121306
For non-coplanar vectors \(a, b\) and \(c\), if the point of intersection of the line \(r=a+t(b-c)\) and the plane \(\mathbf{r}=\mathbf{b}+\mathbf{c}+\mathbf{x}(\mathbf{a}-\mathbf{b})+\mathbf{y}(\mathbf{c}+\mathbf{a})\) is \(l a+m b+n c\), then \(3 l+4 m+2 n=\)
1 0
2 \(\frac{1}{2}\)
3 2
4 1
Explanation:
C Given, equation of plane \(r =b+c+x(a-b)+y(c+a)\) \(=(1-x) b+(1+y) c+(x+y) a\) And equation of line \(r=a+t(b-c)\) From equation (i) and (ii), we get- \(\mathrm{a}+\mathrm{t}(\mathrm{b}-\mathrm{c}) =(\mathrm{x}+\mathrm{y}) \mathrm{a}+1(1-\mathrm{x}) \mathrm{b}+(1+\mathrm{y}) \mathrm{c}\) \(\mathrm{x}+\mathrm{y} =1\) \(1-\mathrm{x} =\mathrm{t}\) \(1+\mathrm{y} =-\mathrm{t}\) Solving equation (iii), (iv) and (v), we get- \(\mathrm{t}=\frac{-1}{2}, \mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{-1}{2}\) \(\mathrm{r}=\mathrm{a}-\frac{1}{2}(\mathrm{~b}-\mathrm{c})\) \(\mathrm{r}=\mathrm{a} \frac{1}{2} \mathrm{~b}+\frac{1}{2} \mathrm{c}=\mathrm{a}+\mathrm{mb}+\mathrm{nc}\) \(l=1, \mathrm{~m}=\frac{-1}{2} \text { and } \mathrm{n}=\frac{1}{2}\) Now, \(3 \mathrm{l}+4 \mathrm{~m}+2 \mathrm{n}\) \(=3(1)+4\left(-\frac{1}{2}\right)+2\left(\frac{1}{2}\right)=3-2+1=2\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121290
The equation of the plane through the points \((2,2,1)\) and \((9,3,6)\) and perpendicular to the plane \(2 x+6 y+6 z-1=0\) is
1 \(3 x+4 y+5 z+9=0\)
2 \(3 \mathrm{x}+4 \mathrm{y}-5 \mathrm{z}+9=0\)
3 \(3 x-4 y+5 z-9=0\)
4 \(3 x+4 y-5 z-9=0\)
Explanation:
D Let, equation of plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=0\) Equation of plane passing through \((2,2,1)\)
The \(a(x-2)+b(y-2)+c(z-1)=0\)
The point \((9,3,6)\) lies on it
\(\mathrm{a}(9-2)+\mathrm{b}(3-2)+\mathrm{c}(6-1)=0\)
\(7 \mathrm{a}+\mathrm{b}+5 \mathrm{c}=0\)
And perpendicular to the plane,
\(2 x+6 y+6 z-1=0\)
\(2 a+6 b+6 c=0\)
From equation (i) (ii) and (iii)-
\(\left \vert\begin{array}{ccc}x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6\end{array}\right \vert=0\)
\((x-2)(6-30)-(y-2)(42-10)+(z-1)(42-2)\)
\(=0\)
\(-24(x-2)-32(y-2)+40(z-1)=0\)
\(-24 x-32 y+40 z+48+64-40=0\)
\(24 x+32 y-40 z-72=0\)
\(3 x+4 y-5 z-9=0\)
AMU-2012
Three Dimensional Geometry
121298
The equation of the plane mid-parallel to the planes \(2 x-3 y+6 z+21=0\) and \(2 x-3 y+6 z-14=0\) is given by .......
