121104
The in center of the triangle formed by the points \((0,0,0),(3,0,0),(0,4,0)\) is \(\qquad\)
1 \((1,1,0)\)
2 \((-1,-1,0)\)
3 \((1,0,1)\)
4 \((-1,0,-1)\)
Explanation:
A Let, the co-ordinates of point, \(\mathrm{O}(0,0,0), \mathrm{A}(3,0,0), \mathrm{B}(0,4,0)\) \(\mathrm{OA}=3, \mathrm{OB}=4, \mathrm{AB}=5\) We know that formula of in center - \(x=\frac{a x_1+b x_2+c x_3}{a+b+c}\) \(x=\left(\frac{5(0)+4(3)+3(0)}{3+4+5}\right)\) Similarly - \(y=\left(\frac{5(0)+4(0)+3(4)}{3+4+5}\right)\) \(z=\left(\frac{5(0)+4(0)+3(0)}{3+4+5}\right)\)In center \((\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,1,0)\)
AP EAMCET-23.08.2021
Three Dimensional Geometry
121105
A line passes through the two points \(\mathbf{A}(2,-3\), 1 ) and \(B(8,-1,2)\). The coordinates of a point on this line at a distance of 14 units from \(A\) are
1 \((10,7,7)\)
2 \((14,0,5)\)
3 \((86,25,41)\)
4 None of these
Explanation:
D The equation of line passing through two points A \((2,-3,-1)\) and \(\mathrm{B}(8,-1,2)\) is \(\lambda=\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{z}_2-\mathrm{z}_1}\) \(\Rightarrow \quad \frac{\mathrm{x}-2}{6}=\frac{\mathrm{y}+3}{2}=\frac{\mathrm{z}+1}{3}=\lambda\) \(\Rightarrow \quad \mathrm{x}=6 \lambda+2, \quad \mathrm{y}=2 \lambda-3, \mathrm{z}=3 \lambda-1\) Distance between point A and general point of line. \(14^2=(6 \lambda+2-2)^2+(2 \lambda-3+3)^2+(3 \lambda-1+1)^2\) Solving about equation \(49 \lambda^2=196\) \(\lambda^2=4\) \(\lambda= \pm 2\) \(\because \quad \mathrm{x}=6 \lambda+2=14 \text { or }-10\) \(\mathrm{y} =2 \lambda-3=1 \text { or }-7\) \(\mathrm{z}=3 \lambda-1=5 \text { or }-7\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121106
The projection of a line on a co-ordinate axes are \(2,3,6\). Then the length of the line is
1 7
2 5
3 1
4 11
Explanation:
A \((x, y, z) \equiv(2,3,6)\) \(\text { Length of line segment } =\sqrt{x^2+y^2+z^2}\) \(=\sqrt{2^2+3^2+6^2}\) \(=\sqrt{4+9+36}\) \(=\sqrt{49}\) \(=7\)
Manipal UGET-2019
Three Dimensional Geometry
121107
If \(a, b, c\) are the position vectors of \(A, B, C\) respectively such that \(3 a+4 b-7 c=0\) then \(C\) divides \(A B\) in the ratio
1 \(4: 3\)
2 \(4: 7\)
3 \(3: 7\)
4 \(3: 4\)
Explanation:
A Given, Let \(\mathrm{C}\) divides \(\mathrm{AB}\) in the ratio \(\mathrm{m}: 1\). \(\Rightarrow \mathrm{c}=\frac{\mathrm{mb}+\mathrm{a}}{\mathrm{m}+1}\) \(\Rightarrow \mathrm{mc}+\mathrm{c}=\mathrm{mb}+\mathrm{a}\) \(\Rightarrow \mathrm{a}+\mathrm{mb}-\mathrm{mc}-\mathrm{c}=0\) \(3 \mathrm{a}+4 \mathrm{~b}-7 \mathrm{c}=0\) On comparing equation (i) and (ii), we get \(\Rightarrow \quad \mathrm{m}=4 / 3\) \(\therefore\) Required ratio is \(4: 3\)
