121444
Distance between two parallel planes \(2 x+y+\) \(2 z=8\) and \(4 x+2 y+4 z+5=0\) is
1 \(\frac{5}{2}\)
2 \(\frac{7}{2}\)
3 \(\frac{9}{2}\)
4 \(\frac{3}{2}\)
Explanation:
B The planes are \(4 x+2 y+4 z=16,4 x+2 y+4 z=-5\) \(\Rightarrow \text { Distance between planes }\) \(=\frac{16-(-5)}{\sqrt{4^2+2^2+4^2}}=\frac{21}{6}=\frac{7}{2}\)
COMEDK-2013
Three Dimensional Geometry
121410
From a point \(P(a, b, c)\) perpendicular \(P A, P B\) and drawn to \(y z\) and \(\mathrm{zx}\) planes. Find the equation of the plane \(O A B\), where \(O\) is the origin.
1 \(b c x+c a y+a b z=0\)
2 \(b c x+c a y-a b z=0\)
3 \(b c x-c a y+a b z=0\)
4 \(-\mathrm{bcx}+\mathrm{cay}+\mathrm{abz}=0\)
Explanation:
B P(a, b, c) and PA and PB are perpendicular to \(\mathrm{YZ}\) and \(\mathrm{ZX}\) planes. Hence, co-ordinate of \(\mathrm{A}\) and \(\mathrm{B}\) are \((0, b, c)\) and \((a, 0, c)\) respectively. Equation of plane passing through \((0,0,0)\),
\((0, b, c)\) and \((a, 0, c)\) is \(\left \vert\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 0 & \mathrm{~b} & \mathrm{c} \\ \mathrm{a} & 0 & \mathrm{c}\end{array}\right \vert=0\)
\(\Rightarrow x(b c-0)-y(0-a c)+z(0-a b)=0\)
\(\Rightarrow b c x+a c y-a b z=0\)
WB JEE-2021
Three Dimensional Geometry
121404
Equation of line passing through the point \((2,3,1)\) and parallel to the line of intersection of the plane \(x-2 y-z+5=0\) and \(x+y+3 z=6\) is
1 \(\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
2 \(\frac{x-2}{4}=\frac{y-3}{3}=\frac{z-1}{2}\)
3 \(\frac{x-2}{5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
4 \(\frac{x-2}{5}=\frac{y-3}{4}=\frac{z-1}{3}\)
Explanation:
A Given \(\text { Point }(2,3,1)\)
\(P_1=x-2 y-z+5=0, \vec{n}_1=\hat{i}-2 \hat{j}-\hat{k}\)
\(P_2=x+y+3 z-6=0, \vec{n}_2=\hat{i}+\hat{j}+3 \hat{k}\)
Intersection of the plane-
\(\text { Line, } L=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 1 & -2 & -1 \\ 1 & 1 & 3\end{array}\right \vert\)
\(=\hat{i}(-6+1)-\hat{j}(3+1)+\hat{k}(1+2)=-5 \hat{i}-4 \hat{j}+3 \hat{k}\)
Direction ratios of line \((-5,-4,3)\) i.e. \((a, b, c)\)
\(\therefore\) Points \((2,3,1)\) i.e. \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\)
Equation of line, \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
\(=\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
Karnataka CET-2015
Three Dimensional Geometry
121419
The foot of the perpendicular from the point (7, \(14,5)\) to the plane \(2 x+4 y-z=2\) are
1 \((1,2,8)\)
2 \((3,2,8)\)
3 \((5,10,6)\)
4 \((9,18,4)\)
Explanation:
A We know that the length of the perpendicular from the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) to the plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\frac{\left \vert\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\) and the co-ordinate \((\alpha, \beta, \gamma)\) or the foot of the \(\perp\) are given by- \(\frac{\alpha-\mathrm{x}_1}{\mathrm{a}}=\frac{\beta-\mathrm{y}_1}{\mathrm{~b}}=\frac{\gamma-\mathrm{z}_1}{\mathrm{c}}=-\left(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\right) \ldots\)
