121080
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane \(5 y+8=0\).
1 \(\left(0,-\frac{18}{5}, 2\right)\)
2 \(\left(0, \frac{8}{5}, 2\right)\)
3 \(\left(\frac{8}{25}, 0,0\right)\)
4 \(\left(0,-\frac{8}{5}, 0\right)\)
Explanation:
D Given- Origin \(=(0,0)\) Plane \(=5 \mathrm{y}+8=0\) Direction ratio of the plane \(=(0,5,0)\) \(\frac{\mathrm{x}-0}{0}=\frac{\mathrm{y}-0}{5}=\frac{\mathrm{z}-0}{0}=\lambda\) Then, Point Q lies on the plane, so coordinates of the point \(\mathrm{Q}\) Satisfies the equation of the plane. \(\mathrm{Q}(0,5 \lambda, 0)\) \(5 \mathrm{y}+8=0\) \(5(5 \lambda)+8=0\) \(25 \lambda+8=0\) \(\lambda=\frac{-8}{25}\) Now, point \(\mathrm{Q}=\left[0,5 \times\left(\frac{-8}{25}\right), 0\right]\) \(=\left(0, \frac{-8}{5}, 0\right)\)
Karnataka CET-2016
Three Dimensional Geometry
121081
The distance of the point \(P(a, b, c)\) from the \(x-\) axis is
1 \(\sqrt{a^2+b^2}\)
2 \(\sqrt{\mathrm{b}^2+\mathrm{c}^2}\)
3 a
4 \(\sqrt{a^2+c^2}\)
Explanation:
B Given- Point \((a, b, c)\) is perpendicular on \(x\)-axis \((a, 0,0)\) Distance \(P(a, b, c)\) from \(x\)-axis is \(\sqrt{(a-a)^2+(b-0)^2+(c-0)^2}\) \(=\sqrt{0+b^2+c^2}\) \(=\sqrt{b^2+c^2}\)
Karnataka CET-2014
Three Dimensional Geometry
121082
Find the coordinates of the point where the line joining the points \((2,-3,1)\) and \((3,-4,-5)\) cuts the plane \(2 x+y+z=7\).
1 \((1,2,-7)\)
2 \((1,-2,7)\)
3 \((-1,-2,7)\)
4 \((1,2,7)\)
Explanation:
B Given, The two points are \(\mathrm{P}(2,-3,1)\) and \(\mathrm{Q}(3,-4,-5)\). Thus the direction ratios of the line joining the points \(P\) and \(\mathrm{Q}\) \(=\left(\mathrm{x}_2-\mathrm{x}_1\right),\left(\mathrm{y}_2-\mathrm{y}_1\right),\left(\mathrm{z}_2-\mathrm{z}_1\right)\) \(=(3-2),(-4-(-3)),(-5-1)\) \(=(1,-1,-6)\) With the above direction ratios line passes through (2, \(-3,1)\), thus equation of the \(P Q\) will be given as- \(\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+3}{-1}=\frac{\mathrm{z}-1}{-6}=\mathrm{r}\) Any Co-ordinates on the line \(P Q\) will be in the form of- \(x=r+2, y=-r-3, z=-6 r+1\) and the equation of plane \(2 x+y+z=7\) Then, \(2(r+2)+(-r-3)+(-6 r+1)=7\) \(2 r+4-r-3-6 r+1=7\) \(-5 r+2=7\) \(-5 r=7-2=5\) \(r=-\frac{5}{5}\) \(r=-1\) Therefore, coordinates of the required points are \((-1+2),(-(-1)-3),(-6(-1)+1)=(1,-2,7)\)
BITSAT-2012
Three Dimensional Geometry
121083
The coordinates of the point where the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses the \(\mathrm{XY}\)-plane are
