121182
A line with direction cosines proportional to 2 , 1,2 meets the line \(L_1\) passing through \((0,-1,0)\) with direction ratios \(1,1,1\) at \(A(x, y, z)\) and another line \(L_2\) at \(B(1,1,1)\) then \(x+y+z=\)
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Explanation:
B Equation of line \(\mathrm{L}_1\) passing through \((0,-1,0)\) and with direction ratio \((1,1,1)\) is \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}+1}{1}=\frac{\mathrm{z}}{1}\) \(\therefore \quad \mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv(\lambda, \lambda-1, \lambda)\) A line intersect \(\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) and \(\mathrm{B}(1,1,1)\) \(\therefore\) Direction ratio of \(\mathrm{AB}\) \((\lambda-1, \lambda-2, \lambda-1)\) Given direction ratio of line intersect \(A B\) is \(2,1,2\) \(\therefore \lambda-1=2, \lambda-2=1, \lambda-1=2 \Rightarrow \lambda=3\) \(\because \quad \mathrm{x}=3, \mathrm{y}=2, \mathrm{z}=3\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=3+2+3=8\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121183
The direction cosines of the normal to the plane containing the lines having direction ratios 1,2 , 1 and \(4,5,-3\) are
A Let \(b_1=\hat{i}+2 \hat{j}+\hat{k}\) And, \(\mathrm{b}_2=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
Normal vector to plane
\(\mathrm{b}_1 \times \mathrm{b}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & 1 \\ 4 & 5 & -3\end{array}\right \vert\)
\(=\mathrm{i}(-6-5)-\hat{\mathrm{j}}(-3-4)+\hat{\mathrm{k}}(5-8)\)
\(=11 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
\(\therefore\) direction cosines of normal to the plane
\(=\frac{-11}{\sqrt{(-11)^2+7^2+(-3)^2}}, \frac{7}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-3}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121185
The direction cosines of the line making angles \(\frac{\pi}{4}, \frac{\pi}{3}\) And \(\theta\left(0\lt \theta\lt \frac{\pi}{2}\right)\) respectively with \(x, y\) and \(\mathrm{z}\) axes, are
121112
The direction cosines of a line which lies in ZOX plane and makes an angle of \(30^{\circ}\) with \(\mathrm{Z}-\) axis are
1 \(\frac{\sqrt{3}}{2}, 0, \pm \frac{1}{2}\)
2 \(0, \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\)
3 \(\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\)
4 \(0, \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\)
Explanation:
C At \(\mathrm{Y}\)-axis, \(\cos 90^{\circ}=0\) At \(Z-\) axis, \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
Let, the line make \(\theta\) with \(\mathrm{X}\) - axis
\(\therefore \quad \cos ^2 \theta+\cos \left(\frac{\pi}{2}\right)^2+\left(\cos 30^{\circ}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+0+\left(\frac{\sqrt{3}}{2}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+\frac{3}{4}=1\)
\(\Rightarrow \quad \cos ^2 \theta=1-\frac{3}{4}=\frac{1}{4}\)
or \(\quad \cos \theta= \pm \frac{1}{2}\)
So, \(\quad \cos \theta=\left \vert\frac{1}{2}\right \vert\)
Therefore, direction of line is \(\left( \pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\right)\)
121182
A line with direction cosines proportional to 2 , 1,2 meets the line \(L_1\) passing through \((0,-1,0)\) with direction ratios \(1,1,1\) at \(A(x, y, z)\) and another line \(L_2\) at \(B(1,1,1)\) then \(x+y+z=\)
1 7
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Explanation:
B Equation of line \(\mathrm{L}_1\) passing through \((0,-1,0)\) and with direction ratio \((1,1,1)\) is \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}+1}{1}=\frac{\mathrm{z}}{1}\) \(\therefore \quad \mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv(\lambda, \lambda-1, \lambda)\) A line intersect \(\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) and \(\mathrm{B}(1,1,1)\) \(\therefore\) Direction ratio of \(\mathrm{AB}\) \((\lambda-1, \lambda-2, \lambda-1)\) Given direction ratio of line intersect \(A B\) is \(2,1,2\) \(\therefore \lambda-1=2, \lambda-2=1, \lambda-1=2 \Rightarrow \lambda=3\) \(\because \quad \mathrm{x}=3, \mathrm{y}=2, \mathrm{z}=3\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=3+2+3=8\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121183
