Explanation
D.\mathrm{C} of the line are \displaystyle \dfrac{1}{\sqrt{3}} , \displaystyle \dfrac{1}{\sqrt{3}} , \displaystyle \dfrac{1}{\sqrt{3}}
Any point on the line at a distance t from {P}(2, -1,2) is \left (2+\displaystyle \dfrac{{t}}{\sqrt{3}}, -1+\dfrac{{t}}{\sqrt{3}}, 2+\dfrac{{t}}{\sqrt{3}}\right)
which lies on 2x+y+z=9.
\Rightarrow t=\sqrt 3
Let \overrightarrow { { m }_{ 1 } } and \overrightarrow { { m }_{ 2 } } be vectors parallel to the two given lines.
Then, angle between the two given lines is same as the angle between \overrightarrow { { m }_{ 1 } } and \overrightarrow { { m }_{ 2 } }.
\overrightarrow { { m }_{ 1 } } = Vector parallel to the line with direction ratios (1,1,2)=i+j+2k and
\overrightarrow { { m }_{ 2 } } = Vector parallel to the line with direction ratio \left( \left( \sqrt { 3 } -1 \right) ,\left( -\sqrt { 3 } -1 \right) ,4 \right) =\left( \sqrt { 3 } -1 \right) i+\left( -\sqrt { 3 } -1 \right) j+4k
Let \theta be the angle between the given lines.
Then \displaystyle \cos { \theta } =\dfrac { \overrightarrow { { m }_{ 1 } } .\overrightarrow { { m }_{ 2 } } }{ \left| \overrightarrow { { m }_{ 1 } } \right| \left| \overrightarrow { { m }_{ 2 } } \right| } =\dfrac { \left( \sqrt { 3 } -1 \right) -\left( \sqrt { 3 } +1 \right) +8 }{ \sqrt { 1+1+4 } \sqrt { { \left( \sqrt { 3 } -1 \right) }^{ 2 }+{ \left( \sqrt { 3 } +1 \right) }^{ 2 }+16 } }
\Rightarrow \displaystyle \cos { \theta } =\dfrac { 6 }{ \sqrt { 6 } \sqrt { 24 } } =\dfrac { 1 }{ 2 }
\Rightarrow \theta =\dfrac { \pi }{ 3 }
The given lines are parallel to the vector.
\overrightarrow { { b }_{ 1 } } =1+2j+2k and \overrightarrow { { b }_{ 2 } } =3i+2j+6k respectively.
So, the angle \theta between them is given by
\displaystyle \cos { \theta } =\dfrac { \overrightarrow { { b }_{ 1 } } .\overrightarrow { { b }_{ 2 } } }{ \left| \overrightarrow { { b }_{ 1 } } \right| \left| \overrightarrow { { b }_{ 1 } } \right| } =\dfrac { \left( i+2j+2k \right) \left( 3i+2j+6k \right) }{ \left| i+2j+2k \right| \left| 3i+2j+6k \right| }
\displaystyle =\dfrac { 3+4+12 }{ \sqrt { 1+4+4 } \sqrt { 9+4+36 } } =\dfrac { 19 }{ 21 }
\Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 } \right) }
\left (\displaystyle \dfrac{2}{7},\dfrac{3}{7},\dfrac{-6}{7}\right)
\left (\displaystyle \dfrac{3}{7}\dfrac{-2}{7},\dfrac{6}{7}\right)
\left (\displaystyle \dfrac{2}{3}\dfrac{-1}{3},\dfrac{2}{3}\right)
Since, the plane is perpendicular to the z axis, a = b=0
The plane passes through (2,-3,4).
So d = 4c
\dfrac{a+b+c}{d} = \dfrac{c}{4c} = \dfrac{1}{4}
lf \theta is the angle between two lines whose d.cs are l_{1},m_{\mathrm{1}},n_{\mathrm{1}} and l_{2},m_{2},n_{2}, then
\displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{\mathrm{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}=
Let \vec{a_1} and \vec{a_2} be unit vectors along the lines.
\vec{a_1} = l_1 \hat{i} + m_1 \hat{j} + n_1 \hat{k}
\vec{a_2} = l_2 \hat{i} + m_2 \hat{j} + n_2 \hat{k}
\Rightarrow l_1^2+m_1^2+n_1^2 = 1 and l_2^2+m_2^2+n_2^2 = 1
Consider the dot product of \vec{a_1} and \vec{a_2}
\vec{a_1}.\vec{a_2} = (l_1l_2)+(m_1m_2)+(n_1n_2)
\Rightarrow \cos \theta = (l_1l_2)+(m_1m_2)+(n_1n_2)
S = \displaystyle \dfrac{\Sigma(l_{1}+l_{2})^{2}}{4\cos^{2}(\dfrac{\theta}{2})}+\dfrac{\Sigma(l_{{I}}-l_{2})^{2}}{4\sin^{2}(\dfrac{\theta}{2})}
\Rightarrow S = \dfrac{ 2 + 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{ 2 - 2 (l_1l_2 + m_1m_2+ n_1n_2) } {4 \sin^2{\dfrac{\theta}{2}}}
\Rightarrow S = \dfrac{2+2\cos\theta}{ 4 \cos^2{\dfrac{\theta}{2}}} + \dfrac{2-2\cos\theta}{4 \sin^2{\dfrac{\theta}{2}}}
Now, 1+\cos\theta = 2\cos^2\dfrac{\theta}2 and 1-\cos\theta = 2\sin^2\dfrac{\theta}2
\Rightarrow S = \dfrac22 + \dfrac22
\Rightarrow S =2
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