A Given- Points, \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Direction ratio of line, \(A B\) \(\mathrm{AB}=[(5-3),(1-4),(6-1)]=(2,-3,5)\) Then equation of line becomes, \(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=r\) \(\Rightarrow \quad x=2 r+3, y=-3 r+4\) \(\text { and } \quad z=5 r+1\) Line is crossing the \(\mathrm{X}-\mathrm{Y}\) plane, so \(\mathrm{Z}\) - coordinate will be 0 . \(\Rightarrow 5 \mathrm{r}+1=0\) \(\Rightarrow \mathrm{r}=-\frac{1}{5}\) \(\therefore \quad \frac{\mathrm{x}-3}{2}=-\frac{1}{5} \Rightarrow \mathrm{x}=3-\frac{2}{5}=\frac{13}{5}\) Similarly, \(\frac{y-4}{-3}=\frac{-1}{5} \Rightarrow y=4+\frac{3}{5}=\frac{23}{5}\) Therefore required coordinates will be \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
BITSAT-2013
Three Dimensional Geometry
121084
If the plane \(2 x+3 y+5 z=1\) intersect the coordinate axes at the points \(A, B, C\) then the centroid of \(\triangle \mathrm{ABC}\) is
D Given, Plane \(2 x+3 y+5 z=1\) intersects the coordinate axis at the point \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) in \(\triangle \mathrm{ABC}\). \(\therefore\) Equation of plane in intercepted form- \(\frac{\mathrm{x}}{\frac{1}{2}}+\frac{\mathrm{y}}{\frac{1}{3}}+\frac{\mathrm{z}}{5}=1\) \(\therefore \quad \mathrm{A}\left(\frac{1}{2}, 0,0\right), \mathrm{B}\left(0, \frac{1}{3}, 0\right), \mathrm{C}\left(0,0, \frac{1}{5}\right),\)Therefore, centroid of \(\triangle \mathrm{ABC}\) is \(\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)\).
121080
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane \(5 y+8=0\).
1 \(\left(0,-\frac{18}{5}, 2\right)\)
2 \(\left(0, \frac{8}{5}, 2\right)\)
3 \(\left(\frac{8}{25}, 0,0\right)\)
4 \(\left(0,-\frac{8}{5}, 0\right)\)
Explanation:
D Given- Origin \(=(0,0)\) Plane \(=5 \mathrm{y}+8=0\) Direction ratio of the plane \(=(0,5,0)\) \(\frac{\mathrm{x}-0}{0}=\frac{\mathrm{y}-0}{5}=\frac{\mathrm{z}-0}{0}=\lambda\) Then, Point Q lies on the plane, so coordinates of the point \(\mathrm{Q}\) Satisfies the equation of the plane. \(\mathrm{Q}(0,5 \lambda, 0)\) \(5 \mathrm{y}+8=0\) \(5(5 \lambda)+8=0\) \(25 \lambda+8=0\) \(\lambda=\frac{-8}{25}\) Now, point \(\mathrm{Q}=\left[0,5 \times\left(\frac{-8}{25}\right), 0\right]\) \(=\left(0, \frac{-8}{5}, 0\right)\)
Karnataka CET-2016
Three Dimensional Geometry
121081
The distance of the point \(P(a, b, c)\) from the \(x-\) axis is
1 \(\sqrt{a^2+b^2}\)
2 \(\sqrt{\mathrm{b}^2+\mathrm{c}^2}\)
3 a
4 \(\sqrt{a^2+c^2}\)
Explanation:
B Given- Point \((a, b, c)\) is perpendicular on \(x\)-axis \((a, 0,0)\) Distance \(P(a, b, c)\) from \(x\)-axis is \(\sqrt{(a-a)^2+(b-0)^2+(c-0)^2}\) \(=\sqrt{0+b^2+c^2}\) \(=\sqrt{b^2+c^2}\)
Karnataka CET-2014
Three Dimensional Geometry
121082
Find the coordinates of the point where the line joining the points \((2,-3,1)\) and \((3,-4,-5)\) cuts the plane \(2 x+y+z=7\).
