121358
The equations of planes parallel to the plane \(x+2 y+2 z+8=0\), which are at a distance of 2 units from the point \((1,1,2)\) are
1 \(x+2 y+2 z-6=0\) or \(x+2 y+2 z-7=0\)
2 \(x+2 y+2 z-13=0\) or \(x+2 y+2 z-1=0\)
3 \(x+2 y+2 z-5=0\) or \(x+2 y+2 z-3=0\)
4 \(x+2 y+2 z+3=0\) or \(x+2 y+2 z-5=0\)
Explanation:
B Given, The equation of the plane parallel to the plane \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+18=0\) is \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+\lambda=0\).
Now, the distance of this plane from the point \((1,1,2)\) is
\(\therefore\left \vert\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^2+2^2+2^2}}\right \vert=\left \vert\frac{7+\lambda}{3}\right \vert\)
We have \(\left \vert\frac{7+\lambda}{3}\right \vert=2 \Rightarrow \frac{7+\lambda}{3}= \pm 2\)
\(\therefore \lambda= \pm 6-7, \lambda=-1 \text { or } \lambda=-13\)Hence, equations of plane are \(x+2 y+2 z-1=0\) or \(x+2 y+2 z-13=0\)
MHT CET-2020
Three Dimensional Geometry
121340
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}\) and the plane \(2 x-2 y+z=5\) is
121351
If the foot of perpendicular drawn from the origin to the plane is \((3,2,1)\), then the equation of plane is
1 \(3 x+2 y-z=14\)
2 \(3 x+2 y+z=14\)
3 \(3 x-2 y-z=12\)
4 \(3 \mathrm{x}+2 \mathrm{y}-\mathrm{z}=12\)
Explanation:
B Let \(\mathrm{P}(3,2,1)\) be the foot of the perpendicular from the origin to the plane.
\(\overline{\mathrm{OP}}=\overline{\mathrm{n}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { and }\)
\(\mathrm{P}= \vert3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \vert=\sqrt{3^2+2^2+1^2}=\sqrt{14}\)
\(\hat{\mathrm{n}}=\frac{3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\)
The vector equation of the plane is
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot \frac{(3 \hat{i}+2 \hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14}\)
\(\therefore 3 x+2 y+z=14 \Rightarrow 3 x+2 y+z-14=0\)
MHT CET-2020
Three Dimensional Geometry
121365
Let the image of the point \(P(2,-1,3)\) in the plane \(x+2 y-z=0\) be \(Q\). Then the distance of the plane \(3 x+2 y+z+29=0\) from the point \(Q\) is
121358
The equations of planes parallel to the plane \(x+2 y+2 z+8=0\), which are at a distance of 2 units from the point \((1,1,2)\) are
1 \(x+2 y+2 z-6=0\) or \(x+2 y+2 z-7=0\)
2 \(x+2 y+2 z-13=0\) or \(x+2 y+2 z-1=0\)
3 \(x+2 y+2 z-5=0\) or \(x+2 y+2 z-3=0\)
4 \(x+2 y+2 z+3=0\) or \(x+2 y+2 z-5=0\)
Explanation:
B Given, The equation of the plane parallel to the plane \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+18=0\) is \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+\lambda=0\).
Now, the distance of this plane from the point \((1,1,2)\) is
\(\therefore\left \vert\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^2+2^2+2^2}}\right \vert=\left \vert\frac{7+\lambda}{3}\right \vert\)
We have \(\left \vert\frac{7+\lambda}{3}\right \vert=2 \Rightarrow \frac{7+\lambda}{3}= \pm 2\)
\(\therefore \lambda= \pm 6-7, \lambda=-1 \text { or } \lambda=-13\)Hence, equations of plane are \(x+2 y+2 z-1=0\) or \(x+2 y+2 z-13=0\)
MHT CET-2020
Three Dimensional Geometry
121340
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}\) and the plane \(2 x-2 y+z=5\) is
121351
If the foot of perpendicular drawn from the origin to the plane is \((3,2,1)\), then the equation of plane is
1 \(3 x+2 y-z=14\)
2 \(3 x+2 y+z=14\)
3 \(3 x-2 y-z=12\)
4 \(3 \mathrm{x}+2 \mathrm{y}-\mathrm{z}=12\)
Explanation:
B Let \(\mathrm{P}(3,2,1)\) be the foot of the perpendicular from the origin to the plane.
