NEET Test Series from KOTA - 10 Papers In MS WORD
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Three Dimensional Geometry
121152
If the two lines \(l_1: \frac{x-2}{3}=\frac{y+1}{-2}, z=2\) and \(l_2: \frac{\mathrm{x}-1}{1}=\frac{2 \mathrm{y}+3}{\alpha}=\frac{\mathrm{z}+5}{2}\) perpendicular, then an angle between the lines \(l_2\) and \(l_3: \frac{1-x}{3}=\frac{2 y-3}{-4}=\frac{z}{4}\) is:
1 \(\cos ^{-1}\left(\frac{29}{4}\right)\)
2 \(\sec ^{-1}\left(\frac{29}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{2}{29}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}-2}{0}\)
Direction ratios \(-a_1=3, b_1=-2, c_1=0\)
\(l_2: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3 / 2}{\alpha / 2}=\frac{\mathrm{z}+5}{2}\)
Direction ratios \(-\mathrm{a}_2=1, \mathrm{~b}_2=\frac{\alpha}{2}, \mathrm{c}_2=2\)
Given that, \(l_1\) perpendicular to \(l_2\)
\(\therefore \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0\)
\(\Rightarrow 3-\alpha+0=0\)
\(\Rightarrow \alpha=3\)
So, \(\mathrm{a}_2=1, \mathrm{~b}_2=\frac{3}{2}, \mathrm{c}_2=2\)
Now,
\(l_3: \frac{x-1}{-3}=\frac{y-\frac{1}{2}}{-4}=\frac{z-0}{4}\)
\(a_3=-3, b_3=-2, c_3=4\)
Angle between \(l_2\) and \(l_3\) is given by
\(\cos \theta=\frac{\left \vert\mathrm{a}_2 \mathrm{a}_3+\mathrm{b}_2 \mathrm{~b}_3+\mathrm{c}_2 \mathrm{c}_3\right \vert}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2} \sqrt{\mathrm{a}_3^2+\mathrm{b}_3^2+\mathrm{c}_3^2}}\)
\(\Rightarrow \quad \cos \theta=\frac{ \vert-3-6+8 \vert}{\sqrt{1+\frac{9}{4}+4} \times \sqrt{9+4+16}}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{\sqrt{\frac{29}{4}} \cdot \sqrt{29}}=\frac{2}{\sqrt{29} \cdot \sqrt{29}}=\frac{2}{29}\)
\(\Rightarrow=\cos ^{-1}\left(\frac{2}{29}\right)\)
JEE Main-26.06.2022
Three Dimensional Geometry
121154
The acute angle between the lines whose direction cosines are given by the equations \(l+\) \(\mathbf{m}+\mathbf{n}=0\) and \(2 l m+2 l n-m n=0\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{2 \pi}{5}\)
Explanation:
C Given, \(l+\mathrm{m}+\mathrm{n}=0\)
.......(i)
\(2 l m+2 l n-m n=0\)
........(ii)
and \(\quad 2 l m+2 l n-m n=0\)
\(\cdots \cdots . .(\) ii)
From (i) \(\mathrm{n}=-(l+\mathrm{m})\).