In the given question, \(\mathrm{x}_1=7, \mathrm{y}_1=14, \mathrm{z}_1=5, \mathrm{a}=2, \mathrm{~b}=4, \mathrm{c}=-1, \mathrm{~d}=-2\) By putting these values in (1), we get
\(\frac{\alpha-7}{2}=\frac{\beta-14}{4}=\frac{\gamma-5}{-1}=-\frac{63}{21}\)
\(\Rightarrow \alpha=1, \beta=2 \text { and } \gamma=8\)Hence, foot of \(\perp\) from point to the plane \((1,2,8)\)
BITSAT-2012
Three Dimensional Geometry
121421
The equation of the plane in normal form passing through the point \(A(\overline{\mathbf{a}})\), parallel to a vector \(\overline{\mathbf{b}}\) and containing \(\overline{\mathbf{a}}\) vector \(\overline{\mathbf{c}}\) is
121444
Distance between two parallel planes \(2 x+y+\) \(2 z=8\) and \(4 x+2 y+4 z+5=0\) is
1 \(\frac{5}{2}\)
2 \(\frac{7}{2}\)
3 \(\frac{9}{2}\)
4 \(\frac{3}{2}\)
Explanation:
B The planes are \(4 x+2 y+4 z=16,4 x+2 y+4 z=-5\) \(\Rightarrow \text { Distance between planes }\) \(=\frac{16-(-5)}{\sqrt{4^2+2^2+4^2}}=\frac{21}{6}=\frac{7}{2}\)
COMEDK-2013
Three Dimensional Geometry
121410
From a point \(P(a, b, c)\) perpendicular \(P A, P B\) and drawn to \(y z\) and \(\mathrm{zx}\) planes. Find the equation of the plane \(O A B\), where \(O\) is the origin.
1 \(b c x+c a y+a b z=0\)
2 \(b c x+c a y-a b z=0\)
3 \(b c x-c a y+a b z=0\)
4 \(-\mathrm{bcx}+\mathrm{cay}+\mathrm{abz}=0\)
Explanation:
B P(a, b, c) and PA and PB are perpendicular to \(\mathrm{YZ}\) and \(\mathrm{ZX}\) planes. Hence, co-ordinate of \(\mathrm{A}\) and \(\mathrm{B}\) are \((0, b, c)\) and \((a, 0, c)\) respectively. Equation of plane passing through \((0,0,0)\),
\((0, b, c)\) and \((a, 0, c)\) is \(\left \vert\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 0 & \mathrm{~b} & \mathrm{c} \\ \mathrm{a} & 0 & \mathrm{c}\end{array}\right \vert=0\)
\(\Rightarrow x(b c-0)-y(0-a c)+z(0-a b)=0\)
\(\Rightarrow b c x+a c y-a b z=0\)
WB JEE-2021
Three Dimensional Geometry
121404
Equation of line passing through the point \((2,3,1)\) and parallel to the line of intersection of the plane \(x-2 y-z+5=0\) and \(x+y+3 z=6\) is
1 \(\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
2 \(\frac{x-2}{4}=\frac{y-3}{3}=\frac{z-1}{2}\)
3 \(\frac{x-2}{5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
4 \(\frac{x-2}{5}=\frac{y-3}{4}=\frac{z-1}{3}\)
Explanation:
A Given \(\text { Point }(2,3,1)\)
\(P_1=x-2 y-z+5=0, \vec{n}_1=\hat{i}-2 \hat{j}-\hat{k}\)
\(P_2=x+y+3 z-6=0, \vec{n}_2=\hat{i}+\hat{j}+3 \hat{k}\)
Intersection of the plane-
\(\text { Line, } L=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 1 & -2 & -1 \\ 1 & 1 & 3\end{array}\right \vert\)
\(=\hat{i}(-6+1)-\hat{j}(3+1)+\hat{k}(1+2)=-5 \hat{i}-4 \hat{j}+3 \hat{k}\)
Direction ratios of line \((-5,-4,3)\) i.e. \((a, b, c)\)