The direction cosines of the normal to the plane containing the lines having direction ratios 1,2 , 1 and \(4,5,-3\) are
A Let \(b_1=\hat{i}+2 \hat{j}+\hat{k}\) And, \(\mathrm{b}_2=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
Normal vector to plane
\(\mathrm{b}_1 \times \mathrm{b}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & 1 \\ 4 & 5 & -3\end{array}\right \vert\)
\(=\mathrm{i}(-6-5)-\hat{\mathrm{j}}(-3-4)+\hat{\mathrm{k}}(5-8)\)
\(=11 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
\(\therefore\) direction cosines of normal to the plane
\(=\frac{-11}{\sqrt{(-11)^2+7^2+(-3)^2}}, \frac{7}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-3}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121185
The direction cosines of the line making angles \(\frac{\pi}{4}, \frac{\pi}{3}\) And \(\theta\left(0\lt \theta\lt \frac{\pi}{2}\right)\) respectively with \(x, y\) and \(\mathrm{z}\) axes, are
121112
The direction cosines of a line which lies in ZOX plane and makes an angle of \(30^{\circ}\) with \(\mathrm{Z}-\) axis are
1 \(\frac{\sqrt{3}}{2}, 0, \pm \frac{1}{2}\)
2 \(0, \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\)
3 \(\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\)
4 \(0, \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\)
Explanation:
C At \(\mathrm{Y}\)-axis, \(\cos 90^{\circ}=0\) At \(Z-\) axis, \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
Let, the line make \(\theta\) with \(\mathrm{X}\) - axis
\(\therefore \quad \cos ^2 \theta+\cos \left(\frac{\pi}{2}\right)^2+\left(\cos 30^{\circ}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+0+\left(\frac{\sqrt{3}}{2}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+\frac{3}{4}=1\)
\(\Rightarrow \quad \cos ^2 \theta=1-\frac{3}{4}=\frac{1}{4}\)
or \(\quad \cos \theta= \pm \frac{1}{2}\)
So, \(\quad \cos \theta=\left \vert\frac{1}{2}\right \vert\)
Therefore, direction of line is \(\left( \pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\right)\)
121182
A line with direction cosines proportional to 2 , 1,2 meets the line \(L_1\) passing through \((0,-1,0)\) with direction ratios \(1,1,1\) at \(A(x, y, z)\) and another line \(L_2\) at \(B(1,1,1)\) then \(x+y+z=\)
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2 8
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4 10
Explanation:
B Equation of line \(\mathrm{L}_1\) passing through \((0,-1,0)\) and with direction ratio \((1,1,1)\) is \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}+1}{1}=\frac{\mathrm{z}}{1}\) \(\therefore \quad \mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv(\lambda, \lambda-1, \lambda)\) A line intersect \(\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) and \(\mathrm{B}(1,1,1)\) \(\therefore\) Direction ratio of \(\mathrm{AB}\) \((\lambda-1, \lambda-2, \lambda-1)\) Given direction ratio of line intersect \(A B\) is \(2,1,2\) \(\therefore \lambda-1=2, \lambda-2=1, \lambda-1=2 \Rightarrow \lambda=3\) \(\because \quad \mathrm{x}=3, \mathrm{y}=2, \mathrm{z}=3\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=3+2+3=8\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121183
The direction cosines of the normal to the plane containing the lines having direction ratios 1,2 , 1 and \(4,5,-3\) are
A Let \(b_1=\hat{i}+2 \hat{j}+\hat{k}\) And, \(\mathrm{b}_2=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
Normal vector to plane
\(\mathrm{b}_1 \times \mathrm{b}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & 1 \\ 4 & 5 & -3\end{array}\right \vert\)
\(=\mathrm{i}(-6-5)-\hat{\mathrm{j}}(-3-4)+\hat{\mathrm{k}}(5-8)\)
\(=11 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
\(\therefore\) direction cosines of normal to the plane
\(=\frac{-11}{\sqrt{(-11)^2+7^2+(-3)^2}}, \frac{7}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-3}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121185
The direction cosines of the line making angles \(\frac{\pi}{4}, \frac{\pi}{3}\) And \(\theta\left(0\lt \theta\lt \frac{\pi}{2}\right)\) respectively with \(x, y\) and \(\mathrm{z}\) axes, are