(Put in (ii))
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l(l+\mathrm{m})+\mathrm{m}(l+\mathrm{m})=0\)
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l^2-2 l \mathrm{~m}+l \mathrm{~m}+\mathrm{m}^2=0\)
\(\Rightarrow \quad \mathrm{m}^2+l \mathrm{~m}-2 l^2=0\)
\(\Rightarrow \quad\left(\mathrm{m}^2-l^2\right)+\left(l \mathrm{~m}+l^2\right)=0\)
\(\Rightarrow \quad(\mathrm{m}+l)(\mathrm{m}-l)+l(\mathrm{~m}-l)=0\)
\(\Rightarrow \quad(\mathrm{m}-l)(\mathrm{m}+2 l)=0\)
\(\Rightarrow \quad \mathrm{m}=l,-2 l\)
\(\text {.......(iii) }\)
\(\text { Putting } \mathrm{m}=l \text { in equation (i) }\)
\(\text { We get, } \mathrm{n}=-2 l\)
\(\text { So, }\left(a_1, b_1, c_1\right)=(1,1,-2)\)
\(\text { Again putting , } \mathrm{m}=-2 l \text { in equation (i) }\)
\(\text { we get, } \mathrm{n}=l\)
\(\text { So, }\left(\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2\right)=(1,-2,1)\)
\(\text { Now, } \cos \theta=\left \vert\frac{\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2}{\sqrt{\mathrm{a}_1{ }^2+\mathrm{b}_1{ }^2+\mathrm{c}_1{ }^2} \cdot \sqrt{\mathrm{a}_2{ }^2+\mathrm{b}_2{ }^2+\mathrm{c}_2{ }^2}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{1 \times 1+1 \times(-2)+(-2) \times 1}{\sqrt{1+1+4} \cdot \sqrt{1+4+1}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{\sqrt{6} \cdot \sqrt{6}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{6}\right \vert\)
\(\theta=\cos ^{-1}\left(\frac{1}{2}\right)\)
\(\text { So, } \quad \theta=\frac{\pi}{3}\)
AP EAMCET-22.04.2018
Three Dimensional Geometry
121186
The direction cosines \(l, \mathrm{~m}, \mathrm{n}\) of two lines are satisfying \(3 l+\mathrm{m}+5 \mathrm{n}=0\) and \(6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{~m}=0\). If \(\theta\) is the angle between those lines then \( \vert\cos \theta \vert\) \(=\)
1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{6}\)
4 \(\frac{1}{\sqrt{3}}\)
Explanation:
C We have, \(3 l+\mathrm{m}+5 \mathrm{n}=0\)
Here, \(\mathrm{m}=(-3 l-5 \mathrm{n})\)
And, \(6 \mathrm{mn}-2 \mathrm{n} l+5 \mathrm{~m}=0\)
\(\Rightarrow \quad 6 \mathrm{n}(-3 l-5 \mathrm{n})-2 \mathrm{n} l+5 \mathrm{l}(-3 l-5 \mathrm{n})=0\)
\(\Rightarrow \quad-18 \mathrm{n} l-30 \mathrm{n}^2-2 \mathrm{n} l-15 l^2-25 \mathrm{n} l=0\)
\(\Rightarrow \quad-15 l^2-30 \mathrm{n}^2-45 \mathrm{n} l=0\)
\(\Rightarrow \quad l^2-3 \mathrm{n} l-2 \mathrm{n}^2=0\)
\(\Rightarrow \quad(l+\mathrm{n})(l+2 \mathrm{n})=0 \Rightarrow l=-\mathrm{n},-2 \mathrm{n}\)
When \(l=-\mathrm{n}\)
\(-3 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=-2 \mathrm{n}\)
So, direction ratio are
\((-\mathrm{n},-2 \mathrm{n}, \mathrm{n}) \text { or }(1,2,-1)\)
When \(l=-2 \mathrm{n}\)
\(-6 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=\mathrm{n}\)
So, direction ratio are
\((-2 \mathrm{n}, \mathrm{n}, \mathrm{n}) \text { or }(-2,1,1)\)
\(\therefore \cos \theta=\frac{1 \times(-2)+2 \times 1+(-1) \times 1}{\sqrt{1+4+1} \sqrt{4+1+1}}=\frac{-1}{6}\)
\(\therefore \vert\cos \theta \vert=\frac{1}{6}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121120
A line makes angles \(\alpha, \beta, \gamma\) with the coordinate axes, then \(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma\) is equal to
121152
If the two lines \(l_1: \frac{x-2}{3}=\frac{y+1}{-2}, z=2\) and \(l_2: \frac{\mathrm{x}-1}{1}=\frac{2 \mathrm{y}+3}{\alpha}=\frac{\mathrm{z}+5}{2}\) perpendicular, then an angle between the lines \(l_2\) and \(l_3: \frac{1-x}{3}=\frac{2 y-3}{-4}=\frac{z}{4}\) is:
1 \(\cos ^{-1}\left(\frac{29}{4}\right)\)