1 \((1,2,-7)\)
2 \((1,-2,7)\)
3 \((-1,-2,7)\)
4 \((1,2,7)\)
Explanation:
B Given, The two points are \(\mathrm{P}(2,-3,1)\) and \(\mathrm{Q}(3,-4,-5)\). Thus the direction ratios of the line joining the points \(P\) and \(\mathrm{Q}\) \(=\left(\mathrm{x}_2-\mathrm{x}_1\right),\left(\mathrm{y}_2-\mathrm{y}_1\right),\left(\mathrm{z}_2-\mathrm{z}_1\right)\) \(=(3-2),(-4-(-3)),(-5-1)\) \(=(1,-1,-6)\) With the above direction ratios line passes through (2, \(-3,1)\), thus equation of the \(P Q\) will be given as- \(\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+3}{-1}=\frac{\mathrm{z}-1}{-6}=\mathrm{r}\) Any Co-ordinates on the line \(P Q\) will be in the form of- \(x=r+2, y=-r-3, z=-6 r+1\) and the equation of plane \(2 x+y+z=7\) Then, \(2(r+2)+(-r-3)+(-6 r+1)=7\) \(2 r+4-r-3-6 r+1=7\) \(-5 r+2=7\) \(-5 r=7-2=5\) \(r=-\frac{5}{5}\) \(r=-1\) Therefore, coordinates of the required points are \((-1+2),(-(-1)-3),(-6(-1)+1)=(1,-2,7)\)
BITSAT-2012
Three Dimensional Geometry
121083
The coordinates of the point where the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses the \(\mathrm{XY}\)-plane are
A Given- Points, \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Direction ratio of line, \(A B\) \(\mathrm{AB}=[(5-3),(1-4),(6-1)]=(2,-3,5)\) Then equation of line becomes, \(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=r\) \(\Rightarrow \quad x=2 r+3, y=-3 r+4\) \(\text { and } \quad z=5 r+1\) Line is crossing the \(\mathrm{X}-\mathrm{Y}\) plane, so \(\mathrm{Z}\) - coordinate will be 0 . \(\Rightarrow 5 \mathrm{r}+1=0\) \(\Rightarrow \mathrm{r}=-\frac{1}{5}\) \(\therefore \quad \frac{\mathrm{x}-3}{2}=-\frac{1}{5} \Rightarrow \mathrm{x}=3-\frac{2}{5}=\frac{13}{5}\) Similarly, \(\frac{y-4}{-3}=\frac{-1}{5} \Rightarrow y=4+\frac{3}{5}=\frac{23}{5}\) Therefore required coordinates will be \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
BITSAT-2013
Three Dimensional Geometry
121084
If the plane \(2 x+3 y+5 z=1\) intersect the coordinate axes at the points \(A, B, C\) then the centroid of \(\triangle \mathrm{ABC}\) is
D Given, Plane \(2 x+3 y+5 z=1\) intersects the coordinate axis at the point \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) in \(\triangle \mathrm{ABC}\). \(\therefore\) Equation of plane in intercepted form- \(\frac{\mathrm{x}}{\frac{1}{2}}+\frac{\mathrm{y}}{\frac{1}{3}}+\frac{\mathrm{z}}{5}=1\) \(\therefore \quad \mathrm{A}\left(\frac{1}{2}, 0,0\right), \mathrm{B}\left(0, \frac{1}{3}, 0\right), \mathrm{C}\left(0,0, \frac{1}{5}\right),\)Therefore, centroid of \(\triangle \mathrm{ABC}\) is \(\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)\).
121080
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane \(5 y+8=0\).