\(\overline{\mathrm{OP}}=\overline{\mathrm{n}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { and }\)
\(\mathrm{P}= \vert3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \vert=\sqrt{3^2+2^2+1^2}=\sqrt{14}\)
\(\hat{\mathrm{n}}=\frac{3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\)
The vector equation of the plane is
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot \frac{(3 \hat{i}+2 \hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14}\)
\(\therefore 3 x+2 y+z=14 \Rightarrow 3 x+2 y+z-14=0\)
MHT CET-2020
Three Dimensional Geometry
121365
Let the image of the point \(P(2,-1,3)\) in the plane \(x+2 y-z=0\) be \(Q\). Then the distance of the plane \(3 x+2 y+z+29=0\) from the point \(Q\) is
121358
The equations of planes parallel to the plane \(x+2 y+2 z+8=0\), which are at a distance of 2 units from the point \((1,1,2)\) are
1 \(x+2 y+2 z-6=0\) or \(x+2 y+2 z-7=0\)
2 \(x+2 y+2 z-13=0\) or \(x+2 y+2 z-1=0\)
3 \(x+2 y+2 z-5=0\) or \(x+2 y+2 z-3=0\)
4 \(x+2 y+2 z+3=0\) or \(x+2 y+2 z-5=0\)
Explanation:
B Given, The equation of the plane parallel to the plane \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+18=0\) is \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+\lambda=0\).
Now, the distance of this plane from the point \((1,1,2)\) is
\(\therefore\left \vert\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^2+2^2+2^2}}\right \vert=\left \vert\frac{7+\lambda}{3}\right \vert\)
We have \(\left \vert\frac{7+\lambda}{3}\right \vert=2 \Rightarrow \frac{7+\lambda}{3}= \pm 2\)
\(\therefore \lambda= \pm 6-7, \lambda=-1 \text { or } \lambda=-13\)Hence, equations of plane are \(x+2 y+2 z-1=0\) or \(x+2 y+2 z-13=0\)
MHT CET-2020
Three Dimensional Geometry
121340
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}\) and the plane \(2 x-2 y+z=5\) is
121351
If the foot of perpendicular drawn from the origin to the plane is \((3,2,1)\), then the equation of plane is
1 \(3 x+2 y-z=14\)
2 \(3 x+2 y+z=14\)
3 \(3 x-2 y-z=12\)
4 \(3 \mathrm{x}+2 \mathrm{y}-\mathrm{z}=12\)
Explanation:
B Let \(\mathrm{P}(3,2,1)\) be the foot of the perpendicular from the origin to the plane.
\(\overline{\mathrm{OP}}=\overline{\mathrm{n}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { and }\)
\(\mathrm{P}= \vert3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \vert=\sqrt{3^2+2^2+1^2}=\sqrt{14}\)
\(\hat{\mathrm{n}}=\frac{3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\)
The vector equation of the plane is
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot \frac{(3 \hat{i}+2 \hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14}\)
\(\therefore 3 x+2 y+z=14 \Rightarrow 3 x+2 y+z-14=0\)
MHT CET-2020
Three Dimensional Geometry
121365
Let the image of the point \(P(2,-1,3)\) in the plane \(x+2 y-z=0\) be \(Q\). Then the distance of the plane \(3 x+2 y+z+29=0\) from the point \(Q\) is
121358
The equations of planes parallel to the plane \(x+2 y+2 z+8=0\), which are at a distance of 2 units from the point \((1,1,2)\) are
1 \(x+2 y+2 z-6=0\) or \(x+2 y+2 z-7=0\)
2 \(x+2 y+2 z-13=0\) or \(x+2 y+2 z-1=0\)
3 \(x+2 y+2 z-5=0\) or \(x+2 y+2 z-3=0\)
4 \(x+2 y+2 z+3=0\) or \(x+2 y+2 z-5=0\)
Explanation:
B Given, The equation of the plane parallel to the plane \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+18=0\) is \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+\lambda=0\).