1 \(4 x+6 y-12 z+7=0\)
2 \(4 x-6 y-12 z-7=0\)
3 \(4 x-6 y+12 z+7=0\)
4 \(4 x+6 y+12 z-7=0\)
Explanation:
C Given, \(P_1: 2 x-3 y+6 z+21=0\)
\(P_2: 2 x-3 y+6 z-14=0\)
\(\mathrm{P}_2: 2 \mathrm{x}-3 \mathrm{y}+6 \mathrm{z}-14=0\)
Let equation of required plane be -
\(P: 2 x-3 y+6 z+\lambda=0\)
According to question,
Direction between \(P \& P_1=\) Distance between \(P \& P_2\)
\(\frac{\left \vert\mathrm{d}_2-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{\left \vert\mathrm{d}_3-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\)
\(\frac{ \vert\lambda-21 \vert}{\sqrt{2^2+(-3)^2+6^2}}=\frac{ \vert\lambda+14 \vert}{\sqrt{2^2+(-3)^2+6^2}}\)
\( \vert\lambda-21 \vert= \vert\lambda+14 \vert\)
\(2 \lambda=-14+21 \quad \quad \text { (put in (iii)) }\)
\(\lambda=\frac{7}{2} \quad\)
\(2 x-3 y+6 z+\frac{7}{2}=0\)
\(4 x-6 y+12 z+7=0\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121300
The equation of the plane in normal form which passes through the points \((-2,1,3),(1,1\), \(1)\) and \((2,3,4)\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121306
For non-coplanar vectors \(a, b\) and \(c\), if the point of intersection of the line \(r=a+t(b-c)\) and the plane \(\mathbf{r}=\mathbf{b}+\mathbf{c}+\mathbf{x}(\mathbf{a}-\mathbf{b})+\mathbf{y}(\mathbf{c}+\mathbf{a})\) is \(l a+m b+n c\), then \(3 l+4 m+2 n=\)
1 0
2 \(\frac{1}{2}\)
3 2
4 1
Explanation:
C Given, equation of plane \(r =b+c+x(a-b)+y(c+a)\) \(=(1-x) b+(1+y) c+(x+y) a\) And equation of line \(r=a+t(b-c)\) From equation (i) and (ii), we get- \(\mathrm{a}+\mathrm{t}(\mathrm{b}-\mathrm{c}) =(\mathrm{x}+\mathrm{y}) \mathrm{a}+1(1-\mathrm{x}) \mathrm{b}+(1+\mathrm{y}) \mathrm{c}\) \(\mathrm{x}+\mathrm{y} =1\) \(1-\mathrm{x} =\mathrm{t}\) \(1+\mathrm{y} =-\mathrm{t}\) Solving equation (iii), (iv) and (v), we get- \(\mathrm{t}=\frac{-1}{2}, \mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{-1}{2}\) \(\mathrm{r}=\mathrm{a}-\frac{1}{2}(\mathrm{~b}-\mathrm{c})\) \(\mathrm{r}=\mathrm{a} \frac{1}{2} \mathrm{~b}+\frac{1}{2} \mathrm{c}=\mathrm{a}+\mathrm{mb}+\mathrm{nc}\) \(l=1, \mathrm{~m}=\frac{-1}{2} \text { and } \mathrm{n}=\frac{1}{2}\) Now, \(3 \mathrm{l}+4 \mathrm{~m}+2 \mathrm{n}\) \(=3(1)+4\left(-\frac{1}{2}\right)+2\left(\frac{1}{2}\right)=3-2+1=2\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121290
The equation of the plane through the points \((2,2,1)\) and \((9,3,6)\) and perpendicular to the plane \(2 x+6 y+6 z-1=0\) is
1 \(3 x+4 y+5 z+9=0\)
2 \(3 \mathrm{x}+4 \mathrm{y}-5 \mathrm{z}+9=0\)
3 \(3 x-4 y+5 z-9=0\)
4 \(3 x+4 y-5 z-9=0\)
Explanation:
D Let, equation of plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=0\) Equation of plane passing through \((2,2,1)\)
The \(a(x-2)+b(y-2)+c(z-1)=0\)
The point \((9,3,6)\) lies on it
\(\mathrm{a}(9-2)+\mathrm{b}(3-2)+\mathrm{c}(6-1)=0\)
\(7 \mathrm{a}+\mathrm{b}+5 \mathrm{c}=0\)
And perpendicular to the plane,
\(2 x+6 y+6 z-1=0\)
\(2 a+6 b+6 c=0\)
From equation (i) (ii) and (iii)-
\(\left \vert\begin{array}{ccc}x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6\end{array}\right \vert=0\)
\((x-2)(6-30)-(y-2)(42-10)+(z-1)(42-2)\)
\(=0\)
\(-24(x-2)-32(y-2)+40(z-1)=0\)
\(-24 x-32 y+40 z+48+64-40=0\)
\(24 x+32 y-40 z-72=0\)
\(3 x+4 y-5 z-9=0\)
AMU-2012
Three Dimensional Geometry
121298
The equation of the plane mid-parallel to the planes \(2 x-3 y+6 z+21=0\) and \(2 x-3 y+6 z-14=0\) is given by .......