1 \(\left(0,-\frac{18}{5}, 2\right)\)
2 \(\left(0, \frac{8}{5}, 2\right)\)
3 \(\left(\frac{8}{25}, 0,0\right)\)
4 \(\left(0,-\frac{8}{5}, 0\right)\)
Explanation:
D Given- Origin \(=(0,0)\) Plane \(=5 \mathrm{y}+8=0\) Direction ratio of the plane \(=(0,5,0)\) \(\frac{\mathrm{x}-0}{0}=\frac{\mathrm{y}-0}{5}=\frac{\mathrm{z}-0}{0}=\lambda\) Then, Point Q lies on the plane, so coordinates of the point \(\mathrm{Q}\) Satisfies the equation of the plane. \(\mathrm{Q}(0,5 \lambda, 0)\) \(5 \mathrm{y}+8=0\) \(5(5 \lambda)+8=0\) \(25 \lambda+8=0\) \(\lambda=\frac{-8}{25}\) Now, point \(\mathrm{Q}=\left[0,5 \times\left(\frac{-8}{25}\right), 0\right]\) \(=\left(0, \frac{-8}{5}, 0\right)\)
Karnataka CET-2016
Three Dimensional Geometry
121081
The distance of the point \(P(a, b, c)\) from the \(x-\) axis is
1 \(\sqrt{a^2+b^2}\)
2 \(\sqrt{\mathrm{b}^2+\mathrm{c}^2}\)
3 a
4 \(\sqrt{a^2+c^2}\)
Explanation:
B Given- Point \((a, b, c)\) is perpendicular on \(x\)-axis \((a, 0,0)\) Distance \(P(a, b, c)\) from \(x\)-axis is \(\sqrt{(a-a)^2+(b-0)^2+(c-0)^2}\) \(=\sqrt{0+b^2+c^2}\) \(=\sqrt{b^2+c^2}\)
Karnataka CET-2014
Three Dimensional Geometry
121082
Find the coordinates of the point where the line joining the points \((2,-3,1)\) and \((3,-4,-5)\) cuts the plane \(2 x+y+z=7\).
1 \((1,2,-7)\)
2 \((1,-2,7)\)
3 \((-1,-2,7)\)
4 \((1,2,7)\)
Explanation:
B Given, The two points are \(\mathrm{P}(2,-3,1)\) and \(\mathrm{Q}(3,-4,-5)\). Thus the direction ratios of the line joining the points \(P\) and \(\mathrm{Q}\) \(=\left(\mathrm{x}_2-\mathrm{x}_1\right),\left(\mathrm{y}_2-\mathrm{y}_1\right),\left(\mathrm{z}_2-\mathrm{z}_1\right)\) \(=(3-2),(-4-(-3)),(-5-1)\) \(=(1,-1,-6)\) With the above direction ratios line passes through (2, \(-3,1)\), thus equation of the \(P Q\) will be given as- \(\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+3}{-1}=\frac{\mathrm{z}-1}{-6}=\mathrm{r}\) Any Co-ordinates on the line \(P Q\) will be in the form of- \(x=r+2, y=-r-3, z=-6 r+1\) and the equation of plane \(2 x+y+z=7\) Then, \(2(r+2)+(-r-3)+(-6 r+1)=7\) \(2 r+4-r-3-6 r+1=7\) \(-5 r+2=7\) \(-5 r=7-2=5\) \(r=-\frac{5}{5}\) \(r=-1\) Therefore, coordinates of the required points are \((-1+2),(-(-1)-3),(-6(-1)+1)=(1,-2,7)\)
BITSAT-2012
Three Dimensional Geometry
121083
The coordinates of the point where the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses the \(\mathrm{XY}\)-plane are
A Given- Points, \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Direction ratio of line, \(A B\) \(\mathrm{AB}=[(5-3),(1-4),(6-1)]=(2,-3,5)\) Then equation of line becomes, \(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=r\) \(\Rightarrow \quad x=2 r+3, y=-3 r+4\) \(\text { and } \quad z=5 r+1\) Line is crossing the \(\mathrm{X}-\mathrm{Y}\) plane, so \(\mathrm{Z}\) - coordinate will be 0 . \(\Rightarrow 5 \mathrm{r}+1=0\) \(\Rightarrow \mathrm{r}=-\frac{1}{5}\) \(\therefore \quad \frac{\mathrm{x}-3}{2}=-\frac{1}{5} \Rightarrow \mathrm{x}=3-\frac{2}{5}=\frac{13}{5}\) Similarly, \(\frac{y-4}{-3}=\frac{-1}{5} \Rightarrow y=4+\frac{3}{5}=\frac{23}{5}\) Therefore required coordinates will be \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
BITSAT-2013
Three Dimensional Geometry
121084
If the plane \(2 x+3 y+5 z=1\) intersect the coordinate axes at the points \(A, B, C\) then the centroid of \(\triangle \mathrm{ABC}\) is
D Given, Plane \(2 x+3 y+5 z=1\) intersects the coordinate axis at the point \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) in \(\triangle \mathrm{ABC}\). \(\therefore\) Equation of plane in intercepted form- \(\frac{\mathrm{x}}{\frac{1}{2}}+\frac{\mathrm{y}}{\frac{1}{3}}+\frac{\mathrm{z}}{5}=1\) \(\therefore \quad \mathrm{A}\left(\frac{1}{2}, 0,0\right), \mathrm{B}\left(0, \frac{1}{3}, 0\right), \mathrm{C}\left(0,0, \frac{1}{5}\right),\)Therefore, centroid of \(\triangle \mathrm{ABC}\) is \(\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)\).