121104
The in center of the triangle formed by the points \((0,0,0),(3,0,0),(0,4,0)\) is \(\qquad\)
1 \((1,1,0)\)
2 \((-1,-1,0)\)
3 \((1,0,1)\)
4 \((-1,0,-1)\)
Explanation:
A Let, the co-ordinates of point, \(\mathrm{O}(0,0,0), \mathrm{A}(3,0,0), \mathrm{B}(0,4,0)\) \(\mathrm{OA}=3, \mathrm{OB}=4, \mathrm{AB}=5\) We know that formula of in center - \(x=\frac{a x_1+b x_2+c x_3}{a+b+c}\) \(x=\left(\frac{5(0)+4(3)+3(0)}{3+4+5}\right)\) Similarly - \(y=\left(\frac{5(0)+4(0)+3(4)}{3+4+5}\right)\) \(z=\left(\frac{5(0)+4(0)+3(0)}{3+4+5}\right)\)In center \((\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,1,0)\)
AP EAMCET-23.08.2021
Three Dimensional Geometry
121105
A line passes through the two points \(\mathbf{A}(2,-3\), 1 ) and \(B(8,-1,2)\). The coordinates of a point on this line at a distance of 14 units from \(A\) are
1 \((10,7,7)\)
2 \((14,0,5)\)
3 \((86,25,41)\)
4 None of these
Explanation:
D The equation of line passing through two points A \((2,-3,-1)\) and \(\mathrm{B}(8,-1,2)\) is \(\lambda=\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{z}_2-\mathrm{z}_1}\) \(\Rightarrow \quad \frac{\mathrm{x}-2}{6}=\frac{\mathrm{y}+3}{2}=\frac{\mathrm{z}+1}{3}=\lambda\) \(\Rightarrow \quad \mathrm{x}=6 \lambda+2, \quad \mathrm{y}=2 \lambda-3, \mathrm{z}=3 \lambda-1\) Distance between point A and general point of line. \(14^2=(6 \lambda+2-2)^2+(2 \lambda-3+3)^2+(3 \lambda-1+1)^2\) Solving about equation \(49 \lambda^2=196\) \(\lambda^2=4\) \(\lambda= \pm 2\) \(\because \quad \mathrm{x}=6 \lambda+2=14 \text { or }-10\) \(\mathrm{y} =2 \lambda-3=1 \text { or }-7\) \(\mathrm{z}=3 \lambda-1=5 \text { or }-7\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121106
The projection of a line on a co-ordinate axes are \(2,3,6\). Then the length of the line is
1 7
2 5
3 1
4 11
Explanation:
A \((x, y, z) \equiv(2,3,6)\) \(\text { Length of line segment } =\sqrt{x^2+y^2+z^2}\) \(=\sqrt{2^2+3^2+6^2}\) \(=\sqrt{4+9+36}\) \(=\sqrt{49}\) \(=7\)
Manipal UGET-2019
Three Dimensional Geometry
121107
If \(a, b, c\) are the position vectors of \(A, B, C\) respectively such that \(3 a+4 b-7 c=0\) then \(C\) divides \(A B\) in the ratio
1 \(4: 3\)
2 \(4: 7\)
3 \(3: 7\)
4 \(3: 4\)
Explanation:
A Given, Let \(\mathrm{C}\) divides \(\mathrm{AB}\) in the ratio \(\mathrm{m}: 1\). \(\Rightarrow \mathrm{c}=\frac{\mathrm{mb}+\mathrm{a}}{\mathrm{m}+1}\) \(\Rightarrow \mathrm{mc}+\mathrm{c}=\mathrm{mb}+\mathrm{a}\) \(\Rightarrow \mathrm{a}+\mathrm{mb}-\mathrm{mc}-\mathrm{c}=0\) \(3 \mathrm{a}+4 \mathrm{~b}-7 \mathrm{c}=0\) On comparing equation (i) and (ii), we get \(\Rightarrow \quad \mathrm{m}=4 / 3\) \(\therefore\) Required ratio is \(4: 3\)
