121396
If a plane \(\pi\) passes through the point \((-1,6,2)\) is perpendicular to the planes \(x+2 y+2 z-5=\) 0 and \(3 x+3 y+2 z-8=0\), then, the perpendicular distance from the point \((1,-1,1)\) to the plane \(\pi\) is
1 \(\frac{20}{\sqrt{29}}\)
2 \(\frac{21}{\sqrt{29}}\)
3 \(\frac{27}{\sqrt{29}}\)
4 \(\sqrt{29}\)
Explanation:
D Equation of plane passing through the point \((-1,6,2)\) and perpendicular to planes
\(x+2 y+2 z-5=0\) and \(3 x+3 y+2 z-8=0\) is
\(\left \vert\begin{array}{ccc}x+1 &y-6 &z-2 \\ 1& 2& 2 \\ 3& 3& 2\end{array}\right \vert=0\)
\(\Rightarrow(x+1)(4-6)-(y-6)(2-6)+(z-2)(3-6)=0\)
\(\Rightarrow-2 x-2+4 y-24-3 z+6=0\)
\(\Rightarrow 2 x-4 y+3 z+20=0\)
\(\Rightarrow\) Distance from \((1,-1,1)\) to plane is
\(d=\left \vert\frac{2+4+3+20}{\sqrt{2^2+4^2+3^2}}\right \vert=\frac{29}{\sqrt{29}}=\sqrt{29}\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121355
The length of the perpendicular to the plane \(\overrightarrow{\mathbf{r}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=14\) from the origin is
1 7 units
2 \(\sqrt{14}\) units
3 14 units
4 \(\sqrt{7}\) units
Explanation:
B Length of perpendicular from the point \(A(\vec{a})\) to the Plane \(\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{n}}=\mathrm{p}\) is \(\frac{ \vert\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{p} \vert}{ \vert\overrightarrow{\mathrm{n}} \vert}\)
Here, \(\vec{a} . \vec{n}=0\) and \( \vert\vec{n} \vert=\sqrt{1+4+9}=\sqrt{14}\)
Hence, required distance is \(\frac{ \vert0-14 \vert}{\sqrt{14}}=\sqrt{14}\)
MHT CET-2020
Three Dimensional Geometry
121357
The unit vector perpendicular to the plane \(4 x-3 y+12 z=15\) is
A Given, The unit vector perpendicular to plane \(4 x-3 y+12 z=15\) is \(\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{\sqrt{16+9+144}}=\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{13}\)
MHT CET-2020
Three Dimensional Geometry
121361
Distance of the point \((2,3,4)\) from the plane \(3 x-6 y+2 z+11=0\) is
1 0
2 1
3 2
4 3
Explanation:
B The distance of a point \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) from a plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\mathrm{d}=\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right \vert\)
\(\therefore\) Distance of the point \((2,3,4)\) from the plane \(3 \mathrm{x}-6 \mathrm{y}+2 \mathrm{z}+11=0\) is
\(\Rightarrow \quad \mathrm{d}=\left \vert\frac{6-18+8+11}{\sqrt{9+36+4}}\right \vert=\left \vert\frac{7}{\sqrt{49}}\right \vert=\left \vert\frac{7}{7}\right \vert\)
121396
If a plane \(\pi\) passes through the point \((-1,6,2)\) is perpendicular to the planes \(x+2 y+2 z-5=\) 0 and \(3 x+3 y+2 z-8=0\), then, the perpendicular distance from the point \((1,-1,1)\) to the plane \(\pi\) is
1 \(\frac{20}{\sqrt{29}}\)
2 \(\frac{21}{\sqrt{29}}\)
3 \(\frac{27}{\sqrt{29}}\)
4 \(\sqrt{29}\)
Explanation:
D Equation of plane passing through the point \((-1,6,2)\) and perpendicular to planes
\(x+2 y+2 z-5=0\) and \(3 x+3 y+2 z-8=0\) is
\(\left \vert\begin{array}{ccc}x+1 &y-6 &z-2 \\ 1& 2& 2 \\ 3& 3& 2\end{array}\right \vert=0\)
\(\Rightarrow(x+1)(4-6)-(y-6)(2-6)+(z-2)(3-6)=0\)
\(\Rightarrow-2 x-2+4 y-24-3 z+6=0\)
\(\Rightarrow 2 x-4 y+3 z+20=0\)
\(\Rightarrow\) Distance from \((1,-1,1)\) to plane is
\(d=\left \vert\frac{2+4+3+20}{\sqrt{2^2+4^2+3^2}}\right \vert=\frac{29}{\sqrt{29}}=\sqrt{29}\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121355