121080
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane \(5 y+8=0\).
1 \(\left(0,-\frac{18}{5}, 2\right)\)
2 \(\left(0, \frac{8}{5}, 2\right)\)
3 \(\left(\frac{8}{25}, 0,0\right)\)
4 \(\left(0,-\frac{8}{5}, 0\right)\)
Explanation:
D Given- Origin \(=(0,0)\) Plane \(=5 \mathrm{y}+8=0\) Direction ratio of the plane \(=(0,5,0)\) \(\frac{\mathrm{x}-0}{0}=\frac{\mathrm{y}-0}{5}=\frac{\mathrm{z}-0}{0}=\lambda\) Then, Point Q lies on the plane, so coordinates of the point \(\mathrm{Q}\) Satisfies the equation of the plane. \(\mathrm{Q}(0,5 \lambda, 0)\) \(5 \mathrm{y}+8=0\) \(5(5 \lambda)+8=0\) \(25 \lambda+8=0\) \(\lambda=\frac{-8}{25}\) Now, point \(\mathrm{Q}=\left[0,5 \times\left(\frac{-8}{25}\right), 0\right]\) \(=\left(0, \frac{-8}{5}, 0\right)\)
Karnataka CET-2016
Three Dimensional Geometry
121081
The distance of the point \(P(a, b, c)\) from the \(x-\) axis is
1 \(\sqrt{a^2+b^2}\)
2 \(\sqrt{\mathrm{b}^2+\mathrm{c}^2}\)
3 a
4 \(\sqrt{a^2+c^2}\)
Explanation:
B Given- Point \((a, b, c)\) is perpendicular on \(x\)-axis \((a, 0,0)\) Distance \(P(a, b, c)\) from \(x\)-axis is \(\sqrt{(a-a)^2+(b-0)^2+(c-0)^2}\) \(=\sqrt{0+b^2+c^2}\) \(=\sqrt{b^2+c^2}\)
Karnataka CET-2014
Three Dimensional Geometry
121082
Find the coordinates of the point where the line joining the points \((2,-3,1)\) and \((3,-4,-5)\) cuts the plane \(2 x+y+z=7\).
1 \((1,2,-7)\)
2 \((1,-2,7)\)
3 \((-1,-2,7)\)
4 \((1,2,7)\)
Explanation:
B Given, The two points are \(\mathrm{P}(2,-3,1)\) and \(\mathrm{Q}(3,-4,-5)\). Thus the direction ratios of the line joining the points \(P\) and \(\mathrm{Q}\) \(=\left(\mathrm{x}_2-\mathrm{x}_1\right),\left(\mathrm{y}_2-\mathrm{y}_1\right),\left(\mathrm{z}_2-\mathrm{z}_1\right)\) \(=(3-2),(-4-(-3)),(-5-1)\) \(=(1,-1,-6)\) With the above direction ratios line passes through (2, \(-3,1)\), thus equation of the \(P Q\) will be given as- \(\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+3}{-1}=\frac{\mathrm{z}-1}{-6}=\mathrm{r}\) Any Co-ordinates on the line \(P Q\) will be in the form of- \(x=r+2, y=-r-3, z=-6 r+1\) and the equation of plane \(2 x+y+z=7\) Then, \(2(r+2)+(-r-3)+(-6 r+1)=7\) \(2 r+4-r-3-6 r+1=7\) \(-5 r+2=7\) \(-5 r=7-2=5\) \(r=-\frac{5}{5}\) \(r=-1\) Therefore, coordinates of the required points are \((-1+2),(-(-1)-3),(-6(-1)+1)=(1,-2,7)\)
BITSAT-2012
Three Dimensional Geometry
121083
The coordinates of the point where the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses the \(\mathrm{XY}\)-plane are
