121165
The angle between the lines with direction ratio \(4,-3,5\) and \(3,4,5\) is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
A Given that direction ratio \((4,-3,5)\) and \((3,4,5)\)
Then,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
\(\cos \theta =\left \vert\frac{4 \times 3+(-3) \times 4+5 \times 5}{\sqrt{16+9+25} \sqrt{9+16+25}}\right \vert\)
\(=\frac{12-12+25}{\sqrt{50} \times \sqrt{50}}\)
\(=\frac{25}{50}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{2}\)
\(\Rightarrow \quad \cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow\left(\frac{\pi}{3}\right)\)
AMU-2016
Three Dimensional Geometry
121168
The direction cosines of two lines are \(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\) and \(\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\). Then the angle between the lines is equal to
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
B Given, Direction cosines -
\(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right) \text { and }\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
Angle between two lines is -
\(\cos \theta= \left \vertl_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2\right \vert\)
\(\cos \theta=\left \vert\frac{\sqrt{3}}{2} \times\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{4} \times\left(\frac{\sqrt{3}}{4}\right)\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-3}{4}+\frac{1}{16}+\frac{3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-12+1+3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\frac{1}{2}\)
\(\therefore \theta=60^{\circ}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121171
If the direction ratios Lines \(L_1\) and \(L_2\) are 2 , 1,1 and \(3,-3,4\) respectively, then the direction cosines of a line that is perpendicular to both \(\mathrm{L}_1\) and \(\mathrm{L}_2\) are
B Given, \(\overrightarrow{\mathrm{L}}_1=(2,-1,1)\)
\(\overrightarrow{\mathrm{L}}_2=(3,-3,4)\)
\(\text { So, } \overrightarrow{\mathrm{L}}_1=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\text { and } \overrightarrow{\mathrm{L}}_2=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
Perpendicular vector -
\(\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{L}}_1 \times \overrightarrow{\mathrm{L}}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 3 & -3 & 4\end{array}\right \vert\)
\(\vec{n}=-\hat{i}-5 \hat{j}-3 \hat{k}\)
\(\mathrm{DC} \text { 'S of this vector }\)
\(l=\frac{-1}{\sqrt{1^2+(-5)^2+(-3)^2}}=\frac{-1}{\sqrt{35}},\)
\(\mathrm{~m}=\frac{-5}{\sqrt{(-1)+(-5)^2+(-3)^2}}=\frac{-5}{\sqrt{35}} \text { and }\)
\(\mathrm{n}=\frac{-3}{\sqrt{(-1)^2+(-5)^2+(-3)^2}}=\frac{-3}{\sqrt{35}}\)
\((l, \mathrm{~m}, \mathrm{n})=\left( \pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}\right)\)DC'S of this vector
AP EAMCET-23.04.2018
Three Dimensional Geometry
121179
The direction ratios of the line perpendicular to the lines having direction ratios \(2,3,1\) and \(1,2,1\) are
1 \(2,2,-2\)
2 \(1,1,1\)
3 \(-2,1,1\)
4 \(1,-1,1\)
Explanation:
D Let \(\bar{a}\) and \(\bar{b}\) be the vector along the lines whose direction ratios are 2,3,1 and 1,2,1 respectively. \(\bar{a}=2 \hat{i}+3 \hat{j}+\hat{k} \quad \text { and } \quad \bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
A vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by,
\(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{array}\right \vert=\hat{\mathrm{i}}(3-2)-\hat{\mathrm{j}}(2-1)+\hat{\mathrm{k}}(4-3)=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Hence d.r.s. are \(1,-1,1\)
MHT CET-2020
Three Dimensional Geometry
121184
Find the direction of intersecting line of two planes \(2 x+y+z=1\) and \(3 x+2 y-z=3\)
1 \((3,5,1)\)
2 \((3,5,-1)\)
3 \((-3,5,1)\)
4 None of these
Explanation:
C Given, \(2 x+y+z=1\)
\(3 x+2 y-z=3\)
Normal vector are -
\(\vec{n}_1=(2 \hat{i}+\hat{j}+\hat{k}), \vec{n}_2=(3 \hat{i}+2 \hat{j}-\hat{k})\)
\(\vec{a}_1=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & 2 & -1\end{array}\right \vert\)
\(=\hat{i}(-1-2)-\hat{j}(-2-3)+\hat{k}(4-3)\)
Hence, the direction of intersecting line
\(=-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(=(-3,5,1)\)
121165