2 \(\sec ^{-1}\left(\frac{29}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{2}{29}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}-2}{0}\)
Direction ratios \(-a_1=3, b_1=-2, c_1=0\)
\(l_2: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3 / 2}{\alpha / 2}=\frac{\mathrm{z}+5}{2}\)
Direction ratios \(-\mathrm{a}_2=1, \mathrm{~b}_2=\frac{\alpha}{2}, \mathrm{c}_2=2\)
Given that, \(l_1\) perpendicular to \(l_2\)
\(\therefore \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0\)
\(\Rightarrow 3-\alpha+0=0\)
\(\Rightarrow \alpha=3\)
So, \(\mathrm{a}_2=1, \mathrm{~b}_2=\frac{3}{2}, \mathrm{c}_2=2\)
Now,
\(l_3: \frac{x-1}{-3}=\frac{y-\frac{1}{2}}{-4}=\frac{z-0}{4}\)
\(a_3=-3, b_3=-2, c_3=4\)
Angle between \(l_2\) and \(l_3\) is given by
\(\cos \theta=\frac{\left \vert\mathrm{a}_2 \mathrm{a}_3+\mathrm{b}_2 \mathrm{~b}_3+\mathrm{c}_2 \mathrm{c}_3\right \vert}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2} \sqrt{\mathrm{a}_3^2+\mathrm{b}_3^2+\mathrm{c}_3^2}}\)
\(\Rightarrow \quad \cos \theta=\frac{ \vert-3-6+8 \vert}{\sqrt{1+\frac{9}{4}+4} \times \sqrt{9+4+16}}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{\sqrt{\frac{29}{4}} \cdot \sqrt{29}}=\frac{2}{\sqrt{29} \cdot \sqrt{29}}=\frac{2}{29}\)
\(\Rightarrow=\cos ^{-1}\left(\frac{2}{29}\right)\)
JEE Main-26.06.2022
Three Dimensional Geometry
121154
The acute angle between the lines whose direction cosines are given by the equations \(l+\) \(\mathbf{m}+\mathbf{n}=0\) and \(2 l m+2 l n-m n=0\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{2 \pi}{5}\)
Explanation:
C Given, \(l+\mathrm{m}+\mathrm{n}=0\)
.......(i)
\(2 l m+2 l n-m n=0\)
........(ii)
and \(\quad 2 l m+2 l n-m n=0\)
\(\cdots \cdots . .(\) ii)
From (i) \(\mathrm{n}=-(l+\mathrm{m})\).
(Put in (ii))
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l(l+\mathrm{m})+\mathrm{m}(l+\mathrm{m})=0\)
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l^2-2 l \mathrm{~m}+l \mathrm{~m}+\mathrm{m}^2=0\)
\(\Rightarrow \quad \mathrm{m}^2+l \mathrm{~m}-2 l^2=0\)
\(\Rightarrow \quad\left(\mathrm{m}^2-l^2\right)+\left(l \mathrm{~m}+l^2\right)=0\)
\(\Rightarrow \quad(\mathrm{m}+l)(\mathrm{m}-l)+l(\mathrm{~m}-l)=0\)
\(\Rightarrow \quad(\mathrm{m}-l)(\mathrm{m}+2 l)=0\)
\(\Rightarrow \quad \mathrm{m}=l,-2 l\)
\(\text {.......(iii) }\)
\(\text { Putting } \mathrm{m}=l \text { in equation (i) }\)
\(\text { We get, } \mathrm{n}=-2 l\)
\(\text { So, }\left(a_1, b_1, c_1\right)=(1,1,-2)\)
\(\text { Again putting , } \mathrm{m}=-2 l \text { in equation (i) }\)
\(\text { we get, } \mathrm{n}=l\)
\(\text { So, }\left(\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2\right)=(1,-2,1)\)
\(\text { Now, } \cos \theta=\left \vert\frac{\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2}{\sqrt{\mathrm{a}_1{ }^2+\mathrm{b}_1{ }^2+\mathrm{c}_1{ }^2} \cdot \sqrt{\mathrm{a}_2{ }^2+\mathrm{b}_2{ }^2+\mathrm{c}_2{ }^2}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{1 \times 1+1 \times(-2)+(-2) \times 1}{\sqrt{1+1+4} \cdot \sqrt{1+4+1}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{\sqrt{6} \cdot \sqrt{6}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{6}\right \vert\)
\(\theta=\cos ^{-1}\left(\frac{1}{2}\right)\)
\(\text { So, } \quad \theta=\frac{\pi}{3}\)
AP EAMCET-22.04.2018
Three Dimensional Geometry
121186
The direction cosines \(l, \mathrm{~m}, \mathrm{n}\) of two lines are satisfying \(3 l+\mathrm{m}+5 \mathrm{n}=0\) and \(6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{~m}=0\). If \(\theta\) is the angle between those lines then \( \vert\cos \theta \vert\) \(=\)