The length of the perpendicular to the plane \(\overrightarrow{\mathbf{r}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=14\) from the origin is
1 7 units
2 \(\sqrt{14}\) units
3 14 units
4 \(\sqrt{7}\) units
Explanation:
B Length of perpendicular from the point \(A(\vec{a})\) to the Plane \(\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{n}}=\mathrm{p}\) is \(\frac{ \vert\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{p} \vert}{ \vert\overrightarrow{\mathrm{n}} \vert}\)
Here, \(\vec{a} . \vec{n}=0\) and \( \vert\vec{n} \vert=\sqrt{1+4+9}=\sqrt{14}\)
Hence, required distance is \(\frac{ \vert0-14 \vert}{\sqrt{14}}=\sqrt{14}\)
MHT CET-2020
Three Dimensional Geometry
121357
The unit vector perpendicular to the plane \(4 x-3 y+12 z=15\) is
A Given, The unit vector perpendicular to plane \(4 x-3 y+12 z=15\) is \(\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{\sqrt{16+9+144}}=\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{13}\)
MHT CET-2020
Three Dimensional Geometry
121361
Distance of the point \((2,3,4)\) from the plane \(3 x-6 y+2 z+11=0\) is
1 0
2 1
3 2
4 3
Explanation:
B The distance of a point \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) from a plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\mathrm{d}=\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right \vert\)
\(\therefore\) Distance of the point \((2,3,4)\) from the plane \(3 \mathrm{x}-6 \mathrm{y}+2 \mathrm{z}+11=0\) is
\(\Rightarrow \quad \mathrm{d}=\left \vert\frac{6-18+8+11}{\sqrt{9+36+4}}\right \vert=\left \vert\frac{7}{\sqrt{49}}\right \vert=\left \vert\frac{7}{7}\right \vert\)
121396
If a plane \(\pi\) passes through the point \((-1,6,2)\) is perpendicular to the planes \(x+2 y+2 z-5=\) 0 and \(3 x+3 y+2 z-8=0\), then, the perpendicular distance from the point \((1,-1,1)\) to the plane \(\pi\) is
1 \(\frac{20}{\sqrt{29}}\)
2 \(\frac{21}{\sqrt{29}}\)
3 \(\frac{27}{\sqrt{29}}\)
4 \(\sqrt{29}\)
Explanation:
D Equation of plane passing through the point \((-1,6,2)\) and perpendicular to planes
\(x+2 y+2 z-5=0\) and \(3 x+3 y+2 z-8=0\) is
\(\left \vert\begin{array}{ccc}x+1 &y-6 &z-2 \\ 1& 2& 2 \\ 3& 3& 2\end{array}\right \vert=0\)
\(\Rightarrow(x+1)(4-6)-(y-6)(2-6)+(z-2)(3-6)=0\)
\(\Rightarrow-2 x-2+4 y-24-3 z+6=0\)
\(\Rightarrow 2 x-4 y+3 z+20=0\)
\(\Rightarrow\) Distance from \((1,-1,1)\) to plane is
\(d=\left \vert\frac{2+4+3+20}{\sqrt{2^2+4^2+3^2}}\right \vert=\frac{29}{\sqrt{29}}=\sqrt{29}\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121355
The length of the perpendicular to the plane \(\overrightarrow{\mathbf{r}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=14\) from the origin is
1 7 units
2 \(\sqrt{14}\) units
3 14 units
4 \(\sqrt{7}\) units
Explanation:
B Length of perpendicular from the point \(A(\vec{a})\) to the Plane \(\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{n}}=\mathrm{p}\) is \(\frac{ \vert\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{p} \vert}{ \vert\overrightarrow{\mathrm{n}} \vert}\)
Here, \(\vec{a} . \vec{n}=0\) and \( \vert\vec{n} \vert=\sqrt{1+4+9}=\sqrt{14}\)
Hence, required distance is \(\frac{ \vert0-14 \vert}{\sqrt{14}}=\sqrt{14}\)
MHT CET-2020
Three Dimensional Geometry
121357
The unit vector perpendicular to the plane \(4 x-3 y+12 z=15\) is