The angle between the lines with direction ratio \(4,-3,5\) and \(3,4,5\) is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
A Given that direction ratio \((4,-3,5)\) and \((3,4,5)\)
Then,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
\(\cos \theta =\left \vert\frac{4 \times 3+(-3) \times 4+5 \times 5}{\sqrt{16+9+25} \sqrt{9+16+25}}\right \vert\)
\(=\frac{12-12+25}{\sqrt{50} \times \sqrt{50}}\)
\(=\frac{25}{50}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{2}\)
\(\Rightarrow \quad \cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow\left(\frac{\pi}{3}\right)\)
AMU-2016
Three Dimensional Geometry
121168
The direction cosines of two lines are \(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\) and \(\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\). Then the angle between the lines is equal to
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
B Given, Direction cosines -
\(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right) \text { and }\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
Angle between two lines is -
\(\cos \theta= \left \vertl_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2\right \vert\)
\(\cos \theta=\left \vert\frac{\sqrt{3}}{2} \times\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{4} \times\left(\frac{\sqrt{3}}{4}\right)\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-3}{4}+\frac{1}{16}+\frac{3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-12+1+3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\frac{1}{2}\)
\(\therefore \theta=60^{\circ}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121171
If the direction ratios Lines \(L_1\) and \(L_2\) are 2 , 1,1 and \(3,-3,4\) respectively, then the direction cosines of a line that is perpendicular to both \(\mathrm{L}_1\) and \(\mathrm{L}_2\) are
B Given, \(\overrightarrow{\mathrm{L}}_1=(2,-1,1)\)
\(\overrightarrow{\mathrm{L}}_2=(3,-3,4)\)
\(\text { So, } \overrightarrow{\mathrm{L}}_1=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\text { and } \overrightarrow{\mathrm{L}}_2=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
Perpendicular vector -
\(\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{L}}_1 \times \overrightarrow{\mathrm{L}}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 3 & -3 & 4\end{array}\right \vert\)
\(\vec{n}=-\hat{i}-5 \hat{j}-3 \hat{k}\)
\(\mathrm{DC} \text { 'S of this vector }\)
\(l=\frac{-1}{\sqrt{1^2+(-5)^2+(-3)^2}}=\frac{-1}{\sqrt{35}},\)
\(\mathrm{~m}=\frac{-5}{\sqrt{(-1)+(-5)^2+(-3)^2}}=\frac{-5}{\sqrt{35}} \text { and }\)
\(\mathrm{n}=\frac{-3}{\sqrt{(-1)^2+(-5)^2+(-3)^2}}=\frac{-3}{\sqrt{35}}\)
\((l, \mathrm{~m}, \mathrm{n})=\left( \pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}\right)\)DC'S of this vector
AP EAMCET-23.04.2018
Three Dimensional Geometry
121179
The direction ratios of the line perpendicular to the lines having direction ratios \(2,3,1\) and \(1,2,1\) are
1 \(2,2,-2\)
2 \(1,1,1\)
3 \(-2,1,1\)
4 \(1,-1,1\)
Explanation:
D Let \(\bar{a}\) and \(\bar{b}\) be the vector along the lines whose direction ratios are 2,3,1 and 1,2,1 respectively. \(\bar{a}=2 \hat{i}+3 \hat{j}+\hat{k} \quad \text { and } \quad \bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
A vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by,
\(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{array}\right \vert=\hat{\mathrm{i}}(3-2)-\hat{\mathrm{j}}(2-1)+\hat{\mathrm{k}}(4-3)=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Hence d.r.s. are \(1,-1,1\)
MHT CET-2020
Three Dimensional Geometry
121184
Find the direction of intersecting line of two planes \(2 x+y+z=1\) and \(3 x+2 y-z=3\)
1 \((3,5,1)\)
2 \((3,5,-1)\)
3 \((-3,5,1)\)
4 None of these
Explanation:
C Given, \(2 x+y+z=1\)
\(3 x+2 y-z=3\)
Normal vector are -
\(\vec{n}_1=(2 \hat{i}+\hat{j}+\hat{k}), \vec{n}_2=(3 \hat{i}+2 \hat{j}-\hat{k})\)
\(\vec{a}_1=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & 2 & -1\end{array}\right \vert\)
\(=\hat{i}(-1-2)-\hat{j}(-2-3)+\hat{k}(4-3)\)
Hence, the direction of intersecting line
\(=-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(=(-3,5,1)\)
121165
The angle between the lines with direction ratio \(4,-3,5\) and \(3,4,5\) is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