121104
The in center of the triangle formed by the points \((0,0,0),(3,0,0),(0,4,0)\) is \(\qquad\)
1 \((1,1,0)\)
2 \((-1,-1,0)\)
3 \((1,0,1)\)
4 \((-1,0,-1)\)
Explanation:
A Let, the co-ordinates of point, \(\mathrm{O}(0,0,0), \mathrm{A}(3,0,0), \mathrm{B}(0,4,0)\) \(\mathrm{OA}=3, \mathrm{OB}=4, \mathrm{AB}=5\) We know that formula of in center - \(x=\frac{a x_1+b x_2+c x_3}{a+b+c}\) \(x=\left(\frac{5(0)+4(3)+3(0)}{3+4+5}\right)\) Similarly - \(y=\left(\frac{5(0)+4(0)+3(4)}{3+4+5}\right)\) \(z=\left(\frac{5(0)+4(0)+3(0)}{3+4+5}\right)\)In center \((\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,1,0)\)
AP EAMCET-23.08.2021
Three Dimensional Geometry
121105
A line passes through the two points \(\mathbf{A}(2,-3\), 1 ) and \(B(8,-1,2)\). The coordinates of a point on this line at a distance of 14 units from \(A\) are
1 \((10,7,7)\)
2 \((14,0,5)\)
3 \((86,25,41)\)
4 None of these
Explanation:
D The equation of line passing through two points A \((2,-3,-1)\) and \(\mathrm{B}(8,-1,2)\) is \(\lambda=\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{z}_2-\mathrm{z}_1}\) \(\Rightarrow \quad \frac{\mathrm{x}-2}{6}=\frac{\mathrm{y}+3}{2}=\frac{\mathrm{z}+1}{3}=\lambda\) \(\Rightarrow \quad \mathrm{x}=6 \lambda+2, \quad \mathrm{y}=2 \lambda-3, \mathrm{z}=3 \lambda-1\) Distance between point A and general point of line. \(14^2=(6 \lambda+2-2)^2+(2 \lambda-3+3)^2+(3 \lambda-1+1)^2\) Solving about equation \(49 \lambda^2=196\) \(\lambda^2=4\) \(\lambda= \pm 2\) \(\because \quad \mathrm{x}=6 \lambda+2=14 \text { or }-10\) \(\mathrm{y} =2 \lambda-3=1 \text { or }-7\) \(\mathrm{z}=3 \lambda-1=5 \text { or }-7\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121106
The projection of a line on a co-ordinate axes are \(2,3,6\). Then the length of the line is
1 7
2 5
3 1
4 11
Explanation:
A \((x, y, z) \equiv(2,3,6)\) \(\text { Length of line segment } =\sqrt{x^2+y^2+z^2}\) \(=\sqrt{2^2+3^2+6^2}\) \(=\sqrt{4+9+36}\) \(=\sqrt{49}\) \(=7\)
Manipal UGET-2019
Three Dimensional Geometry
121107
If \(a, b, c\) are the position vectors of \(A, B, C\) respectively such that \(3 a+4 b-7 c=0\) then \(C\) divides \(A B\) in the ratio
1 \(4: 3\)
2 \(4: 7\)
3 \(3: 7\)
4 \(3: 4\)
Explanation:
A Given, Let \(\mathrm{C}\) divides \(\mathrm{AB}\) in the ratio \(\mathrm{m}: 1\). \(\Rightarrow \mathrm{c}=\frac{\mathrm{mb}+\mathrm{a}}{\mathrm{m}+1}\) \(\Rightarrow \mathrm{mc}+\mathrm{c}=\mathrm{mb}+\mathrm{a}\) \(\Rightarrow \mathrm{a}+\mathrm{mb}-\mathrm{mc}-\mathrm{c}=0\) \(3 \mathrm{a}+4 \mathrm{~b}-7 \mathrm{c}=0\) On comparing equation (i) and (ii), we get \(\Rightarrow \quad \mathrm{m}=4 / 3\) \(\therefore\) Required ratio is \(4: 3\)