1 \(4 x+6 y-12 z+7=0\)
2 \(4 x-6 y-12 z-7=0\)
3 \(4 x-6 y+12 z+7=0\)
4 \(4 x+6 y+12 z-7=0\)
Explanation:
C Given, \(P_1: 2 x-3 y+6 z+21=0\)
\(P_2: 2 x-3 y+6 z-14=0\)
\(\mathrm{P}_2: 2 \mathrm{x}-3 \mathrm{y}+6 \mathrm{z}-14=0\)
Let equation of required plane be -
\(P: 2 x-3 y+6 z+\lambda=0\)
According to question,
Direction between \(P \& P_1=\) Distance between \(P \& P_2\)
\(\frac{\left \vert\mathrm{d}_2-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{\left \vert\mathrm{d}_3-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\)
\(\frac{ \vert\lambda-21 \vert}{\sqrt{2^2+(-3)^2+6^2}}=\frac{ \vert\lambda+14 \vert}{\sqrt{2^2+(-3)^2+6^2}}\)
\( \vert\lambda-21 \vert= \vert\lambda+14 \vert\)
\(2 \lambda=-14+21 \quad \quad \text { (put in (iii)) }\)
\(\lambda=\frac{7}{2} \quad\)
\(2 x-3 y+6 z+\frac{7}{2}=0\)
\(4 x-6 y+12 z+7=0\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121300
The equation of the plane in normal form which passes through the points \((-2,1,3),(1,1\), \(1)\) and \((2,3,4)\) is
121306
For non-coplanar vectors \(a, b\) and \(c\), if the point of intersection of the line \(r=a+t(b-c)\) and the plane \(\mathbf{r}=\mathbf{b}+\mathbf{c}+\mathbf{x}(\mathbf{a}-\mathbf{b})+\mathbf{y}(\mathbf{c}+\mathbf{a})\) is \(l a+m b+n c\), then \(3 l+4 m+2 n=\)
1 0
2 \(\frac{1}{2}\)
3 2
4 1
Explanation:
C Given, equation of plane \(r =b+c+x(a-b)+y(c+a)\) \(=(1-x) b+(1+y) c+(x+y) a\) And equation of line \(r=a+t(b-c)\) From equation (i) and (ii), we get- \(\mathrm{a}+\mathrm{t}(\mathrm{b}-\mathrm{c}) =(\mathrm{x}+\mathrm{y}) \mathrm{a}+1(1-\mathrm{x}) \mathrm{b}+(1+\mathrm{y}) \mathrm{c}\) \(\mathrm{x}+\mathrm{y} =1\) \(1-\mathrm{x} =\mathrm{t}\) \(1+\mathrm{y} =-\mathrm{t}\) Solving equation (iii), (iv) and (v), we get- \(\mathrm{t}=\frac{-1}{2}, \mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{-1}{2}\) \(\mathrm{r}=\mathrm{a}-\frac{1}{2}(\mathrm{~b}-\mathrm{c})\) \(\mathrm{r}=\mathrm{a} \frac{1}{2} \mathrm{~b}+\frac{1}{2} \mathrm{c}=\mathrm{a}+\mathrm{mb}+\mathrm{nc}\) \(l=1, \mathrm{~m}=\frac{-1}{2} \text { and } \mathrm{n}=\frac{1}{2}\) Now, \(3 \mathrm{l}+4 \mathrm{~m}+2 \mathrm{n}\) \(=3(1)+4\left(-\frac{1}{2}\right)+2\left(\frac{1}{2}\right)=3-2+1=2\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121290
The equation of the plane through the points \((2,2,1)\) and \((9,3,6)\) and perpendicular to the plane \(2 x+6 y+6 z-1=0\) is
1 \(3 x+4 y+5 z+9=0\)
2 \(3 \mathrm{x}+4 \mathrm{y}-5 \mathrm{z}+9=0\)
3 \(3 x-4 y+5 z-9=0\)
4 \(3 x+4 y-5 z-9=0\)
Explanation:
D Let, equation of plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=0\) Equation of plane passing through \((2,2,1)\)
The \(a(x-2)+b(y-2)+c(z-1)=0\)
The point \((9,3,6)\) lies on it
\(\mathrm{a}(9-2)+\mathrm{b}(3-2)+\mathrm{c}(6-1)=0\)
\(7 \mathrm{a}+\mathrm{b}+5 \mathrm{c}=0\)
And perpendicular to the plane,
\(2 x+6 y+6 z-1=0\)
\(2 a+6 b+6 c=0\)
From equation (i) (ii) and (iii)-
\(\left \vert\begin{array}{ccc}x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6\end{array}\right \vert=0\)
\((x-2)(6-30)-(y-2)(42-10)+(z-1)(42-2)\)
\(=0\)
\(-24(x-2)-32(y-2)+40(z-1)=0\)
\(-24 x-32 y+40 z+48+64-40=0\)
\(24 x+32 y-40 z-72=0\)
\(3 x+4 y-5 z-9=0\)
AMU-2012
Three Dimensional Geometry
121298
The equation of the plane mid-parallel to the planes \(2 x-3 y+6 z+21=0\) and \(2 x-3 y+6 z-14=0\) is given by .......
1 \(4 x+6 y-12 z+7=0\)
2 \(4 x-6 y-12 z-7=0\)
3 \(4 x-6 y+12 z+7=0\)
4 \(4 x+6 y+12 z-7=0\)
Explanation:
C Given, \(P_1: 2 x-3 y+6 z+21=0\)
\(P_2: 2 x-3 y+6 z-14=0\)
\(\mathrm{P}_2: 2 \mathrm{x}-3 \mathrm{y}+6 \mathrm{z}-14=0\)
Let equation of required plane be -
\(P: 2 x-3 y+6 z+\lambda=0\)
According to question,
Direction between \(P \& P_1=\) Distance between \(P \& P_2\)
\(\frac{\left \vert\mathrm{d}_2-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{\left \vert\mathrm{d}_3-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\)
\(\frac{ \vert\lambda-21 \vert}{\sqrt{2^2+(-3)^2+6^2}}=\frac{ \vert\lambda+14 \vert}{\sqrt{2^2+(-3)^2+6^2}}\)
\( \vert\lambda-21 \vert= \vert\lambda+14 \vert\)
\(2 \lambda=-14+21 \quad \quad \text { (put in (iii)) }\)
\(\lambda=\frac{7}{2} \quad\)
\(2 x-3 y+6 z+\frac{7}{2}=0\)
\(4 x-6 y+12 z+7=0\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121300
The equation of the plane in normal form which passes through the points \((-2,1,3),(1,1\), \(1)\) and \((2,3,4)\) is
121306
For non-coplanar vectors \(a, b\) and \(c\), if the point of intersection of the line \(r=a+t(b-c)\) and the plane \(\mathbf{r}=\mathbf{b}+\mathbf{c}+\mathbf{x}(\mathbf{a}-\mathbf{b})+\mathbf{y}(\mathbf{c}+\mathbf{a})\) is \(l a+m b+n c\), then \(3 l+4 m+2 n=\)
1 0
2 \(\frac{1}{2}\)
3 2
4 1
Explanation:
C Given, equation of plane \(r =b+c+x(a-b)+y(c+a)\) \(=(1-x) b+(1+y) c+(x+y) a\) And equation of line \(r=a+t(b-c)\) From equation (i) and (ii), we get- \(\mathrm{a}+\mathrm{t}(\mathrm{b}-\mathrm{c}) =(\mathrm{x}+\mathrm{y}) \mathrm{a}+1(1-\mathrm{x}) \mathrm{b}+(1+\mathrm{y}) \mathrm{c}\) \(\mathrm{x}+\mathrm{y} =1\) \(1-\mathrm{x} =\mathrm{t}\) \(1+\mathrm{y} =-\mathrm{t}\) Solving equation (iii), (iv) and (v), we