A Given that direction ratio \((4,-3,5)\) and \((3,4,5)\)
Then,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
\(\cos \theta =\left \vert\frac{4 \times 3+(-3) \times 4+5 \times 5}{\sqrt{16+9+25} \sqrt{9+16+25}}\right \vert\)
\(=\frac{12-12+25}{\sqrt{50} \times \sqrt{50}}\)
\(=\frac{25}{50}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{2}\)
\(\Rightarrow \quad \cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow\left(\frac{\pi}{3}\right)\)
AMU-2016
Three Dimensional Geometry
121168
The direction cosines of two lines are \(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\) and \(\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\). Then the angle between the lines is equal to
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
B Given, Direction cosines -
\(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right) \text { and }\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
Angle between two lines is -
\(\cos \theta= \left \vertl_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2\right \vert\)
\(\cos \theta=\left \vert\frac{\sqrt{3}}{2} \times\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{4} \times\left(\frac{\sqrt{3}}{4}\right)\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-3}{4}+\frac{1}{16}+\frac{3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-12+1+3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\frac{1}{2}\)
\(\therefore \theta=60^{\circ}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121171
If the direction ratios Lines \(L_1\) and \(L_2\) are 2 , 1,1 and \(3,-3,4\) respectively, then the direction cosines of a line that is perpendicular to both \(\mathrm{L}_1\) and \(\mathrm{L}_2\) are
B Given, \(\overrightarrow{\mathrm{L}}_1=(2,-1,1)\)
\(\overrightarrow{\mathrm{L}}_2=(3,-3,4)\)
\(\text { So, } \overrightarrow{\mathrm{L}}_1=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\text { and } \overrightarrow{\mathrm{L}}_2=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
Perpendicular vector -
\(\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{L}}_1 \times \overrightarrow{\mathrm{L}}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 3 & -3 & 4\end{array}\right \vert\)
\(\vec{n}=-\hat{i}-5 \hat{j}-3 \hat{k}\)
\(\mathrm{DC} \text { 'S of this vector }\)
\(l=\frac{-1}{\sqrt{1^2+(-5)^2+(-3)^2}}=\frac{-1}{\sqrt{35}},\)
\(\mathrm{~m}=\frac{-5}{\sqrt{(-1)+(-5)^2+(-3)^2}}=\frac{-5}{\sqrt{35}} \text { and }\)
\(\mathrm{n}=\frac{-3}{\sqrt{(-1)^2+(-5)^2+(-3)^2}}=\frac{-3}{\sqrt{35}}\)
\((l, \mathrm{~m}, \mathrm{n})=\left( \pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}\right)\)DC'S of this vector
AP EAMCET-23.04.2018
Three Dimensional Geometry
121179
The direction ratios of the line perpendicular to the lines having direction ratios \(2,3,1\) and \(1,2,1\) are
1 \(2,2,-2\)
2 \(1,1,1\)
3 \(-2,1,1\)
4 \(1,-1,1\)
Explanation:
D Let \(\bar{a}\) and \(\bar{b}\) be the vector along the lines whose direction ratios are 2,3,1 and 1,2,1 respectively. \(\bar{a}=2 \hat{i}+3 \hat{j}+\hat{k} \quad \text { and } \quad \bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
A vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by,
\(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{array}\right \vert=\hat{\mathrm{i}}(3-2)-\hat{\mathrm{j}}(2-1)+\hat{\mathrm{k}}(4-3)=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Hence d.r.s. are \(1,-1,1\)
MHT CET-2020
Three Dimensional Geometry
121184
Find the direction of intersecting line of two planes \(2 x+y+z=1\) and \(3 x+2 y-z=3\)
1 \((3,5,1)\)
2 \((3,5,-1)\)
3 \((-3,5,1)\)
4 None of these
Explanation:
C Given, \(2 x+y+z=1\)
\(3 x+2 y-z=3\)
Normal vector are -
\(\vec{n}_1=(2 \hat{i}+\hat{j}+\hat{k}), \vec{n}_2=(3 \hat{i}+2 \hat{j}-\hat{k})\)
\(\vec{a}_1=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & 2 & -1\end{array}\right \vert\)
\(=\hat{i}(-1-2)-\hat{j}(-2-3)+\hat{k}(4-3)\)
Hence, the direction of intersecting line
\(=-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(=(-3,5,1)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121165
The angle between the lines with direction ratio \(4,-3,5\) and \(3,4,5\) is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