1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{6}\)
4 \(\frac{1}{\sqrt{3}}\)
Explanation:
C We have, \(3 l+\mathrm{m}+5 \mathrm{n}=0\)
Here, \(\mathrm{m}=(-3 l-5 \mathrm{n})\)
And, \(6 \mathrm{mn}-2 \mathrm{n} l+5 \mathrm{~m}=0\)
\(\Rightarrow \quad 6 \mathrm{n}(-3 l-5 \mathrm{n})-2 \mathrm{n} l+5 \mathrm{l}(-3 l-5 \mathrm{n})=0\)
\(\Rightarrow \quad-18 \mathrm{n} l-30 \mathrm{n}^2-2 \mathrm{n} l-15 l^2-25 \mathrm{n} l=0\)
\(\Rightarrow \quad-15 l^2-30 \mathrm{n}^2-45 \mathrm{n} l=0\)
\(\Rightarrow \quad l^2-3 \mathrm{n} l-2 \mathrm{n}^2=0\)
\(\Rightarrow \quad(l+\mathrm{n})(l+2 \mathrm{n})=0 \Rightarrow l=-\mathrm{n},-2 \mathrm{n}\)
When \(l=-\mathrm{n}\)
\(-3 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=-2 \mathrm{n}\)
So, direction ratio are
\((-\mathrm{n},-2 \mathrm{n}, \mathrm{n}) \text { or }(1,2,-1)\)
When \(l=-2 \mathrm{n}\)
\(-6 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=\mathrm{n}\)
So, direction ratio are
\((-2 \mathrm{n}, \mathrm{n}, \mathrm{n}) \text { or }(-2,1,1)\)
\(\therefore \cos \theta=\frac{1 \times(-2)+2 \times 1+(-1) \times 1}{\sqrt{1+4+1} \sqrt{4+1+1}}=\frac{-1}{6}\)
\(\therefore \vert\cos \theta \vert=\frac{1}{6}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121120
A line makes angles \(\alpha, \beta, \gamma\) with the coordinate axes, then \(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma\) is equal to
121152
If the two lines \(l_1: \frac{x-2}{3}=\frac{y+1}{-2}, z=2\) and \(l_2: \frac{\mathrm{x}-1}{1}=\frac{2 \mathrm{y}+3}{\alpha}=\frac{\mathrm{z}+5}{2}\) perpendicular, then an angle between the lines \(l_2\) and \(l_3: \frac{1-x}{3}=\frac{2 y-3}{-4}=\frac{z}{4}\) is:
1 \(\cos ^{-1}\left(\frac{29}{4}\right)\)
2 \(\sec ^{-1}\left(\frac{29}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{2}{29}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}-2}{0}\)
Direction ratios \(-a_1=3, b_1=-2, c_1=0\)
\(l_2: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3 / 2}{\alpha / 2}=\frac{\mathrm{z}+5}{2}\)
Direction ratios \(-\mathrm{a}_2=1, \mathrm{~b}_2=\frac{\alpha}{2}, \mathrm{c}_2=2\)
Given that, \(l_1\) perpendicular to \(l_2\)
\(\therefore \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0\)
\(\Rightarrow 3-\alpha+0=0\)
\(\Rightarrow \alpha=3\)
So, \(\mathrm{a}_2=1, \mathrm{~b}_2=\frac{3}{2}, \mathrm{c}_2=2\)
Now,
\(l_3: \frac{x-1}{-3}=\frac{y-\frac{1}{2}}{-4}=\frac{z-0}{4}\)
\(a_3=-3, b_3=-2, c_3=4\)
Angle between \(l_2\) and \(l_3\) is given by
\(\cos \theta=\frac{\left \vert\mathrm{a}_2 \mathrm{a}_3+\mathrm{b}_2 \mathrm{~b}_3+\mathrm{c}_2 \mathrm{c}_3\right \vert}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2} \sqrt{\mathrm{a}_3^2+\mathrm{b}_3^2+\mathrm{c}_3^2}}\)
\(\Rightarrow \quad \cos \theta=\frac{ \vert-3-6+8 \vert}{\sqrt{1+\frac{9}{4}+4} \times \sqrt{9+4+16}}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{\sqrt{\frac{29}{4}} \cdot \sqrt{29}}=\frac{2}{\sqrt{29} \cdot \sqrt{29}}=\frac{2}{29}\)
\(\Rightarrow=\cos ^{-1}\left(\frac{2}{29}\right)\)
JEE Main-26.06.2022
Three Dimensional Geometry
121154
The acute angle between the lines whose direction cosines are given by the equations \(l+\) \(\mathbf{m}+\mathbf{n}=0\) and \(2 l m+2 l n-m n=0\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{2 \pi}{5}\)
Explanation:
C Given, \(l+\mathrm{m}+\mathrm{n}=0\)
.......(i)
\(2 l m+2 l n-m n=0\)
........(ii)
and \(\quad 2 l m+2 l n-m n=0\)
\(\cdots \cdots . .(\) ii)
From (i) \(\mathrm{n}=-(l+\mathrm{m})\).
