Three Dimension...

If $$\left(\dfrac {1}{2},\dfrac {1}{3},n\right)$$ are the direction cosines of a line then the value of $$n$$ is

The angle between the pair of lines with direction ratios (1, 1, 2) and $$(\sqrt{3} - 1, -\sqrt{3} - 1, 4)$$ is 

A line makes angles $$\alpha,\beta,\gamma,\delta$$ with the four diagonals of a cube then $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta$$ is equal to

The direction ratios of the line, given by the planes x - y + z - 5 = 0, x - 3y - 6 = 0 are 

The direction cosines of a vector A are $$\cos { \alpha  } =\frac { 4 }{ 5\sqrt { 2 }  } ,$$ $$cos \beta =\frac { 1 }{ \sqrt { 2 }  } ,$$ and $$cos \gamma = \frac{ 3 }{ 5\sqrt { 2 }  } ,$$ then vector A is

If $$\overline { O A } = 3 \overline { i } + \overline { j } - \overline { k }$$, $$| \overline { A B } | = 2 \sqrt { 6 }$$ and AB has the direction ratios 1, -1 , 2 then $$| O B | =$$ 

The vector $$a = \alpha 1 + 2 j + \beta k$$ lies in the plane of the vectors $$b = i + jt$$ and $$c = j + k$$ and bisects the angle between $$b$$ and $$c$$. Then which one of the following gives possible values $$\alpha$$ and $$\beta$$.

Let $$l_{1},\ m_{1},\ n_{1};\ l_{2},\ m_{2},\ n_{2};\ l_{3},\ m_{3},\ n_{3}$$ be the direction cosines of three mutually perpendicular line then $$\begin{vmatrix} { l }_{ 1 } & m_{ 1 } & n_{ 1 } \\ { l }_{ 2 } & m_{ 2 } & n_{ 2 } \\ { l }_{ 3 } & m_{ 3 } & n_{ 3 } \end{vmatrix}$$

The point where $$\vec{ x }$$ which is perpendicular to $$(2,-3,1)$$ and $$(1,-2,3)$$ and which satisfies the condition $$\vec { x } \cdot ( \hat { i } + 2 \hat { j } - 7 \hat{ k } ) = 10$$

The equation of the plane through $$\left(0,-5,1\right)$$ which is perpendicular to the planes $$2x+4y+2z+3=0$$,$$2x+5y+3z+4=0$$ is