A Given that direction ratio \((4,-3,5)\) and \((3,4,5)\)
Then,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
\(\cos \theta =\left \vert\frac{4 \times 3+(-3) \times 4+5 \times 5}{\sqrt{16+9+25} \sqrt{9+16+25}}\right \vert\)
\(=\frac{12-12+25}{\sqrt{50} \times \sqrt{50}}\)
\(=\frac{25}{50}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{2}\)
\(\Rightarrow \quad \cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow\left(\frac{\pi}{3}\right)\)
AMU-2016
Three Dimensional Geometry
121168
The direction cosines of two lines are \(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\) and \(\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\). Then the angle between the lines is equal to
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
B Given, Direction cosines -
\(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right) \text { and }\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
Angle between two lines is -
\(\cos \theta= \left \vertl_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2\right \vert\)
\(\cos \theta=\left \vert\frac{\sqrt{3}}{2} \times\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{4} \times\left(\frac{\sqrt{3}}{4}\right)\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-3}{4}+\frac{1}{16}+\frac{3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-12+1+3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\frac{1}{2}\)
\(\therefore \theta=60^{\circ}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121171
If the direction ratios Lines \(L_1\) and \(L_2\) are 2 , 1,1 and \(3,-3,4\) respectively, then the direction cosines of a line that is perpendicular to both \(\mathrm{L}_1\) and \(\mathrm{L}_2\) are
B Given, \(\overrightarrow{\mathrm{L}}_1=(2,-1,1)\)
\(\overrightarrow{\mathrm{L}}_2=(3,-3,4)\)
\(\text { So, } \overrightarrow{\mathrm{L}}_1=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\text { and } \overrightarrow{\mathrm{L}}_2=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
Perpendicular vector -
\(\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{L}}_1 \times \overrightarrow{\mathrm{L}}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 3 & -3 & 4\end{array}\right \vert\)
\(\vec{n}=-\hat{i}-5 \hat{j}-3 \hat{k}\)
\(\mathrm{DC} \text { 'S of this vector }\)
\(l=\frac{-1}{\sqrt{1^2+(-5)^2+(-3)^2}}=\frac{-1}{\sqrt{35}},\)
\(\mathrm{~m}=\frac{-5}{\sqrt{(-1)+(-5)^2+(-3)^2}}=\frac{-5}{\sqrt{35}} \text { and }\)
\(\mathrm{n}=\frac{-3}{\sqrt{(-1)^2+(-5)^2+(-3)^2}}=\frac{-3}{\sqrt{35}}\)
\((l, \mathrm{~m}, \mathrm{n})=\left( \pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}\right)\)DC'S of this vector
AP EAMCET-23.04.2018
Three Dimensional Geometry
121179
The direction ratios of the line perpendicular to the lines having direction ratios \(2,3,1\) and \(1,2,1\) are
1 \(2,2,-2\)
2 \(1,1,1\)
3 \(-2,1,1\)
4 \(1,-1,1\)
Explanation:
D Let \(\bar{a}\) and \(\bar{b}\) be the vector along the lines whose direction ratios are 2,3,1 and 1,2,1 respectively. \(\bar{a}=2 \hat{i}+3 \hat{j}+\hat{k} \quad \text { and } \quad \bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
A vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by,
\(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{array}\right \vert=\hat{\mathrm{i}}(3-2)-\hat{\mathrm{j}}(2-1)+\hat{\mathrm{k}}(4-3)=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Hence d.r.s. are \(1,-1,1\)
MHT CET-2020
Three Dimensional Geometry
121184
Find the direction of intersecting line of two planes \(2 x+y+z=1\) and \(3 x+2 y-z=3\)
1 \((3,5,1)\)
2 \((3,5,-1)\)
3 \((-3,5,1)\)
4 None of these
Explanation:
C Given, \(2 x+y+z=1\)
\(3 x+2 y-z=3\)
Normal vector are -
\(\vec{n}_1=(2 \hat{i}+\hat{j}+\hat{k}), \vec{n}_2=(3 \hat{i}+2 \hat{j}-\hat{k})\)
\(\vec{a}_1=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & 2 & -1\end{array}\right \vert\)
\(=\hat{i}(-1-2)-\hat{j}(-2-3)+\hat{k}(4-3)\)
Hence, the direction of intersecting line
\(=-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(=(-3,5,1)\)
121165
The angle between the lines with direction ratio \(4,-3,5\) and \(3,4,5\) is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
A Given that direction ratio \((4,-3,5)\) and \((3,4,5)\)
