121163
If \(A(3,4,5), B(4,6,3) C(-1,2,4)\) and \(D(1,0,5)\) are such that the angle between the lines \(D C\) and \(A B\) is \(\theta\), then \(\cos \theta\) is equal to
C Given, If a line makes angle \(\alpha\) with \(\mathrm{x}\)-axis, \(\beta\) with \(\mathrm{y}\)-axis \& \(\gamma\) with z-axis. So, direction cosine of the line be\(\ell=\cos \alpha, \quad \mathrm{m}=\cos \beta, \mathrm{n}=\cos \gamma\) Given, \(\alpha=\beta=\gamma \Rightarrow \ell=m=n=\cos \alpha\) We know that: \(\ell^2+\mathrm{m}^2+\mathrm{n}^2=1\) \(\cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1\) \(3 \cos ^2 \alpha=1\) \(\cos ^2 \alpha=\frac{1}{3}\) \(\cos \alpha= \pm \frac{1}{\sqrt{3}}\) Similarly, \(\cos \beta= \pm \frac{1}{\sqrt{3}}, \cos \gamma= \pm \frac{1}{\sqrt{3}}\) So, direction cosine are, \(\ell= \pm \frac{1}{\sqrt{3}}, \mathrm{~m}= \pm \frac{1}{\sqrt{3}}, \mathrm{n}= \pm \frac{1}{\sqrt{3}}\)
AP EAMCET-19.08.2021
Three Dimensional Geometry
121166
The direction ratios of a normal to a plane are \((3,12,4)\). Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?
1 \((2,2,1)\)
2 \((2,1,2)\)
3 \((2,1,1)\)
4 \((3,-1,2)\)
Explanation:
B Let plane be \(a x+b y+c z+d=0\) Direction ratio of normal to the plane \((3,12,4)\) So, equation (i) is- \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z}+\mathrm{d}=0\) Now, distance of (ii) from origin is 2 unit \(\frac{3 \times 0+12 \times 0+4 \times 0+d}{\sqrt{3^2+12^2+4^2}}=2\) \(d= \pm(2 \times 13)= \pm 26\) \(\therefore\) Equation of plane is \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z} \pm 26=0\)This plane is passes through \((2,1,2)\)
J and K CET-2018
Three Dimensional Geometry
121169
The direction cosines of a line are \(\left\langle\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right\rangle\) respectively. Then its direction ratios are
1 \((9,6,-2)\)
2 \((-9,-6,2)\)
3 \((-9,6,-2)\)
4 \((9,-6,-2)\)
Explanation:
C Given, DC's \(\left(\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right)\) Direction cosine formula \(\mathrm{l}=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-9}{11}\) \(\mathrm{~m}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{6}{11}\) \(\mathrm{n}=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-2}{11}\)Here, \((a, b, c)\) is the direction ratio i.e. \((-9,6,-2)\)
121163
If \(A(3,4,5), B(4,6,3) C(-1,2,4)\) and \(D(1,0,5)\) are such that the angle between the lines \(D C\) and \(A B\) is \(\theta\), then \(\cos \theta\) is equal to
C Given, If a line makes angle \(\alpha\) with \(\mathrm{x}\)-axis, \(\beta\) with \(\mathrm{y}\)-axis \& \(\gamma\) with z-axis. So, direction cosine of the line be\(\ell=\cos \alpha, \quad \mathrm{m}=\cos \beta, \mathrm{n}=\cos \gamma\) Given, \(\alpha=\beta=\gamma \Rightarrow \ell=m=n=\cos \alpha\) We know that: \(\ell^2+\mathrm{m}^2+\mathrm{n}^2=1\) \(\cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1\) \(3 \cos ^2 \alpha=1\) \(\cos ^2 \alpha=\frac{1}{3}\) \(\cos \alpha= \pm \frac{1}{\sqrt{3}}\) Similarly, \(\cos \beta= \pm \frac{1}{\sqrt{3}}, \cos \gamma= \pm \frac{1}{\sqrt{3}}\) So, direction cosine are, \(\ell= \pm \frac{1}{\sqrt{3}}, \mathrm{~m}= \pm \frac{1}{\sqrt{3}}, \mathrm{n}= \pm \frac{1}{\sqrt{3}}\)
AP EAMCET-19.08.2021
Three Dimensional Geometry
121166
The direction ratios of a normal to a plane are \((3,12,4)\). Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?