\(\therefore\) Points \((2,3,1)\) i.e. \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\)
Equation of line, \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
\(=\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
Karnataka CET-2015
Three Dimensional Geometry
121419
The foot of the perpendicular from the point (7, \(14,5)\) to the plane \(2 x+4 y-z=2\) are
1 \((1,2,8)\)
2 \((3,2,8)\)
3 \((5,10,6)\)
4 \((9,18,4)\)
Explanation:
A We know that the length of the perpendicular from the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) to the plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\frac{\left \vert\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\) and the co-ordinate \((\alpha, \beta, \gamma)\) or the foot of the \(\perp\) are given by- \(\frac{\alpha-\mathrm{x}_1}{\mathrm{a}}=\frac{\beta-\mathrm{y}_1}{\mathrm{~b}}=\frac{\gamma-\mathrm{z}_1}{\mathrm{c}}=-\left(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\right) \ldots\)
In the given question, \(\mathrm{x}_1=7, \mathrm{y}_1=14, \mathrm{z}_1=5, \mathrm{a}=2, \mathrm{~b}=4, \mathrm{c}=-1, \mathrm{~d}=-2\) By putting these values in (1), we get
\(\frac{\alpha-7}{2}=\frac{\beta-14}{4}=\frac{\gamma-5}{-1}=-\frac{63}{21}\)
\(\Rightarrow \alpha=1, \beta=2 \text { and } \gamma=8\)Hence, foot of \(\perp\) from point to the plane \((1,2,8)\)
BITSAT-2012
Three Dimensional Geometry
121421
The equation of the plane in normal form passing through the point \(A(\overline{\mathbf{a}})\), parallel to a vector \(\overline{\mathbf{b}}\) and containing \(\overline{\mathbf{a}}\) vector \(\overline{\mathbf{c}}\) is
121444
Distance between two parallel planes \(2 x+y+\) \(2 z=8\) and \(4 x+2 y+4 z+5=0\) is
1 \(\frac{5}{2}\)
2 \(\frac{7}{2}\)
3 \(\frac{9}{2}\)
4 \(\frac{3}{2}\)
Explanation:
B The planes are \(4 x+2 y+4 z=16,4 x+2 y+4 z=-5\) \(\Rightarrow \text { Distance between planes }\) \(=\frac{16-(-5)}{\sqrt{4^2+2^2+4^2}}=\frac{21}{6}=\frac{7}{2}\)
COMEDK-2013
Three Dimensional Geometry
121410
From a point \(P(a, b, c)\) perpendicular \(P A, P B\) and drawn to \(y z\) and \(\mathrm{zx}\) planes. Find the equation of the plane \(O A B\), where \(O\) is the origin.
1 \(b c x+c a y+a b z=0\)
2 \(b c x+c a y-a b z=0\)
3 \(b c x-c a y+a b z=0\)
4 \(-\mathrm{bcx}+\mathrm{cay}+\mathrm{abz}=0\)
Explanation:
B P(a, b, c) and PA and PB are perpendicular to \(\mathrm{YZ}\) and \(\mathrm{ZX}\) planes. Hence, co-ordinate of \(\mathrm{A}\) and \(\mathrm{B}\) are \((0, b, c)\) and \((a, 0, c)\) respectively. Equation of plane passing through \((0,0,0)\),
\((0, b, c)\) and \((a, 0, c)\) is \(\left \vert\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 0 & \mathrm{~b} & \mathrm{c} \\ \mathrm{a} & 0 & \mathrm{c}\end{array}\right \vert=0\)
\(\Rightarrow x(b c-0)-y(0-a c)+z(0-a b)=0\)
\(\Rightarrow b c x+a c y-a b z=0\)
WB JEE-2021
Three Dimensional Geometry
121404
Equation of line passing through the point \((2,3,1)\) and parallel to the line of intersection of the plane \(x-2 y-z+5=0\) and \(x+y+3 z=6\) is