Then,
\(\cos \theta=\left \vert\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right \vert\)
\(\cos \theta =\left \vert\frac{4 \times 3+(-3) \times 4+5 \times 5}{\sqrt{16+9+25} \sqrt{9+16+25}}\right \vert\)
\(=\frac{12-12+25}{\sqrt{50} \times \sqrt{50}}\)
\(=\frac{25}{50}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{2}\)
\(\Rightarrow \quad \cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow\left(\frac{\pi}{3}\right)\)
AMU-2016
Three Dimensional Geometry
121168
The direction cosines of two lines are \(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\) and \(\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\). Then the angle between the lines is equal to
1 \(30^{\circ}\)
2 \(60^{\circ}\)
3 \(45^{\circ}\)
4 \(90^{\circ}\)
Explanation:
B Given, Direction cosines -
\(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right) \text { and }\left(\frac{-\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
Angle between two lines is -
\(\cos \theta= \left \vertl_1 l_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2\right \vert\)
\(\cos \theta=\left \vert\frac{\sqrt{3}}{2} \times\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{4} \times\left(\frac{\sqrt{3}}{4}\right)\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-3}{4}+\frac{1}{16}+\frac{3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\left \vert\frac{-12+1+3}{16}\right \vert\)
\(\Rightarrow \cos \theta=\frac{1}{2}\)
\(\therefore \theta=60^{\circ}\)
AP EAMCET-22.09.2020
Three Dimensional Geometry
121171
If the direction ratios Lines \(L_1\) and \(L_2\) are 2 , 1,1 and \(3,-3,4\) respectively, then the direction cosines of a line that is perpendicular to both \(\mathrm{L}_1\) and \(\mathrm{L}_2\) are
B Given, \(\overrightarrow{\mathrm{L}}_1=(2,-1,1)\)
\(\overrightarrow{\mathrm{L}}_2=(3,-3,4)\)
\(\text { So, } \overrightarrow{\mathrm{L}}_1=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(\text { and } \overrightarrow{\mathrm{L}}_2=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
Perpendicular vector -
\(\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{L}}_1 \times \overrightarrow{\mathrm{L}}_2=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 3 & -3 & 4\end{array}\right \vert\)
\(\vec{n}=-\hat{i}-5 \hat{j}-3 \hat{k}\)
\(\mathrm{DC} \text { 'S of this vector }\)
\(l=\frac{-1}{\sqrt{1^2+(-5)^2+(-3)^2}}=\frac{-1}{\sqrt{35}},\)
\(\mathrm{~m}=\frac{-5}{\sqrt{(-1)+(-5)^2+(-3)^2}}=\frac{-5}{\sqrt{35}} \text { and }\)
\(\mathrm{n}=\frac{-3}{\sqrt{(-1)^2+(-5)^2+(-3)^2}}=\frac{-3}{\sqrt{35}}\)
\((l, \mathrm{~m}, \mathrm{n})=\left( \pm \frac{1}{\sqrt{35}}, \pm \frac{5}{\sqrt{35}}, \pm \frac{3}{\sqrt{35}}\right)\)DC'S of this vector
AP EAMCET-23.04.2018
Three Dimensional Geometry
121179
The direction ratios of the line perpendicular to the lines having direction ratios \(2,3,1\) and \(1,2,1\) are
1 \(2,2,-2\)
2 \(1,1,1\)
3 \(-2,1,1\)
4 \(1,-1,1\)
Explanation:
D Let \(\bar{a}\) and \(\bar{b}\) be the vector along the lines whose direction ratios are 2,3,1 and 1,2,1 respectively. \(\bar{a}=2 \hat{i}+3 \hat{j}+\hat{k} \quad \text { and } \quad \bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
A vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by,
\(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left \vert\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{array}\right \vert=\hat{\mathrm{i}}(3-2)-\hat{\mathrm{j}}(2-1)+\hat{\mathrm{k}}(4-3)=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Hence d.r.s. are \(1,-1,1\)
MHT CET-2020
Three Dimensional Geometry
121184
Find the direction of intersecting line of two planes \(2 x+y+z=1\) and \(3 x+2 y-z=3\)
1 \((3,5,1)\)
2 \((3,5,-1)\)
3 \((-3,5,1)\)
4 None of these
Explanation:
C Given, \(2 x+y+z=1\)
\(3 x+2 y-z=3\)
Normal vector are -
\(\vec{n}_1=(2 \hat{i}+\hat{j}+\hat{k}), \vec{n}_2=(3 \hat{i}+2 \hat{j}-\hat{k})\)
\(\vec{a}_1=\vec{n}_1 \times \vec{n}_2\)
\(=\left \vert\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 3 & 2 & -1\end{array}\right \vert\)
\(=\hat{i}(-1-2)-\hat{j}(-2-3)+\hat{k}(4-3)\)
Hence, the direction of intersecting line
\(=-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\hat{\mathrm{k}}\)
\(=(-3,5,1)\)