(Put in (ii))
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l(l+\mathrm{m})+\mathrm{m}(l+\mathrm{m})=0\)
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l^2-2 l \mathrm{~m}+l \mathrm{~m}+\mathrm{m}^2=0\)
\(\Rightarrow \quad \mathrm{m}^2+l \mathrm{~m}-2 l^2=0\)
\(\Rightarrow \quad\left(\mathrm{m}^2-l^2\right)+\left(l \mathrm{~m}+l^2\right)=0\)
\(\Rightarrow \quad(\mathrm{m}+l)(\mathrm{m}-l)+l(\mathrm{~m}-l)=0\)
\(\Rightarrow \quad(\mathrm{m}-l)(\mathrm{m}+2 l)=0\)
\(\Rightarrow \quad \mathrm{m}=l,-2 l\)
\(\text {.......(iii) }\)
\(\text { Putting } \mathrm{m}=l \text { in equation (i) }\)
\(\text { We get, } \mathrm{n}=-2 l\)
\(\text { So, }\left(a_1, b_1, c_1\right)=(1,1,-2)\)
\(\text { Again putting , } \mathrm{m}=-2 l \text { in equation (i) }\)
\(\text { we get, } \mathrm{n}=l\)
\(\text { So, }\left(\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2\right)=(1,-2,1)\)
\(\text { Now, } \cos \theta=\left \vert\frac{\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2}{\sqrt{\mathrm{a}_1{ }^2+\mathrm{b}_1{ }^2+\mathrm{c}_1{ }^2} \cdot \sqrt{\mathrm{a}_2{ }^2+\mathrm{b}_2{ }^2+\mathrm{c}_2{ }^2}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{1 \times 1+1 \times(-2)+(-2) \times 1}{\sqrt{1+1+4} \cdot \sqrt{1+4+1}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{\sqrt{6} \cdot \sqrt{6}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{6}\right \vert\)
\(\theta=\cos ^{-1}\left(\frac{1}{2}\right)\)
\(\text { So, } \quad \theta=\frac{\pi}{3}\)
AP EAMCET-22.04.2018
Three Dimensional Geometry
121186
The direction cosines \(l, \mathrm{~m}, \mathrm{n}\) of two lines are satisfying \(3 l+\mathrm{m}+5 \mathrm{n}=0\) and \(6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{~m}=0\). If \(\theta\) is the angle between those lines then \( \vert\cos \theta \vert\) \(=\)
1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{6}\)
4 \(\frac{1}{\sqrt{3}}\)
Explanation:
C We have, \(3 l+\mathrm{m}+5 \mathrm{n}=0\)
Here, \(\mathrm{m}=(-3 l-5 \mathrm{n})\)
And, \(6 \mathrm{mn}-2 \mathrm{n} l+5 \mathrm{~m}=0\)
\(\Rightarrow \quad 6 \mathrm{n}(-3 l-5 \mathrm{n})-2 \mathrm{n} l+5 \mathrm{l}(-3 l-5 \mathrm{n})=0\)
\(\Rightarrow \quad-18 \mathrm{n} l-30 \mathrm{n}^2-2 \mathrm{n} l-15 l^2-25 \mathrm{n} l=0\)
\(\Rightarrow \quad-15 l^2-30 \mathrm{n}^2-45 \mathrm{n} l=0\)
\(\Rightarrow \quad l^2-3 \mathrm{n} l-2 \mathrm{n}^2=0\)
\(\Rightarrow \quad(l+\mathrm{n})(l+2 \mathrm{n})=0 \Rightarrow l=-\mathrm{n},-2 \mathrm{n}\)
When \(l=-\mathrm{n}\)
\(-3 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=-2 \mathrm{n}\)
So, direction ratio are
\((-\mathrm{n},-2 \mathrm{n}, \mathrm{n}) \text { or }(1,2,-1)\)
When \(l=-2 \mathrm{n}\)