A Given, The unit vector perpendicular to plane \(4 x-3 y+12 z=15\) is \(\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{\sqrt{16+9+144}}=\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{13}\)
MHT CET-2020
Three Dimensional Geometry
121361
Distance of the point \((2,3,4)\) from the plane \(3 x-6 y+2 z+11=0\) is
1 0
2 1
3 2
4 3
Explanation:
B The distance of a point \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) from a plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\mathrm{d}=\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right \vert\)
\(\therefore\) Distance of the point \((2,3,4)\) from the plane \(3 \mathrm{x}-6 \mathrm{y}+2 \mathrm{z}+11=0\) is
\(\Rightarrow \quad \mathrm{d}=\left \vert\frac{6-18+8+11}{\sqrt{9+36+4}}\right \vert=\left \vert\frac{7}{\sqrt{49}}\right \vert=\left \vert\frac{7}{7}\right \vert\)
121396
If a plane \(\pi\) passes through the point \((-1,6,2)\) is perpendicular to the planes \(x+2 y+2 z-5=\) 0 and \(3 x+3 y+2 z-8=0\), then, the perpendicular distance from the point \((1,-1,1)\) to the plane \(\pi\) is
1 \(\frac{20}{\sqrt{29}}\)
2 \(\frac{21}{\sqrt{29}}\)
3 \(\frac{27}{\sqrt{29}}\)
4 \(\sqrt{29}\)
Explanation:
D Equation of plane passing through the point \((-1,6,2)\) and perpendicular to planes
\(x+2 y+2 z-5=0\) and \(3 x+3 y+2 z-8=0\) is
\(\left \vert\begin{array}{ccc}x+1 &y-6 &z-2 \\ 1& 2& 2 \\ 3& 3& 2\end{array}\right \vert=0\)
\(\Rightarrow(x+1)(4-6)-(y-6)(2-6)+(z-2)(3-6)=0\)
\(\Rightarrow-2 x-2+4 y-24-3 z+6=0\)
\(\Rightarrow 2 x-4 y+3 z+20=0\)
\(\Rightarrow\) Distance from \((1,-1,1)\) to plane is
\(d=\left \vert\frac{2+4+3+20}{\sqrt{2^2+4^2+3^2}}\right \vert=\frac{29}{\sqrt{29}}=\sqrt{29}\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121355
The length of the perpendicular to the plane \(\overrightarrow{\mathbf{r}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=14\) from the origin is
1 7 units
2 \(\sqrt{14}\) units
3 14 units
4 \(\sqrt{7}\) units
Explanation:
B Length of perpendicular from the point \(A(\vec{a})\) to the Plane \(\overrightarrow{\mathrm{r}} . \overrightarrow{\mathrm{n}}=\mathrm{p}\) is \(\frac{ \vert\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{p} \vert}{ \vert\overrightarrow{\mathrm{n}} \vert}\)
Here, \(\vec{a} . \vec{n}=0\) and \( \vert\vec{n} \vert=\sqrt{1+4+9}=\sqrt{14}\)
Hence, required distance is \(\frac{ \vert0-14 \vert}{\sqrt{14}}=\sqrt{14}\)
MHT CET-2020
Three Dimensional Geometry
121357
The unit vector perpendicular to the plane \(4 x-3 y+12 z=15\) is
A Given, The unit vector perpendicular to plane \(4 x-3 y+12 z=15\) is \(\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{\sqrt{16+9+144}}=\frac{4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{13}\)
MHT CET-2020
Three Dimensional Geometry
121361
Distance of the point \((2,3,4)\) from the plane \(3 x-6 y+2 z+11=0\) is
1 0
2 1
3 2
4 3
Explanation:
B The distance of a point \(\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) from a plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\mathrm{d}=\left \vert\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\right \vert\)
\(\therefore\) Distance of the point \((2,3,4)\) from the plane \(3 \mathrm{x}-6 \mathrm{y}+2 \mathrm{z}+11=0\) is
\(\Rightarrow \quad \mathrm{d}=\left \vert\frac{6-18+8+11}{\sqrt{9+36+4}}\right \vert=\left \vert\frac{7}{\sqrt{49}}\right \vert=\left \vert\frac{7}{7}\right \vert\)
