NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121395
The image (reflection) of the point \((1,2,-1)\) in the plane \(\overrightarrow{\mathbf{r}} \cdot(\mathbf{3} \hat{\mathbf{i}}-\mathbf{5} \hat{\mathbf{j}}+\mathbf{4} \hat{\mathbf{k}})=5\) is
A We know that, The image of the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) in the plane \(\mathrm{ax}+\mathrm{by}+\) \(\mathrm{cz}+\mathrm{d}=0\) is given by \(\frac{z-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-2\left(\frac{a x_1+b y_1+c z_1+d}{a^2+b^2+c^2}\right)\) The equation of the plane is \(\overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=5\) or , \(3 x-5 y+4 z=5\) The image of \((1,2-1)\) in this plane is given by \(\frac{x-1}{3}=\frac{y-2}{-5}=\frac{z+1}{4}=-2\left(\frac{3-10-4-5}{9+25+16}\right)\) \(\Rightarrow x=\frac{73}{25}, y=\frac{-6}{5}, z=\frac{39}{25}\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121397
The distance of the point \((-1,9,-16)\) from the plane \(2 x+3 y-z=5\) measured parallel to the line \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\) is
1 \(20 \sqrt{2}\)
2 \(13 \sqrt{2}\)
3 31
4 26
Explanation:
C Distance of \((-1,9,-16)\) from plane \(2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}-5=0\) measured parallel to line
\(\frac{x+4}{3}=\frac{y-2}{-4}=\frac{z-3}{12}\)
\(\text { is }\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a} l+\mathrm{bm}+\mathrm{cn}}\right \vert\)
Where \(l, \mathrm{~m}, \mathrm{n}\) are d.c's of line
\(\left. \vert \vert \frac{2(-1)+3(9)-1(-16)-5}{2\left(\frac{3}{13}\right)+3\left(\frac{-4}{13}\right)-1\left(\frac{12}{13}\right)} \vert= \vert \frac{36 \times 13}{-18} \right\rvert\,=26\)
JEE Main-24.01.2023
Three Dimensional Geometry
121398
The distance of the point \((7,-3,-4)\) from plane passing through the points \((2,-3,1),(-1,1,-2)\) and \((3,-4,2)\) is:
1 4
2 5
3 \(5 \sqrt{2}\)
4 \(4 \sqrt{2}\)
Explanation:
C \(=\left \vert\begin{array}{ccc}x-2& y+3 &z-1 \\ -3 &4 &-3 \\ 4& -5 &4\end{array}\right \vert=0\) \((x-2)[(16-15)-(y+3)][-12+12]+(2-1)[15-16\)
\(=0]\)
\(\mathrm{x}-2+0-2+1=0\)
\(\mathrm{x}-2=1\)
\(\mathrm{x}-\mathrm{z}-1=0\)
Distance of \(\mathrm{P}(7,-3,-4)\) from Plane is
\(\mathrm{d}=\left \vert\frac{7+4-1}{\sqrt{2}}\right \vert=5 \sqrt{2}\)
JEE Main-24.01.2023
Three Dimensional Geometry
121399
The image of the point \(P(1,3,4)\) in the plane \(2 x\) \(-\mathbf{y}+\mathrm{z}+\mathbf{3}=0\), is
1 \((3,5,-2)\)
2 \((-3,5,2)\)
3 \((3,-5,2)\)
4 \((3,5,2)\)
Explanation:
B Image of point in a plane is \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a_1+b y_1+\left(z_1+d\right)\right.}{a^2+b^2+c^2}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-y}{-1}=\frac{-2(2-3+4+3)}{4+1+1}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=-2\) \(\frac{x-1}{2}=-2, \frac{y-3}{-1}=-2, \frac{z-4}{1}=-2\) \(\quad x=-3, y=5, z=2\) \(\text { Image }(-3,5,2)\)
121395
The image (reflection) of the point \((1,2,-1)\) in the plane \(\overrightarrow{\mathbf{r}} \cdot(\mathbf{3} \hat{\mathbf{i}}-\mathbf{5} \hat{\mathbf{j}}+\mathbf{4} \hat{\mathbf{k}})=5\) is
A We know that, The image of the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) in the plane \(\mathrm{ax}+\mathrm{by}+\) \(\mathrm{cz}+\mathrm{d}=0\) is given by \(\frac{z-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-2\left(\frac{a x_1+b y_1+c z_1+d}{a^2+b^2+c^2}\right)\) The equation of the plane is \(\overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=5\) or , \(3 x-5 y+4 z=5\) The image of \((1,2-1)\) in this plane is given by \(\frac{x-1}{3}=\frac{y-2}{-5}=\frac{z+1}{4}=-2\left(\frac{3-10-4-5}{9+25+16}\right)\) \(\Rightarrow x=\frac{73}{25}, y=\frac{-6}{5}, z=\frac{39}{25}\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121397