get- \(\mathrm{t}=\frac{-1}{2}, \mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{-1}{2}\) \(\mathrm{r}=\mathrm{a}-\frac{1}{2}(\mathrm{~b}-\mathrm{c})\) \(\mathrm{r}=\mathrm{a} \frac{1}{2} \mathrm{~b}+\frac{1}{2} \mathrm{c}=\mathrm{a}+\mathrm{mb}+\mathrm{nc}\) \(l=1, \mathrm{~m}=\frac{-1}{2} \text { and } \mathrm{n}=\frac{1}{2}\) Now, \(3 \mathrm{l}+4 \mathrm{~m}+2 \mathrm{n}\) \(=3(1)+4\left(-\frac{1}{2}\right)+2\left(\frac{1}{2}\right)=3-2+1=2\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121290
The equation of the plane through the points \((2,2,1)\) and \((9,3,6)\) and perpendicular to the plane \(2 x+6 y+6 z-1=0\) is
1 \(3 x+4 y+5 z+9=0\)
2 \(3 \mathrm{x}+4 \mathrm{y}-5 \mathrm{z}+9=0\)
3 \(3 x-4 y+5 z-9=0\)
4 \(3 x+4 y-5 z-9=0\)
Explanation:
D Let, equation of plane \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=0\) Equation of plane passing through \((2,2,1)\)
The \(a(x-2)+b(y-2)+c(z-1)=0\)
The point \((9,3,6)\) lies on it
\(\mathrm{a}(9-2)+\mathrm{b}(3-2)+\mathrm{c}(6-1)=0\)
\(7 \mathrm{a}+\mathrm{b}+5 \mathrm{c}=0\)
And perpendicular to the plane,
\(2 x+6 y+6 z-1=0\)
\(2 a+6 b+6 c=0\)
From equation (i) (ii) and (iii)-
\(\left \vert\begin{array}{ccc}x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6\end{array}\right \vert=0\)
\((x-2)(6-30)-(y-2)(42-10)+(z-1)(42-2)\)
\(=0\)
\(-24(x-2)-32(y-2)+40(z-1)=0\)
\(-24 x-32 y+40 z+48+64-40=0\)
\(24 x+32 y-40 z-72=0\)
\(3 x+4 y-5 z-9=0\)
AMU-2012
Three Dimensional Geometry
121298
The equation of the plane mid-parallel to the planes \(2 x-3 y+6 z+21=0\) and \(2 x-3 y+6 z-14=0\) is given by .......
1 \(4 x+6 y-12 z+7=0\)
2 \(4 x-6 y-12 z-7=0\)
3 \(4 x-6 y+12 z+7=0\)
4 \(4 x+6 y+12 z-7=0\)
Explanation:
C Given, \(P_1: 2 x-3 y+6 z+21=0\)
\(P_2: 2 x-3 y+6 z-14=0\)
\(\mathrm{P}_2: 2 \mathrm{x}-3 \mathrm{y}+6 \mathrm{z}-14=0\)
Let equation of required plane be -
\(P: 2 x-3 y+6 z+\lambda=0\)
According to question,
Direction between \(P \& P_1=\) Distance between \(P \& P_2\)
\(\frac{\left \vert\mathrm{d}_2-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{\left \vert\mathrm{d}_3-\mathrm{d}_1\right \vert}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}\)
\(\frac{ \vert\lambda-21 \vert}{\sqrt{2^2+(-3)^2+6^2}}=\frac{ \vert\lambda+14 \vert}{\sqrt{2^2+(-3)^2+6^2}}\)
\( \vert\lambda-21 \vert= \vert\lambda+14 \vert\)
\(2 \lambda=-14+21 \quad \quad \text { (put in (iii)) }\)
\(\lambda=\frac{7}{2} \quad\)
\(2 x-3 y+6 z+\frac{7}{2}=0\)
\(4 x-6 y+12 z+7=0\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121300
The equation of the plane in normal form which passes through the points \((-2,1,3),(1,1\), \(1)\) and \((2,3,4)\) is