121104
The in center of the triangle formed by the points \((0,0,0),(3,0,0),(0,4,0)\) is \(\qquad\)
1 \((1,1,0)\)
2 \((-1,-1,0)\)
3 \((1,0,1)\)
4 \((-1,0,-1)\)
Explanation:
A Let, the co-ordinates of point, \(\mathrm{O}(0,0,0), \mathrm{A}(3,0,0), \mathrm{B}(0,4,0)\) \(\mathrm{OA}=3, \mathrm{OB}=4, \mathrm{AB}=5\) We know that formula of in center - \(x=\frac{a x_1+b x_2+c x_3}{a+b+c}\) \(x=\left(\frac{5(0)+4(3)+3(0)}{3+4+5}\right)\) Similarly - \(y=\left(\frac{5(0)+4(0)+3(4)}{3+4+5}\right)\) \(z=\left(\frac{5(0)+4(0)+3(0)}{3+4+5}\right)\)In center \((\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,1,0)\)
AP EAMCET-23.08.2021
Three Dimensional Geometry
121105
A line passes through the two points \(\mathbf{A}(2,-3\), 1 ) and \(B(8,-1,2)\). The coordinates of a point on this line at a distance of 14 units from \(A\) are
1 \((10,7,7)\)
2 \((14,0,5)\)
3 \((86,25,41)\)
4 None of these
Explanation:
D The equation of line passing through two points A \((2,-3,-1)\) and \(\mathrm{B}(8,-1,2)\) is \(\lambda=\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{z}_2-\mathrm{z}_1}\) \(\Rightarrow \quad \frac{\mathrm{x}-2}{6}=\frac{\mathrm{y}+3}{2}=\frac{\mathrm{z}+1}{3}=\lambda\) \(\Rightarrow \quad \mathrm{x}=6 \lambda+2, \quad \mathrm{y}=2 \lambda-3, \mathrm{z}=3 \lambda-1\) Distance between point A and general point of line. \(14^2=(6 \lambda+2-2)^2+(2 \lambda-3+3)^2+(3 \lambda-1+1)^2\) Solving about equation \(49 \lambda^2=196\) \(\lambda^2=4\) \(\lambda= \pm 2\) \(\because \quad \mathrm{x}=6 \lambda+2=14 \text { or }-10\) \(\mathrm{y} =2 \lambda-3=1 \text { or }-7\) \(\mathrm{z}=3 \lambda-1=5 \text { or }-7\)
Jamia Millia Islamia-2013
Three Dimensional Geometry
121106
The projection of a line on a co-ordinate axes are \(2,3,6\). Then the length of the line is
1 7
2 5
3 1
4 11
Explanation:
A \((x, y, z) \equiv(2,3,6)\) \(\text { Length of line segment } =\sqrt{x^2+y^2+z^2}\) \(=\sqrt{2^2+3^2+6^2}\) \(=\sqrt{4+9+36}\) \(=\sqrt{49}\) \(=7\)
Manipal UGET-2019
Three Dimensional Geometry
121107
If \(a, b, c\) are the position vectors of \(A, B, C\) respectively such that \(3 a+4 b-7 c=0\) then \(C\) divides \(A B\) in the ratio
1 \(4: 3\)
2 \(4: 7\)
3 \(3: 7\)
4 \(3: 4\)
Explanation:
A Given, Let \(\mathrm{C}\) divides \(\mathrm{AB}\) in the ratio \(\mathrm{m}: 1\). \(\Rightarrow \mathrm{c}=\frac{\mathrm{mb}+\mathrm{a}}{\mathrm{m}+1}\) \(\Rightarrow \mathrm{mc}+\mathrm{c}=\mathrm{mb}+\mathrm{a}\) \(\Rightarrow \mathrm{a}+\mathrm{mb}-\mathrm{mc}-\mathrm{c}=0\) \(3 \mathrm{a}+4 \mathrm{~b}-7 \mathrm{c}=0\) On comparing equation (i) and (ii), we get \(\Rightarrow \quad \mathrm{m}=4 / 3\) \(\therefore\) Required ratio is \(4: 3\)