1 \(\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
2 \(\frac{x-2}{4}=\frac{y-3}{3}=\frac{z-1}{2}\)
3 \(\frac{x-2}{5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
4 \(\frac{x-2}{5}=\frac{y-3}{4}=\frac{z-1}{3}\)
Explanation:
A Given \(\text { Point }(2,3,1)\)
\(P_1=x-2 y-z+5=0, \vec{n}_1=\hat{i}-2 \hat{j}-\hat{k}\)
\(P_2=x+y+3 z-6=0, \vec{n}_2=\hat{i}+\hat{j}+3 \hat{k}\)
Intersection of the plane-
\(\text { Line, } L=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 1 & -2 & -1 \\ 1 & 1 & 3\end{array}\right \vert\)
\(=\hat{i}(-6+1)-\hat{j}(3+1)+\hat{k}(1+2)=-5 \hat{i}-4 \hat{j}+3 \hat{k}\)
Direction ratios of line \((-5,-4,3)\) i.e. \((a, b, c)\)
\(\therefore\) Points \((2,3,1)\) i.e. \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\)
Equation of line, \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
\(=\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
Karnataka CET-2015
Three Dimensional Geometry
121419
The foot of the perpendicular from the point (7, \(14,5)\) to the plane \(2 x+4 y-z=2\) are
1 \((1,2,8)\)
2 \((3,2,8)\)
3 \((5,10,6)\)
4 \((9,18,4)\)
Explanation:
A We know that the length of the perpendicular from the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) to the plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\frac{\left \vert\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\) and the co-ordinate \((\alpha, \beta, \gamma)\) or the foot of the \(\perp\) are given by- \(\frac{\alpha-\mathrm{x}_1}{\mathrm{a}}=\frac{\beta-\mathrm{y}_1}{\mathrm{~b}}=\frac{\gamma-\mathrm{z}_1}{\mathrm{c}}=-\left(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\right) \ldots\)
In the given question, \(\mathrm{x}_1=7, \mathrm{y}_1=14, \mathrm{z}_1=5, \mathrm{a}=2, \mathrm{~b}=4, \mathrm{c}=-1, \mathrm{~d}=-2\) By putting these values in (1), we get
\(\frac{\alpha-7}{2}=\frac{\beta-14}{4}=\frac{\gamma-5}{-1}=-\frac{63}{21}\)
\(\Rightarrow \alpha=1, \beta=2 \text { and } \gamma=8\)Hence, foot of \(\perp\) from point to the plane \((1,2,8)\)
BITSAT-2012
Three Dimensional Geometry
121421
The equation of the plane in normal form passing through the point \(A(\overline{\mathbf{a}})\), parallel to a vector \(\overline{\mathbf{b}}\) and containing \(\overline{\mathbf{a}}\) vector \(\overline{\mathbf{c}}\) is
121444
Distance between two parallel planes \(2 x+y+\) \(2 z=8\) and \(4 x+2 y+4 z+5=0\) is
1 \(\frac{5}{2}\)
2 \(\frac{7}{2}\)
3 \(\frac{9}{2}\)
4 \(\frac{3}{2}\)
Explanation:
B The planes are \(4 x+2 y+4 z=16,4 x+2 y+4 z=-5\) \(\Rightarrow \text { Distance between planes }\) \(=\frac{16-(-5)}{\sqrt{4^2+2^2+4^2}}=\frac{21}{6}=\frac{7}{2}\)
COMEDK-2013
Three Dimensional Geometry
121410
From a point \(P(a, b, c)\) perpendicular \(P A, P B\) and drawn to \(y z\) and \(\mathrm{zx}\) planes. Find the equation of the plane \(O A B\), where \(O\) is the origin.