Now, the distance of this plane from the point \((1,1,2)\) is
\(\therefore\left \vert\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^2+2^2+2^2}}\right \vert=\left \vert\frac{7+\lambda}{3}\right \vert\)
We have \(\left \vert\frac{7+\lambda}{3}\right \vert=2 \Rightarrow \frac{7+\lambda}{3}= \pm 2\)
\(\therefore \lambda= \pm 6-7, \lambda=-1 \text { or } \lambda=-13\)Hence, equations of plane are \(x+2 y+2 z-1=0\) or \(x+2 y+2 z-13=0\)
MHT CET-2020
Three Dimensional Geometry
121340
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}\) and the plane \(2 x-2 y+z=5\) is
121351
If the foot of perpendicular drawn from the origin to the plane is \((3,2,1)\), then the equation of plane is
1 \(3 x+2 y-z=14\)
2 \(3 x+2 y+z=14\)
3 \(3 x-2 y-z=12\)
4 \(3 \mathrm{x}+2 \mathrm{y}-\mathrm{z}=12\)
Explanation:
B Let \(\mathrm{P}(3,2,1)\) be the foot of the perpendicular from the origin to the plane.
\(\overline{\mathrm{OP}}=\overline{\mathrm{n}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { and }\)
\(\mathrm{P}= \vert3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \vert=\sqrt{3^2+2^2+1^2}=\sqrt{14}\)
\(\hat{\mathrm{n}}=\frac{3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\)
The vector equation of the plane is
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot \frac{(3 \hat{i}+2 \hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14}\)
\(\therefore 3 x+2 y+z=14 \Rightarrow 3 x+2 y+z-14=0\)
MHT CET-2020
Three Dimensional Geometry
121365
Let the image of the point \(P(2,-1,3)\) in the plane \(x+2 y-z=0\) be \(Q\). Then the distance of the plane \(3 x+2 y+z+29=0\) from the point \(Q\) is
121358
The equations of planes parallel to the plane \(x+2 y+2 z+8=0\), which are at a distance of 2 units from the point \((1,1,2)\) are
1 \(x+2 y+2 z-6=0\) or \(x+2 y+2 z-7=0\)
2 \(x+2 y+2 z-13=0\) or \(x+2 y+2 z-1=0\)
3 \(x+2 y+2 z-5=0\) or \(x+2 y+2 z-3=0\)
4 \(x+2 y+2 z+3=0\) or \(x+2 y+2 z-5=0\)
Explanation:
B Given, The equation of the plane parallel to the plane \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+18=0\) is \(\mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+\lambda=0\).
Now, the distance of this plane from the point \((1,1,2)\) is
\(\therefore\left \vert\frac{1(1)+2(1)+2(2)+\lambda}{\sqrt{1^2+2^2+2^2}}\right \vert=\left \vert\frac{7+\lambda}{3}\right \vert\)
We have \(\left \vert\frac{7+\lambda}{3}\right \vert=2 \Rightarrow \frac{7+\lambda}{3}= \pm 2\)
\(\therefore \lambda= \pm 6-7, \lambda=-1 \text { or } \lambda=-13\)Hence, equations of plane are \(x+2 y+2 z-1=0\) or \(x+2 y+2 z-13=0\)
MHT CET-2020
Three Dimensional Geometry
121340
The sine of the angle between the straight line \(\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}\) and the plane \(2 x-2 y+z=5\) is
121351
If the foot of perpendicular drawn from the origin to the plane is \((3,2,1)\), then the equation of plane is
1 \(3 x+2 y-z=14\)
2 \(3 x+2 y+z=14\)
3 \(3 x-2 y-z=12\)
4 \(3 \mathrm{x}+2 \mathrm{y}-\mathrm{z}=12\)
Explanation:
B Let \(\mathrm{P}(3,2,1)\) be the foot of the perpendicular from the origin to the plane.
\(\overline{\mathrm{OP}}=\overline{\mathrm{n}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { and }\)
\(\mathrm{P}= \vert3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \vert=\sqrt{3^2+2^2+1^2}=\sqrt{14}\)
\(\hat{\mathrm{n}}=\frac{3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\)
The vector equation of the plane is
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot \frac{(3 \hat{i}+2 \hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14}\)
\(\therefore 3 x+2 y+z=14 \Rightarrow 3 x+2 y+z-14=0\)
MHT CET-2020
Three Dimensional Geometry
121365
Let the image of the point \(P(2,-1,3)\) in the plane \(x+2 y-z=0\) be \(Q\). Then the distance of the plane \(3 x+2 y+z+29=0\) from the point \(Q\) is
