& \vertB C \vert=\sqrt{4^2+6^2+8^2}=\sqrt{116}=2 \sqrt{29} \\
& \text { and } \vertC A \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29}
\end{aligned}
\]
So, \( \vertB C \vert= \vertA B \vert+ \vertC A \vert\)
\(\Rightarrow\) The points \(A, B, C\) are collinear.
121110
The point on the plane \(2 x-2 y+4 z+5=0\) that is nearer to \(\left(1, \frac{3}{2}, 2\right)\) is
1 \(\left(0, \frac{5}{2}, 0\right)\)
2 \(\left(-5, \frac{-5}{2}, 0\right)\)
3 \(\left(0,0, \frac{-5}{2}\right)\)
4 \(\left(-, \frac{1}{2}, 0,-1\right)\)
Explanation:
A The direction ratios of the normal to the plane are \(2,-2,4\). So, equation of line passing through \(\left(1, \frac{3}{2}, 2\right)\) and direction ratios are equal to the direction ratio of the normal to the plane i.e. 2, -2 , 4 is, \(\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda\) So, \((x, y, z)=(2 \lambda+1),\left(-2 \lambda+\frac{3}{2}\right),(4 \lambda+2)\) Above point also line on the given plane, \(2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5=0\) \(24 \lambda+12=0\) \(\lambda=-\frac{1}{2}\) So, the co-ordinates of the point in the plane are \(=\left(0, \frac{5}{2}, 0\right)\)
TS EAMCET-04.08.2021
Three Dimensional Geometry
121111
If \(M\) is the foot of the perpendicular drawn from the origin \(O\) on to the variable line \(L\), passing through a fixed point \((a, b)\), then the locus of the mid-point of \(O M\) is
B \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is mid-point of \(\mathrm{OM}\). \(\therefore\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}\right)=(\mathrm{h}, \mathrm{k}) \Rightarrow \alpha=2 \mathrm{~h}, \beta=2 \mathrm{k}\) Coordinates of \(\mathrm{M}\) are ( \(2 \mathrm{~h}, 2 \mathrm{k}\) ). Now, slope of \(\mathrm{OM}=\frac{2 \mathrm{k}-0}{2 \mathrm{~h}-0}=\frac{\mathrm{k}}{\mathrm{h}}\) and slope of MQ \(=\frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}\) According to question \(\mathrm{OM} \perp \mathrm{MQ}\) \(\therefore \frac{\mathrm{k}}{\mathrm{h}} \times \frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}=-1 \Rightarrow 2 \mathrm{k}^2-\mathrm{bk}=-2 \mathrm{~h}^2+\mathrm{ah}\) \(2 \mathrm{~h}^2+2 \mathrm{k}^2-\mathrm{ah}-\mathrm{bk}=0\) So, locus of \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is, \(2 x^2+2 y^2-a x-b y=0\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121087
The points \((1,2,3),(-1,-1,-1)\) and \((3,5,7)\) are the vertices of
1 an equilateral triangle
2 an isosceles triangle
3 a right triangle
4 None of these
Explanation:
D Here, the points are \(A(1,2,3), B(-1,-1,-1)\) and \(C(3,5,7)\).
\[
\begin{aligned}
& \vertA B \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29} \\
& \vertB C \vert=\sqrt{4^2+6^2+8^2}=\sqrt{116}=2 \sqrt{29} \\
& \text { and } \vertC A \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29}
\end{aligned}
\]
So, \( \vertB C \vert= \vertA B \vert+ \vertC A \vert\)
\(\Rightarrow\) The points \(A, B, C\) are collinear.
121110
The point on the plane \(2 x-2 y+4 z+5=0\) that is nearer to \(\left(1, \frac{3}{2}, 2\right)\) is
1 \(\left(0, \frac{5}{2}, 0\right)\)
2 \(\left(-5, \frac{-5}{2}, 0\right)\)
3 \(\left(0,0, \frac{-5}{2}\right)\)
4 \(\left(-, \frac{1}{2}, 0,-1\right)\)
Explanation:
A The direction ratios of the normal to the plane are \(2,-2,4\). So, equation of line passing through \(\left(1, \frac{3}{2}, 2\right)\) and direction ratios are equal to the direction ratio of the normal to the plane i.e. 2, -2 , 4 is, \(\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda\) So, \((x, y, z)=(2 \lambda+1),\left(-2 \lambda+\frac{3}{2}\right),(4 \lambda+2)\) Above point also line on the given plane, \(2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5=0\) \(24 \lambda+12=0\) \(\lambda=-\frac{1}{2}\) So, the co-ordinates of the point in the plane are \(=\left(0, \frac{5}{2}, 0\right)\)
TS EAMCET-04.08.2021
Three Dimensional Geometry
121111
If \(M\) is the foot of the perpendicular drawn from the origin \(O\) on to the variable line \(L\), passing through a fixed point \((a, b)\), then the locus of the mid-point of \(O M\) is
B \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is mid-point of \(\mathrm{OM}\). \(\therefore\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}\right)=(\mathrm{h}, \mathrm{k}) \Rightarrow \alpha=2 \mathrm{~h}, \beta=2 \mathrm{k}\) Coordinates of \(\mathrm{M}\) are ( \(2 \mathrm{~h}, 2 \mathrm{k}\) ). Now, slope of \(\mathrm{OM}=\frac{2 \mathrm{k}-0}{2 \mathrm{~h}-0}=\frac{\mathrm{k}}{\mathrm{h}}\) and slope of MQ \(=\frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}\) According to question \(\mathrm{OM} \perp \mathrm{MQ}\) \(\therefore \frac{\mathrm{k}}{\mathrm{h}} \times \frac{2 \mathrm{k}-\mathrm{b}}{2 \mathrm{~h}-\mathrm{a}}=-1 \Rightarrow 2 \mathrm{k}^2-\mathrm{bk}=-2 \mathrm{~h}^2+\mathrm{ah}\) \(2 \mathrm{~h}^2+2 \mathrm{k}^2-\mathrm{ah}-\mathrm{bk}=0\) So, locus of \(\mathrm{R}(\mathrm{h}, \mathrm{k})\) is, \(2 x^2+2 y^2-a x-b y=0\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121087
The points \((1,2,3),(-1,-1,-1)\) and \((3,5,7)\) are the vertices of
1 an equilateral triangle
2 an isosceles triangle
3 a right triangle
4 None of these
Explanation:
D Here, the points are \(A(1,2,3), B(-1,-1,-1)\) and \(C(3,5,7)\).
\[
\begin{aligned}
& \vertA B \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29} \\
& \vertB C \vert=\sqrt{4^2+6^2+8^2}=\sqrt{116}=2 \sqrt{29} \\
& \text { and } \vertC A \vert=\sqrt{2^2+3^2+4^2}=\sqrt{29}
\end{aligned}
\]
So, \( \vertB C \vert= \vertA B \vert+ \vertC A \vert\)
\(\Rightarrow\) The points \(A, B, C\) are collinear.