1 \(b c x+c a y+a b z=0\)
2 \(b c x+c a y-a b z=0\)
3 \(b c x-c a y+a b z=0\)
4 \(-\mathrm{bcx}+\mathrm{cay}+\mathrm{abz}=0\)
Explanation:
B P(a, b, c) and PA and PB are perpendicular to \(\mathrm{YZ}\) and \(\mathrm{ZX}\) planes. Hence, co-ordinate of \(\mathrm{A}\) and \(\mathrm{B}\) are \((0, b, c)\) and \((a, 0, c)\) respectively. Equation of plane passing through \((0,0,0)\),
\((0, b, c)\) and \((a, 0, c)\) is \(\left \vert\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 0 & \mathrm{~b} & \mathrm{c} \\ \mathrm{a} & 0 & \mathrm{c}\end{array}\right \vert=0\)
\(\Rightarrow x(b c-0)-y(0-a c)+z(0-a b)=0\)
\(\Rightarrow b c x+a c y-a b z=0\)
WB JEE-2021
Three Dimensional Geometry
121404
Equation of line passing through the point \((2,3,1)\) and parallel to the line of intersection of the plane \(x-2 y-z+5=0\) and \(x+y+3 z=6\) is
1 \(\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
2 \(\frac{x-2}{4}=\frac{y-3}{3}=\frac{z-1}{2}\)
3 \(\frac{x-2}{5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
4 \(\frac{x-2}{5}=\frac{y-3}{4}=\frac{z-1}{3}\)
Explanation:
A Given \(\text { Point }(2,3,1)\)
\(P_1=x-2 y-z+5=0, \vec{n}_1=\hat{i}-2 \hat{j}-\hat{k}\)
\(P_2=x+y+3 z-6=0, \vec{n}_2=\hat{i}+\hat{j}+3 \hat{k}\)
Intersection of the plane-
\(\text { Line, } L=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 1 & -2 & -1 \\ 1 & 1 & 3\end{array}\right \vert\)
\(=\hat{i}(-6+1)-\hat{j}(3+1)+\hat{k}(1+2)=-5 \hat{i}-4 \hat{j}+3 \hat{k}\)
Direction ratios of line \((-5,-4,3)\) i.e. \((a, b, c)\)
\(\therefore\) Points \((2,3,1)\) i.e. \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\)
Equation of line, \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
\(=\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
Karnataka CET-2015
Three Dimensional Geometry
121419
The foot of the perpendicular from the point (7, \(14,5)\) to the plane \(2 x+4 y-z=2\) are
1 \((1,2,8)\)
2 \((3,2,8)\)
3 \((5,10,6)\)
4 \((9,18,4)\)
Explanation:
A We know that the length of the perpendicular from the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) to the plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\frac{\left \vert\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\) and the co-ordinate \((\alpha, \beta, \gamma)\) or the foot of the \(\perp\) are given by- \(\frac{\alpha-\mathrm{x}_1}{\mathrm{a}}=\frac{\beta-\mathrm{y}_1}{\mathrm{~b}}=\frac{\gamma-\mathrm{z}_1}{\mathrm{c}}=-\left(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\right) \ldots\)
In the given question, \(\mathrm{x}_1=7, \mathrm{y}_1=14, \mathrm{z}_1=5, \mathrm{a}=2, \mathrm{~b}=4, \mathrm{c}=-1, \mathrm{~d}=-2\) By putting these values in (1), we get
\(\frac{\alpha-7}{2}=\frac{\beta-14}{4}=\frac{\gamma-5}{-1}=-\frac{63}{21}\)
\(\Rightarrow \alpha=1, \beta=2 \text { and } \gamma=8\)Hence, foot of \(\perp\) from point to the plane \((1,2,8)\)
BITSAT-2012
Three Dimensional Geometry
121421
The equation of the plane in normal form passing through the point \(A(\overline{\mathbf{a}})\), parallel to a vector \(\overline{\mathbf{b}}\) and containing \(\overline{\mathbf{a}}\) vector \(\overline{\mathbf{c}}\) is
121444
Distance between two parallel planes \(2 x+y+\) \(2 z=8\) and \(4 x+2 y+4 z+5=0\) is
1 \(\frac{5}{2}\)
2 \(\frac{7}{2}\)
3 \(\frac{9}{2}\)
4 \(\frac{3}{2}\)
Explanation:
B The planes are \(4 x+2 y+4 z=16,4 x+2 y+4 z=-5\) \(\Rightarrow \text { Distance between planes }\) \(=\frac{16-(-5)}{\sqrt{4^2+2^2+4^2}}=\frac{21}{6}=\frac{7}{2}\)
COMEDK-2013
Three Dimensional Geometry
121410
From a point \(P(a, b, c)\) perpendicular \(P A, P B\) and drawn to \(y z\) and \(\mathrm{zx}\) planes. Find the equation of the plane \(O A B\), where \(O\) is the origin.