\(-6 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=\mathrm{n}\)
So, direction ratio are
\((-2 \mathrm{n}, \mathrm{n}, \mathrm{n}) \text { or }(-2,1,1)\)
\(\therefore \cos \theta=\frac{1 \times(-2)+2 \times 1+(-1) \times 1}{\sqrt{1+4+1} \sqrt{4+1+1}}=\frac{-1}{6}\)
\(\therefore \vert\cos \theta \vert=\frac{1}{6}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121120
A line makes angles \(\alpha, \beta, \gamma\) with the coordinate axes, then \(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma\) is equal to
121152
If the two lines \(l_1: \frac{x-2}{3}=\frac{y+1}{-2}, z=2\) and \(l_2: \frac{\mathrm{x}-1}{1}=\frac{2 \mathrm{y}+3}{\alpha}=\frac{\mathrm{z}+5}{2}\) perpendicular, then an angle between the lines \(l_2\) and \(l_3: \frac{1-x}{3}=\frac{2 y-3}{-4}=\frac{z}{4}\) is:
1 \(\cos ^{-1}\left(\frac{29}{4}\right)\)
2 \(\sec ^{-1}\left(\frac{29}{4}\right)\)
3 \(\cos ^{-1}\left(\frac{2}{29}\right)\)
4 \(\cos ^{-1}\left(\frac{2}{\sqrt{29}}\right)\)
Explanation:
C Given, \(l_1: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}-2}{0}\)
Direction ratios \(-a_1=3, b_1=-2, c_1=0\)
\(l_2: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+3 / 2}{\alpha / 2}=\frac{\mathrm{z}+5}{2}\)
Direction ratios \(-\mathrm{a}_2=1, \mathrm{~b}_2=\frac{\alpha}{2}, \mathrm{c}_2=2\)
Given that, \(l_1\) perpendicular to \(l_2\)
\(\therefore \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0\)
\(\Rightarrow 3-\alpha+0=0\)
\(\Rightarrow \alpha=3\)
So, \(\mathrm{a}_2=1, \mathrm{~b}_2=\frac{3}{2}, \mathrm{c}_2=2\)
Now,
\(l_3: \frac{x-1}{-3}=\frac{y-\frac{1}{2}}{-4}=\frac{z-0}{4}\)
\(a_3=-3, b_3=-2, c_3=4\)
Angle between \(l_2\) and \(l_3\) is given by
\(\cos \theta=\frac{\left \vert\mathrm{a}_2 \mathrm{a}_3+\mathrm{b}_2 \mathrm{~b}_3+\mathrm{c}_2 \mathrm{c}_3\right \vert}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2} \sqrt{\mathrm{a}_3^2+\mathrm{b}_3^2+\mathrm{c}_3^2}}\)
\(\Rightarrow \quad \cos \theta=\frac{ \vert-3-6+8 \vert}{\sqrt{1+\frac{9}{4}+4} \times \sqrt{9+4+16}}\)
\(\Rightarrow \quad \cos \theta=\frac{1}{\sqrt{\frac{29}{4}} \cdot \sqrt{29}}=\frac{2}{\sqrt{29} \cdot \sqrt{29}}=\frac{2}{29}\)
\(\Rightarrow=\cos ^{-1}\left(\frac{2}{29}\right)\)
JEE Main-26.06.2022
Three Dimensional Geometry
121154
The acute angle between the lines whose direction cosines are given by the equations \(l+\) \(\mathbf{m}+\mathbf{n}=0\) and \(2 l m+2 l n-m n=0\) is
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{2 \pi}{5}\)
Explanation:
C Given, \(l+\mathrm{m}+\mathrm{n}=0\)
.......(i)
\(2 l m+2 l n-m n=0\)
........(ii)
and \(\quad 2 l m+2 l n-m n=0\)
\(\cdots \cdots . .(\) ii)
From (i) \(\mathrm{n}=-(l+\mathrm{m})\).