121112
The direction cosines of a line which lies in ZOX plane and makes an angle of \(30^{\circ}\) with \(\mathrm{Z}-\) axis are
1 \(\frac{\sqrt{3}}{2}, 0, \pm \frac{1}{2}\)
2 \(0, \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\)
3 \(\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\)
4 \(0, \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\)
Explanation:
C At \(\mathrm{Y}\)-axis, \(\cos 90^{\circ}=0\) At \(Z-\) axis, \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
Let, the line make \(\theta\) with \(\mathrm{X}\) - axis
\(\therefore \quad \cos ^2 \theta+\cos \left(\frac{\pi}{2}\right)^2+\left(\cos 30^{\circ}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+0+\left(\frac{\sqrt{3}}{2}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+\frac{3}{4}=1\)
\(\Rightarrow \quad \cos ^2 \theta=1-\frac{3}{4}=\frac{1}{4}\)
or \(\quad \cos \theta= \pm \frac{1}{2}\)
So, \(\quad \cos \theta=\left \vert\frac{1}{2}\right \vert\)
Therefore, direction of line is \(\left( \pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\right)\)
121182
A line with direction cosines proportional to 2 , 1,2 meets the line \(L_1\) passing through \((0,-1,0)\) with direction ratios \(1,1,1\) at \(A(x, y, z)\) and another line \(L_2\) at \(B(1,1,1)\) then \(x+y+z=\)
1 7
2 8
3 9
4 10
Explanation:
B Equation of line \(\mathrm{L}_1\) passing through \((0,-1,0)\) and with direction ratio \((1,1,1)\) is \(\frac{\mathrm{x}}{1}=\frac{\mathrm{y}+1}{1}=\frac{\mathrm{z}}{1}\) \(\therefore \quad \mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \equiv(\lambda, \lambda-1, \lambda)\) A line intersect \(\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) and \(\mathrm{B}(1,1,1)\) \(\therefore\) Direction ratio of \(\mathrm{AB}\) \((\lambda-1, \lambda-2, \lambda-1)\) Given direction ratio of line intersect \(A B\) is \(2,1,2\) \(\therefore \lambda-1=2, \lambda-2=1, \lambda-1=2 \Rightarrow \lambda=3\) \(\because \quad \mathrm{x}=3, \mathrm{y}=2, \mathrm{z}=3\) \(\mathrm{x}+\mathrm{y}+\mathrm{z}=3+2+3=8\)
TS EAMCET-14.09.2020
Three Dimensional Geometry
121183
The direction cosines of the normal to the plane containing the lines having direction ratios 1,2 , 1 and \(4,5,-3\) are
A Let \(b_1=\hat{i}+2 \hat{j}+\hat{k}\) And, \(\mathrm{b}_2=4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
Normal vector to plane
\(\mathrm{b}_1 \times \mathrm{b}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & 1 \\ 4 & 5 & -3\end{array}\right \vert\)
\(=\mathrm{i}(-6-5)-\hat{\mathrm{j}}(-3-4)+\hat{\mathrm{k}}(5-8)\)
\(=11 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
\(\therefore\) direction cosines of normal to the plane
\(=\frac{-11}{\sqrt{(-11)^2+7^2+(-3)^2}}, \frac{7}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-3}{\sqrt{(-11)^2+7^2+(-3)^2}}\)
\(\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121185
The direction cosines of the line making angles \(\frac{\pi}{4}, \frac{\pi}{3}\) And \(\theta\left(0\lt \theta\lt \frac{\pi}{2}\right)\) respectively with \(x, y\) and \(\mathrm{z}\) axes, are
121112
The direction cosines of a line which lies in ZOX plane and makes an angle of \(30^{\circ}\) with \(\mathrm{Z}-\) axis are
1 \(\frac{\sqrt{3}}{2}, 0, \pm \frac{1}{2}\)
2 \(0, \frac{1}{2}, \pm \frac{\sqrt{3}}{2}\)
3 \(\pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\)
4 \(0, \frac{\sqrt{3}}{2}, \pm \frac{1}{2}\)
Explanation:
C At \(\mathrm{Y}\)-axis, \(\cos 90^{\circ}=0\) At \(Z-\) axis, \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
Let, the line make \(\theta\) with \(\mathrm{X}\) - axis
\(\therefore \quad \cos ^2 \theta+\cos \left(\frac{\pi}{2}\right)^2+\left(\cos 30^{\circ}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+0+\left(\frac{\sqrt{3}}{2}\right)^2=1\)
\(\Rightarrow \quad \cos ^2 \theta+\frac{3}{4}=1\)
\(\Rightarrow \quad \cos ^2 \theta=1-\frac{3}{4}=\frac{1}{4}\)
or \(\quad \cos \theta= \pm \frac{1}{2}\)
So, \(\quad \cos \theta=\left \vert\frac{1}{2}\right \vert\)
Therefore, direction of line is \(\left( \pm \frac{1}{2}, 0, \frac{\sqrt{3}}{2}\right)\)