1 \(b c x+c a y+a b z=0\)
2 \(b c x+c a y-a b z=0\)
3 \(b c x-c a y+a b z=0\)
4 \(-\mathrm{bcx}+\mathrm{cay}+\mathrm{abz}=0\)
Explanation:
B P(a, b, c) and PA and PB are perpendicular to \(\mathrm{YZ}\) and \(\mathrm{ZX}\) planes. Hence, co-ordinate of \(\mathrm{A}\) and \(\mathrm{B}\) are \((0, b, c)\) and \((a, 0, c)\) respectively. Equation of plane passing through \((0,0,0)\),
\((0, b, c)\) and \((a, 0, c)\) is \(\left \vert\begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z} \\ 0 & \mathrm{~b} & \mathrm{c} \\ \mathrm{a} & 0 & \mathrm{c}\end{array}\right \vert=0\)
\(\Rightarrow x(b c-0)-y(0-a c)+z(0-a b)=0\)
\(\Rightarrow b c x+a c y-a b z=0\)
WB JEE-2021
Three Dimensional Geometry
121404
Equation of line passing through the point \((2,3,1)\) and parallel to the line of intersection of the plane \(x-2 y-z+5=0\) and \(x+y+3 z=6\) is
1 \(\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
2 \(\frac{x-2}{4}=\frac{y-3}{3}=\frac{z-1}{2}\)
3 \(\frac{x-2}{5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
4 \(\frac{x-2}{5}=\frac{y-3}{4}=\frac{z-1}{3}\)
Explanation:
A Given \(\text { Point }(2,3,1)\)
\(P_1=x-2 y-z+5=0, \vec{n}_1=\hat{i}-2 \hat{j}-\hat{k}\)
\(P_2=x+y+3 z-6=0, \vec{n}_2=\hat{i}+\hat{j}+3 \hat{k}\)
Intersection of the plane-
\(\text { Line, } L=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{\mathrm{k}} \\ 1 & -2 & -1 \\ 1 & 1 & 3\end{array}\right \vert\)
\(=\hat{i}(-6+1)-\hat{j}(3+1)+\hat{k}(1+2)=-5 \hat{i}-4 \hat{j}+3 \hat{k}\)
Direction ratios of line \((-5,-4,3)\) i.e. \((a, b, c)\)
\(\therefore\) Points \((2,3,1)\) i.e. \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\)
Equation of line, \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\)
\(=\frac{x-2}{-5}=\frac{y-3}{-4}=\frac{z-1}{3}\)
Karnataka CET-2015
Three Dimensional Geometry
121419
The foot of the perpendicular from the point (7, \(14,5)\) to the plane \(2 x+4 y-z=2\) are
1 \((1,2,8)\)
2 \((3,2,8)\)
3 \((5,10,6)\)
4 \((9,18,4)\)
Explanation:
A We know that the length of the perpendicular from the point \(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)\) to the plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0\) is \(\frac{\left \vert\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\) and the co-ordinate \((\alpha, \beta, \gamma)\) or the foot of the \(\perp\) are given by- \(\frac{\alpha-\mathrm{x}_1}{\mathrm{a}}=\frac{\beta-\mathrm{y}_1}{\mathrm{~b}}=\frac{\gamma-\mathrm{z}_1}{\mathrm{c}}=-\left(\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}\right) \ldots\)
In the given question, \(\mathrm{x}_1=7, \mathrm{y}_1=14, \mathrm{z}_1=5, \mathrm{a}=2, \mathrm{~b}=4, \mathrm{c}=-1, \mathrm{~d}=-2\) By putting these values in (1), we get
\(\frac{\alpha-7}{2}=\frac{\beta-14}{4}=\frac{\gamma-5}{-1}=-\frac{63}{21}\)
\(\Rightarrow \alpha=1, \beta=2 \text { and } \gamma=8\)Hence, foot of \(\perp\) from point to the plane \((1,2,8)\)
BITSAT-2012
Three Dimensional Geometry
121421
The equation of the plane in normal form passing through the point \(A(\overline{\mathbf{a}})\), parallel to a vector \(\overline{\mathbf{b}}\) and containing \(\overline{\mathbf{a}}\) vector \(\overline{\mathbf{c}}\) is