(Put in (ii))
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l(l+\mathrm{m})+\mathrm{m}(l+\mathrm{m})=0\)
\(\Rightarrow \quad 2 l \mathrm{~m}-2 l^2-2 l \mathrm{~m}+l \mathrm{~m}+\mathrm{m}^2=0\)
\(\Rightarrow \quad \mathrm{m}^2+l \mathrm{~m}-2 l^2=0\)
\(\Rightarrow \quad\left(\mathrm{m}^2-l^2\right)+\left(l \mathrm{~m}+l^2\right)=0\)
\(\Rightarrow \quad(\mathrm{m}+l)(\mathrm{m}-l)+l(\mathrm{~m}-l)=0\)
\(\Rightarrow \quad(\mathrm{m}-l)(\mathrm{m}+2 l)=0\)
\(\Rightarrow \quad \mathrm{m}=l,-2 l\)
\(\text {.......(iii) }\)
\(\text { Putting } \mathrm{m}=l \text { in equation (i) }\)
\(\text { We get, } \mathrm{n}=-2 l\)
\(\text { So, }\left(a_1, b_1, c_1\right)=(1,1,-2)\)
\(\text { Again putting , } \mathrm{m}=-2 l \text { in equation (i) }\)
\(\text { we get, } \mathrm{n}=l\)
\(\text { So, }\left(\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2\right)=(1,-2,1)\)
\(\text { Now, } \cos \theta=\left \vert\frac{\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2}{\sqrt{\mathrm{a}_1{ }^2+\mathrm{b}_1{ }^2+\mathrm{c}_1{ }^2} \cdot \sqrt{\mathrm{a}_2{ }^2+\mathrm{b}_2{ }^2+\mathrm{c}_2{ }^2}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{1 \times 1+1 \times(-2)+(-2) \times 1}{\sqrt{1+1+4} \cdot \sqrt{1+4+1}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{\sqrt{6} \cdot \sqrt{6}}\right \vert\)
\(\Rightarrow \quad \cos \theta=\left \vert\frac{-3}{6}\right \vert\)
\(\theta=\cos ^{-1}\left(\frac{1}{2}\right)\)
\(\text { So, } \quad \theta=\frac{\pi}{3}\)
AP EAMCET-22.04.2018
Three Dimensional Geometry
121186
The direction cosines \(l, \mathrm{~m}, \mathrm{n}\) of two lines are satisfying \(3 l+\mathrm{m}+5 \mathrm{n}=0\) and \(6 \mathrm{mn}-2 \mathrm{n} l+5 l \mathrm{~m}=0\). If \(\theta\) is the angle between those lines then \( \vert\cos \theta \vert\) \(=\)
1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{2}}\)
3 \(\frac{1}{6}\)
4 \(\frac{1}{\sqrt{3}}\)
Explanation:
C We have, \(3 l+\mathrm{m}+5 \mathrm{n}=0\)
Here, \(\mathrm{m}=(-3 l-5 \mathrm{n})\)
And, \(6 \mathrm{mn}-2 \mathrm{n} l+5 \mathrm{~m}=0\)
\(\Rightarrow \quad 6 \mathrm{n}(-3 l-5 \mathrm{n})-2 \mathrm{n} l+5 \mathrm{l}(-3 l-5 \mathrm{n})=0\)
\(\Rightarrow \quad-18 \mathrm{n} l-30 \mathrm{n}^2-2 \mathrm{n} l-15 l^2-25 \mathrm{n} l=0\)
\(\Rightarrow \quad-15 l^2-30 \mathrm{n}^2-45 \mathrm{n} l=0\)
\(\Rightarrow \quad l^2-3 \mathrm{n} l-2 \mathrm{n}^2=0\)
\(\Rightarrow \quad(l+\mathrm{n})(l+2 \mathrm{n})=0 \Rightarrow l=-\mathrm{n},-2 \mathrm{n}\)
When \(l=-\mathrm{n}\)
\(-3 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=-2 \mathrm{n}\)
So, direction ratio are
\((-\mathrm{n},-2 \mathrm{n}, \mathrm{n}) \text { or }(1,2,-1)\)
When \(l=-2 \mathrm{n}\)
\(-6 \mathrm{n}+\mathrm{m}+5 \mathrm{n}=0 \Rightarrow \mathrm{m}=\mathrm{n}\)
So, direction ratio are
\((-2 \mathrm{n}, \mathrm{n}, \mathrm{n}) \text { or }(-2,1,1)\)
\(\therefore \cos \theta=\frac{1 \times(-2)+2 \times 1+(-1) \times 1}{\sqrt{1+4+1} \sqrt{4+1+1}}=\frac{-1}{6}\)
\(\therefore \vert\cos \theta \vert=\frac{1}{6}\)
TS EAMCET-10.09.2020
Three Dimensional Geometry
121120
A line makes angles \(\alpha, \beta, \gamma\) with the coordinate axes, then \(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma\) is equal to