A Given- Points, \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Direction ratio of line, \(A B\) \(\mathrm{AB}=[(5-3),(1-4),(6-1)]=(2,-3,5)\) Then equation of line becomes, \(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=r\) \(\Rightarrow \quad x=2 r+3, y=-3 r+4\) \(\text { and } \quad z=5 r+1\) Line is crossing the \(\mathrm{X}-\mathrm{Y}\) plane, so \(\mathrm{Z}\) - coordinate will be 0 . \(\Rightarrow 5 \mathrm{r}+1=0\) \(\Rightarrow \mathrm{r}=-\frac{1}{5}\) \(\therefore \quad \frac{\mathrm{x}-3}{2}=-\frac{1}{5} \Rightarrow \mathrm{x}=3-\frac{2}{5}=\frac{13}{5}\) Similarly, \(\frac{y-4}{-3}=\frac{-1}{5} \Rightarrow y=4+\frac{3}{5}=\frac{23}{5}\) Therefore required coordinates will be \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
BITSAT-2013
Three Dimensional Geometry
121084
If the plane \(2 x+3 y+5 z=1\) intersect the coordinate axes at the points \(A, B, C\) then the centroid of \(\triangle \mathrm{ABC}\) is
D Given, Plane \(2 x+3 y+5 z=1\) intersects the coordinate axis at the point \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) in \(\triangle \mathrm{ABC}\). \(\therefore\) Equation of plane in intercepted form- \(\frac{\mathrm{x}}{\frac{1}{2}}+\frac{\mathrm{y}}{\frac{1}{3}}+\frac{\mathrm{z}}{5}=1\) \(\therefore \quad \mathrm{A}\left(\frac{1}{2}, 0,0\right), \mathrm{B}\left(0, \frac{1}{3}, 0\right), \mathrm{C}\left(0,0, \frac{1}{5}\right),\)Therefore, centroid of \(\triangle \mathrm{ABC}\) is \(\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)\).
121080
Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane \(5 y+8=0\).
1 \(\left(0,-\frac{18}{5}, 2\right)\)
2 \(\left(0, \frac{8}{5}, 2\right)\)
3 \(\left(\frac{8}{25}, 0,0\right)\)
4 \(\left(0,-\frac{8}{5}, 0\right)\)
Explanation:
D Given- Origin \(=(0,0)\) Plane \(=5 \mathrm{y}+8=0\) Direction ratio of the plane \(=(0,5,0)\) \(\frac{\mathrm{x}-0}{0}=\frac{\mathrm{y}-0}{5}=\frac{\mathrm{z}-0}{0}=\lambda\) Then, Point Q lies on the plane, so coordinates of the point \(\mathrm{Q}\) Satisfies the equation of the plane. \(\mathrm{Q}(0,5 \lambda, 0)\) \(5 \mathrm{y}+8=0\) \(5(5 \lambda)+8=0\) \(25 \lambda+8=0\) \(\lambda=\frac{-8}{25}\) Now, point \(\mathrm{Q}=\left[0,5 \times\left(\frac{-8}{25}\right), 0\right]\) \(=\left(0, \frac{-8}{5}, 0\right)\)
Karnataka CET-2016
Three Dimensional Geometry
121081
The distance of the point \(P(a, b, c)\) from the \(x-\) axis is
1 \(\sqrt{a^2+b^2}\)
2 \(\sqrt{\mathrm{b}^2+\mathrm{c}^2}\)
3 a
4 \(\sqrt{a^2+c^2}\)
Explanation:
B Given- Point \((a, b, c)\) is perpendicular on \(x\)-axis \((a, 0,0)\) Distance \(P(a, b, c)\) from \(x\)-axis is \(\sqrt{(a-a)^2+(b-0)^2+(c-0)^2}\) \(=\sqrt{0+b^2+c^2}\) \(=\sqrt{b^2+c^2}\)
Karnataka CET-2014
Three Dimensional Geometry
121082
Find the coordinates of the point where the line joining the points \((2,-3,1)\) and \((3,-4,-5)\) cuts the plane \(2 x+y+z=7\).
