Explanation
If a ray makes angles $$\alpha, \beta, \gamma$$ and $$\delta$$ with the four diagonals of a cube.
Then we know that $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta=\displaystyle\dfrac{4}{3}$$
$$\mathrm{A}:$$
$$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta=\displaystyle\dfrac{4}{3}$$
$$\mathrm{B}:$$
$$\sin^{2}\alpha+\sin^{2}\beta+\sin^{2}\gamma+\sin^{2}\delta$$
$$=1-\cos^{2}\alpha+1-\cos^{2}\beta+1-\cos^{2}\gamma+1-\cos^{2}\delta=4-\displaystyle\dfrac{4}{3}=\displaystyle\dfrac{8}{3}$$
$$\mathrm{C}:$$
$$\cos 2\alpha+\cos 2\beta+\cos 2\gamma+\cos 2\delta$$
$$=1-2\sin^{2}\alpha+1-2\sin^{2}\beta+1-2\sin^{2}\gamma+1-2\sin^{2}\delta=4-\displaystyle\dfrac{16}{3}=\displaystyle-\dfrac{4}{3}$$
Descending order is $$B,A,C$$
Hence, option A is correct answer.
Given lines are $$\overrightarrow { r } =3i+2j-4k+\lambda \left( i+2j+2k \right) $$
and $$\overrightarrow { r } =5i-2k+\mu \left( 3i+2j+6k \right) $$
We know, angle between $$\overrightarrow { r } =\overrightarrow { { a }_{ 1 } } +\lambda \overrightarrow { { b }_{ 1 } } $$ and $$\overrightarrow { r } =\overrightarrow { { a }_{ 2 } } +\lambda \overrightarrow { { b }_{ 2 } } $$ is given by,
$$\displaystyle \cos { \theta } =\dfrac { \overrightarrow { { b }_{ 1 } } .\overrightarrow { { b }_{ 2 } } }{ \left| \overrightarrow { { b }_{ 1 } } \right| \left| \overrightarrow { { b }_{ 2 } } \right| } $$
$$\Rightarrow \cos { \theta } =\dfrac { \left( i+2j+2k \right) .\left( 3i+2j+6k \right) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 2 }^{ 2 } } \sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ 6 }^{ 2 } } } $$
$$\displaystyle =\dfrac { 3+4+12 }{ \sqrt { 9 } \sqrt { 49 } } =\dfrac { 19 }{ 21 } $$
$$\Rightarrow \theta =\cos ^{ -1 }{ \left( \dfrac { 19 }{ 21 } \right) } $$
Let l,m,n be the direction ratio of a line making angles ,,, with four diagonals of a cube then as four diagonals have direction ratio,$$( a, a, a)$$;$$(a, a, -a)$$;$$( a, -a, a,)$$ ;$$(-a, a, a) $$(where $$a$$ is the side of the cube)
Now, $$\displaystyle \cos\alpha =\left| \dfrac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left|\dfrac{l+m+n}{\sqrt3 \sqrt{l^2+m^2+n^2}} \right| $$
$$\displaystyle \cos\beta =\left| \dfrac{al+am-an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left| \dfrac{l+m-n}{\sqrt{3} \sqrt{l^2+m^2+n^2}} \right| $$
$$\displaystyle \cos\gamma = \left|\dfrac{al-am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|=\left|\dfrac{l-m+n}{\sqrt3 \sqrt{l^2+m^2+n^2}} \right|$$
$$\displaystyle \cos\delta = \left|\dfrac{al+am+an}{\sqrt{a^2+a^2+a^2}\sqrt{l^2+m^2+n^2}} \right|= \left| \dfrac{-l+m+n}{\sqrt{3} \sqrt{l^2+m^2+n^2}} \right|$$
and $$\cos^2\alpha +\cos^2\beta+\cos^2\gamma+\cos^2\delta$$
$$\displaystyle =\dfrac{1}{3}\times \dfrac{(4l^2+4m^2+4n^2)}{l^2+m^2+n^2}=\dfrac{4}{3}$$
So statement 1 is true and statement-2 is not true .
Hence, option 'C' is correct.
If the straight lines $$\displaystyle \dfrac{x-1}{2}=\dfrac{y+1}{K}=\dfrac{z}{2}$$ and $$\displaystyle \dfrac{x+1}{5}=\dfrac{y+1}{2}=\dfrac{z}{K}$$ are coplanar, then the plane(s) containing these two lines is(are)
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