The distance of the point \((-1,9,-16)\) from the plane \(2 x+3 y-z=5\) measured parallel to the line \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\) is
1 \(20 \sqrt{2}\)
2 \(13 \sqrt{2}\)
3 31
4 26
Explanation:
C Distance of \((-1,9,-16)\) from plane \(2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}-5=0\) measured parallel to line
\(\frac{x+4}{3}=\frac{y-2}{-4}=\frac{z-3}{12}\)
\(\text { is }\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a} l+\mathrm{bm}+\mathrm{cn}}\right \vert\)
Where \(l, \mathrm{~m}, \mathrm{n}\) are d.c's of line
\(\left. \vert \vert \frac{2(-1)+3(9)-1(-16)-5}{2\left(\frac{3}{13}\right)+3\left(\frac{-4}{13}\right)-1\left(\frac{12}{13}\right)} \vert= \vert \frac{36 \times 13}{-18} \right\rvert\,=26\)
JEE Main-24.01.2023
Three Dimensional Geometry
121398
The distance of the point \((7,-3,-4)\) from plane passing through the points \((2,-3,1),(-1,1,-2)\) and \((3,-4,2)\) is:
1 4
2 5
3 \(5 \sqrt{2}\)
4 \(4 \sqrt{2}\)
Explanation:
C \(=\left \vert\begin{array}{ccc}x-2& y+3 &z-1 \\ -3 &4 &-3 \\ 4& -5 &4\end{array}\right \vert=0\) \((x-2)[(16-15)-(y+3)][-12+12]+(2-1)[15-16\)
\(=0]\)
\(\mathrm{x}-2+0-2+1=0\)
\(\mathrm{x}-2=1\)
\(\mathrm{x}-\mathrm{z}-1=0\)
Distance of \(\mathrm{P}(7,-3,-4)\) from Plane is
\(\mathrm{d}=\left \vert\frac{7+4-1}{\sqrt{2}}\right \vert=5 \sqrt{2}\)
JEE Main-24.01.2023
Three Dimensional Geometry
121399
The image of the point \(P(1,3,4)\) in the plane \(2 x\) \(-\mathbf{y}+\mathrm{z}+\mathbf{3}=0\), is
1 \((3,5,-2)\)
2 \((-3,5,2)\)
3 \((3,-5,2)\)
4 \((3,5,2)\)
Explanation:
B Image of point in a plane is \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a_1+b y_1+\left(z_1+d\right)\right.}{a^2+b^2+c^2}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-y}{-1}=\frac{-2(2-3+4+3)}{4+1+1}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=-2\) \(\frac{x-1}{2}=-2, \frac{y-3}{-1}=-2, \frac{z-4}{1}=-2\) \(\quad x=-3, y=5, z=2\) \(\text { Image }(-3,5,2)\)
121395
The image (reflection) of the point \((1,2,-1)\) in the plane \(\overrightarrow{\mathbf{r}} \cdot(\mathbf{3} \hat{\mathbf{i}}-\mathbf{5} \hat{\mathbf{j}}+\mathbf{4} \hat{\mathbf{k}})=5\) is
A We know that, The image of the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) in the plane \(\mathrm{ax}+\mathrm{by}+\) \(\mathrm{cz}+\mathrm{d}=0\) is given by \(\frac{z-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-2\left(\frac{a x_1+b y_1+c z_1+d}{a^2+b^2+c^2}\right)\) The equation of the plane is \(\overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=5\) or , \(3 x-5 y+4 z=5\) The image of \((1,2-1)\) in this plane is given by \(\frac{x-1}{3}=\frac{y-2}{-5}=\frac{z+1}{4}=-2\left(\frac{3-10-4-5}{9+25+16}\right)\) \(\Rightarrow x=\frac{73}{25}, y=\frac{-6}{5}, z=\frac{39}{25}\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121397
The distance of the point \((-1,9,-16)\) from the plane \(2 x+3 y-z=5\) measured parallel to the line \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\) is
1 \(20 \sqrt{2}\)
2 \(13 \sqrt{2}\)
3 31
4 26
Explanation:
C Distance of \((-1,9,-16)\) from plane \(2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}-5=0\) measured parallel to line
\(\frac{x+4}{3}=\frac{y-2}{-4}=\frac{z-3}{12}\)
\(\text { is }\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a} l+\mathrm{bm}+\mathrm{cn}}\right \vert\)
Where \(l, \mathrm{~m}, \mathrm{n}\) are d.c's of line
\(\left. \vert \vert \frac{2(-1)+3(9)-1(-16)-5}{2\left(\frac{3}{13}\right)+3\left(\frac{-4}{13}\right)-1\left(\frac{12}{13}\right)} \vert= \vert \frac{36 \times 13}{-18} \right\rvert\,=26\)
JEE Main-24.01.2023
Three Dimensional Geometry
121398
The distance of the point \((7,-3,-4)\) from plane passing through the points \((2,-3,1),(-1,1,-2)\) and \((3,-4,2)\) is:
1 4
2 5
3 \(5 \sqrt{2}\)
4 \(4 \sqrt{2}\)
Explanation:
C \(=\left \vert\begin{array}{ccc}x-2& y+3 &z-1 \\ -3 &4 &-3 \\ 4& -5 &4\end{array}\right \vert=0\) \((x-2)[(16-15)-(y+3)][-12+12]+(2-1)[15-16\)
\(=0]\)
\(\mathrm{x}-2+0-2+1=0\)
\(\mathrm{x}-2=1\)
\(\mathrm{x}-\mathrm{z}-1=0\)
Distance of \(\mathrm{P}(7,-3,-4)\) from Plane is
\(\mathrm{d}=\left \vert\frac{7+4-1}{\sqrt{2}}\right \vert=5 \sqrt{2}\)
JEE Main-24.01.2023
Three Dimensional Geometry
121399
The image of the point \(P(1,3,4)\) in the plane \(2 x\) \(-\mathbf{y}+\mathrm{z}+\mathbf{3}=0\), is
1 \((3,5,-2)\)
2 \((-3,5,2)\)
3 \((3,-5,2)\)
4 \((3,5,2)\)
Explanation:
B Image of point in a plane is \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a_1+b y_1+\left(z_1+d\right)\right.}{a^2+b^2+c^2}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-y}{-1}=\frac{-2(2-3+4+3)}{4+1+1}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=-2\) \(\frac{x-1}{2}=-2, \frac{y-3}{-1}=-2, \frac{z-4}{1}=-2\) \(\quad x=-3, y=5, z=2\) \(\text { Image }(-3,5,2)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Three Dimensional Geometry
121395
The image (reflection) of the point \((1,2,-1)\) in the plane \(\overrightarrow{\mathbf{r}} \cdot(\mathbf{3} \hat{\mathbf{i}}-\mathbf{5} \hat{\mathbf{j}}+\mathbf{4} \hat{\mathbf{k}})=5\) is
A We know that, The image of the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) in the plane \(\mathrm{ax}+\mathrm{by}+\) \(\mathrm{cz}+\mathrm{d}=0\) is given by \(\frac{z-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-2\left(\frac{a x_1+b y_1+c z_1+d}{a^2+b^2+c^2}\right)\) The equation of the plane is \(\overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=5\) or , \(3 x-5 y+4 z=5\) The image of \((1,2-1)\) in this plane is given by \(\frac{x-1}{3}=\frac{y-2}{-5}=\frac{z+1}{4}=-2\left(\frac{3-10-4-5}{9+25+16}\right)\) \(\Rightarrow x=\frac{73}{25}, y=\frac{-6}{5}, z=\frac{39}{25}\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121397
The distance of the point \((-1,9,-16)\) from the plane \(2 x+3 y-z=5\) measured parallel to the line \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\) is
1 \(20 \sqrt{2}\)
2 \(13 \sqrt{2}\)
3 31
4 26
Explanation:
C Distance of \((-1,9,-16)\) from plane \(2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}-5=0\) measured parallel to line
\(\frac{x+4}{3}=\frac{y-2}{-4}=\frac{z-3}{12}\)
\(\text { is }\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a} l+\mathrm{bm}+\mathrm{cn}}\right \vert\)
Where \(l, \mathrm{~m}, \mathrm{n}\) are d.c's of line
\(\left. \vert \vert \frac{2(-1)+3(9)-1(-16)-5}{2\left(\frac{3}{13}\right)+3\left(\frac{-4}{13}\right)-1\left(\frac{12}{13}\right)} \vert= \vert \frac{36 \times 13}{-18} \right\rvert\,=26\)
JEE Main-24.01.2023
Three Dimensional Geometry
121398
The distance of the point \((7,-3,-4)\) from plane passing through the points \((2,-3,1),(-1,1,-2)\) and \((3,-4,2)\) is:
1 4
2 5
3 \(5 \sqrt{2}\)
4 \(4 \sqrt{2}\)
Explanation:
C \(=\left \vert\begin{array}{ccc}x-2& y+3 &z-1 \\ -3 &4 &-3 \\ 4& -5 &4\end{array}\right \vert=0\) \((x-2)[(16-15)-(y+3)][-12+12]+(2-1)[15-16\)
\(=0]\)
\(\mathrm{x}-2+0-2+1=0\)
\(\mathrm{x}-2=1\)
\(\mathrm{x}-\mathrm{z}-1=0\)
Distance of \(\mathrm{P}(7,-3,-4)\) from Plane is
\(\mathrm{d}=\left \vert\frac{7+4-1}{\sqrt{2}}\right \vert=5 \sqrt{2}\)
JEE Main-24.01.2023
Three Dimensional Geometry
121399
The image of the point \(P(1,3,4)\) in the plane \(2 x\) \(-\mathbf{y}+\mathrm{z}+\mathbf{3}=0\), is
1 \((3,5,-2)\)
2 \((-3,5,2)\)
3 \((3,-5,2)\)
4 \((3,5,2)\)
Explanation:
B Image of point in a plane is \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-2\left(a_1+b y_1+\left(z_1+d\right)\right.}{a^2+b^2+c^2}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-y}{-1}=\frac{-2(2-3+4+3)}{4+1+1}\) \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=-2\) \(\frac{x-1}{2}=-2, \frac{y-3}{-1}=-2, \frac{z-4}{1}=-2\) \(\quad x=-3, y=5, z=2\) \(\text { Image }(-3,5,2)\)