1 \((1,2,-7)\)
2 \((1,-2,7)\)
3 \((-1,-2,7)\)
4 \((1,2,7)\)
Explanation:
B Given, The two points are \(\mathrm{P}(2,-3,1)\) and \(\mathrm{Q}(3,-4,-5)\). Thus the direction ratios of the line joining the points \(P\) and \(\mathrm{Q}\) \(=\left(\mathrm{x}_2-\mathrm{x}_1\right),\left(\mathrm{y}_2-\mathrm{y}_1\right),\left(\mathrm{z}_2-\mathrm{z}_1\right)\) \(=(3-2),(-4-(-3)),(-5-1)\) \(=(1,-1,-6)\) With the above direction ratios line passes through (2, \(-3,1)\), thus equation of the \(P Q\) will be given as- \(\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+3}{-1}=\frac{\mathrm{z}-1}{-6}=\mathrm{r}\) Any Co-ordinates on the line \(P Q\) will be in the form of- \(x=r+2, y=-r-3, z=-6 r+1\) and the equation of plane \(2 x+y+z=7\) Then, \(2(r+2)+(-r-3)+(-6 r+1)=7\) \(2 r+4-r-3-6 r+1=7\) \(-5 r+2=7\) \(-5 r=7-2=5\) \(r=-\frac{5}{5}\) \(r=-1\) Therefore, coordinates of the required points are \((-1+2),(-(-1)-3),(-6(-1)+1)=(1,-2,7)\)
BITSAT-2012
Three Dimensional Geometry
121083
The coordinates of the point where the line through the points \(A(3,4,1)\) and \(B(5,1,6)\) crosses the \(\mathrm{XY}\)-plane are
A Given- Points, \(\mathrm{A}(3,4,1)\) and \(\mathrm{B}(5,1,6)\) Direction ratio of line, \(A B\) \(\mathrm{AB}=[(5-3),(1-4),(6-1)]=(2,-3,5)\) Then equation of line becomes, \(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}=r\) \(\Rightarrow \quad x=2 r+3, y=-3 r+4\) \(\text { and } \quad z=5 r+1\) Line is crossing the \(\mathrm{X}-\mathrm{Y}\) plane, so \(\mathrm{Z}\) - coordinate will be 0 . \(\Rightarrow 5 \mathrm{r}+1=0\) \(\Rightarrow \mathrm{r}=-\frac{1}{5}\) \(\therefore \quad \frac{\mathrm{x}-3}{2}=-\frac{1}{5} \Rightarrow \mathrm{x}=3-\frac{2}{5}=\frac{13}{5}\) Similarly, \(\frac{y-4}{-3}=\frac{-1}{5} \Rightarrow y=4+\frac{3}{5}=\frac{23}{5}\) Therefore required coordinates will be \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\)
BITSAT-2013
Three Dimensional Geometry
121084
If the plane \(2 x+3 y+5 z=1\) intersect the coordinate axes at the points \(A, B, C\) then the centroid of \(\triangle \mathrm{ABC}\) is
D Given, Plane \(2 x+3 y+5 z=1\) intersects the coordinate axis at the point \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) in \(\triangle \mathrm{ABC}\). \(\therefore\) Equation of plane in intercepted form- \(\frac{\mathrm{x}}{\frac{1}{2}}+\frac{\mathrm{y}}{\frac{1}{3}}+\frac{\mathrm{z}}{5}=1\) \(\therefore \quad \mathrm{A}\left(\frac{1}{2}, 0,0\right), \mathrm{B}\left(0, \frac{1}{3}, 0\right), \mathrm{C}\left(0,0, \frac{1}{5}\right),\)Therefore, centroid of \(\triangle \mathrm{ABC}\) is \(\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)\).