1 \((2,2,1)\)
2 \((2,1,2)\)
3 \((2,1,1)\)
4 \((3,-1,2)\)
Explanation:
B Let plane be \(a x+b y+c z+d=0\) Direction ratio of normal to the plane \((3,12,4)\) So, equation (i) is- \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z}+\mathrm{d}=0\) Now, distance of (ii) from origin is 2 unit \(\frac{3 \times 0+12 \times 0+4 \times 0+d}{\sqrt{3^2+12^2+4^2}}=2\) \(d= \pm(2 \times 13)= \pm 26\) \(\therefore\) Equation of plane is \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z} \pm 26=0\)This plane is passes through \((2,1,2)\)
J and K CET-2018
Three Dimensional Geometry
121169
The direction cosines of a line are \(\left\langle\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right\rangle\) respectively. Then its direction ratios are
1 \((9,6,-2)\)
2 \((-9,-6,2)\)
3 \((-9,6,-2)\)
4 \((9,-6,-2)\)
Explanation:
C Given, DC's \(\left(\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right)\) Direction cosine formula \(\mathrm{l}=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-9}{11}\) \(\mathrm{~m}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{6}{11}\) \(\mathrm{n}=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-2}{11}\)Here, \((a, b, c)\) is the direction ratio i.e. \((-9,6,-2)\)
121163
If \(A(3,4,5), B(4,6,3) C(-1,2,4)\) and \(D(1,0,5)\) are such that the angle between the lines \(D C\) and \(A B\) is \(\theta\), then \(\cos \theta\) is equal to
C Given, If a line makes angle \(\alpha\) with \(\mathrm{x}\)-axis, \(\beta\) with \(\mathrm{y}\)-axis \& \(\gamma\) with z-axis. So, direction cosine of the line be\(\ell=\cos \alpha, \quad \mathrm{m}=\cos \beta, \mathrm{n}=\cos \gamma\) Given, \(\alpha=\beta=\gamma \Rightarrow \ell=m=n=\cos \alpha\) We know that: \(\ell^2+\mathrm{m}^2+\mathrm{n}^2=1\) \(\cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1\) \(3 \cos ^2 \alpha=1\) \(\cos ^2 \alpha=\frac{1}{3}\) \(\cos \alpha= \pm \frac{1}{\sqrt{3}}\) Similarly, \(\cos \beta= \pm \frac{1}{\sqrt{3}}, \cos \gamma= \pm \frac{1}{\sqrt{3}}\) So, direction cosine are, \(\ell= \pm \frac{1}{\sqrt{3}}, \mathrm{~m}= \pm \frac{1}{\sqrt{3}}, \mathrm{n}= \pm \frac{1}{\sqrt{3}}\)
AP EAMCET-19.08.2021
Three Dimensional Geometry
121166
The direction ratios of a normal to a plane are \((3,12,4)\). Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?
1 \((2,2,1)\)
2 \((2,1,2)\)
3 \((2,1,1)\)
4 \((3,-1,2)\)
Explanation:
B Let plane be \(a x+b y+c z+d=0\) Direction ratio of normal to the plane \((3,12,4)\) So, equation (i) is- \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z}+\mathrm{d}=0\) Now, distance of (ii) from origin is 2 unit \(\frac{3 \times 0+12 \times 0+4 \times 0+d}{\sqrt{3^2+12^2+4^2}}=2\) \(d= \pm(2 \times 13)= \pm 26\) \(\therefore\) Equation of plane is \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z} \pm 26=0\)This plane is passes through \((2,1,2)\)
J and K CET-2018
Three Dimensional Geometry
121169
The direction cosines of a line are \(\left\langle\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right\rangle\) respectively. Then its direction ratios are
1 \((9,6,-2)\)
2 \((-9,-6,2)\)
3 \((-9,6,-2)\)
4 \((9,-6,-2)\)
Explanation:
C Given, DC's \(\left(\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right)\) Direction cosine formula \(\mathrm{l}=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-9}{11}\) \(\mathrm{~m}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{6}{11}\) \(\mathrm{n}=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-2}{11}\)Here, \((a, b, c)\) is the direction ratio i.e. \((-9,6,-2)\)
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Three Dimensional Geometry
121163
If \(A(3,4,5), B(4,6,3) C(-1,2,4)\) and \(D(1,0,5)\) are such that the angle between the lines \(D C\) and \(A B\) is \(\theta\), then \(\cos \theta\) is equal to
C Given, If a line makes angle \(\alpha\) with \(\mathrm{x}\)-axis, \(\beta\) with \(\mathrm{y}\)-axis \& \(\gamma\) with z-axis. So, direction cosine of the line be\(\ell=\cos \alpha, \quad \mathrm{m}=\cos \beta, \mathrm{n}=\cos \gamma\) Given, \(\alpha=\beta=\gamma \Rightarrow \ell=m=n=\cos \alpha\) We know that: \(\ell^2+\mathrm{m}^2+\mathrm{n}^2=1\) \(\cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1\) \(3 \cos ^2 \alpha=1\) \(\cos ^2 \alpha=\frac{1}{3}\) \(\cos \alpha= \pm \frac{1}{\sqrt{3}}\) Similarly, \(\cos \beta= \pm \frac{1}{\sqrt{3}}, \cos \gamma= \pm \frac{1}{\sqrt{3}}\) So, direction cosine are, \(\ell= \pm \frac{1}{\sqrt{3}}, \mathrm{~m}= \pm \frac{1}{\sqrt{3}}, \mathrm{n}= \pm \frac{1}{\sqrt{3}}\)
AP EAMCET-19.08.2021
Three Dimensional Geometry
121166
The direction ratios of a normal to a plane are \((3,12,4)\). Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?
1 \((2,2,1)\)
2 \((2,1,2)\)
3 \((2,1,1)\)
4 \((3,-1,2)\)
Explanation:
B Let plane be \(a x+b y+c z+d=0\) Direction ratio of normal to the plane \((3,12,4)\) So, equation (i) is- \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z}+\mathrm{d}=0\) Now, distance of (ii) from origin is 2 unit \(\frac{3 \times 0+12 \times 0+4 \times 0+d}{\sqrt{3^2+12^2+4^2}}=2\) \(d= \pm(2 \times 13)= \pm 26\) \(\therefore\) Equation of plane is \(3 \mathrm{x}+12 \mathrm{y}+4 \mathrm{z} \pm 26=0\)This plane is passes through \((2,1,2)\)
J and K CET-2018
Three Dimensional Geometry
121169
The direction cosines of a line are \(\left\langle\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right\rangle\) respectively. Then its direction ratios are
1 \((9,6,-2)\)
2 \((-9,-6,2)\)
3 \((-9,6,-2)\)
4 \((9,-6,-2)\)
Explanation:
C Given, DC's \(\left(\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\right)\) Direction cosine formula \(\mathrm{l}=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-9}{11}\) \(\mathrm{~m}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{6}{11}\) \(\mathrm{n}=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}}=\frac{-2}{11}\)Here, \((a, b, c)\) is the direction ratio i.e. \((-9,6,-2)\)
