Interaction between Two Lines in 3D

The lines $\frac{x-a+d}{\mathrm{\alpha }-\mathrm{\delta }}=\frac{y-a}{\mathrm{\alpha }}=\frac{z-a-d}{\mathrm{\alpha }+\mathrm{\delta }}$ and $\frac{x-b+c}{\mathrm{\beta }-\mathrm{\gamma }}=\frac{y-b}{\mathrm{\beta }}=\frac{z-b-c}{\mathrm{\beta }+\mathrm{\gamma }}$ are coplanar and then equation to the plane in which they lie, is

If three mutually perpendicular lines have direction cosines $\left(l_1,m_1,n_1\right),\left(l_2,m_2,n_2\right)$ and $\left(l_3,m_3,n_3\right)$ , then the line having direction cosines $l_1+l_2+l_3,m_1+m_2+m_3$ and $n_1+n_2+n_3$ make an angle of .... with each other

The angle between the straight lines $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+1}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$ is

If $A\left(3,4,5\right),B\left(4,6,3\right),C\left(-1,2,4\right)$ and $D\left(1,0,5\right)$ are such that the angle between the lines $DC$ and $AB$ is $\theta$$,$ then $\cos \theta$ is equal to

The line $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-\mathrm{k}}$ and $\frac{\mathrm{x}-1}{\mathrm{k}}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-5}{1}$ are coplanar if

If lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are mutually perpendicular, then $k$ is equal to

If the lines $\frac{1-x}{3}=\frac{y-2}{2\mathrm{\alpha }}=\frac{z-3}{2}\mathrm{ }$and  $\frac{x-1}{3\mathrm{\alpha }}=y-1=\frac{6-z}{5}$ are perpendicular, then the value of $\alpha$ is

The angle between the line

$\frac{x}{1}=\frac{y}{0}=\frac{z}{-1} \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\frac{x}{3}=\frac{y}{4}=\frac{z}{5} \mathrm{i}\mathrm{s}\mathrm{ }\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{t}\mathrm{o}$

The point of intersection of the lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and  $\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$ is

The angle between the lines, $\frac{x+4}{1}=\frac{y-3}{2}=\frac{z+2}{3} \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}\mathrm{ }\mathrm{i}\mathrm{s}$

The point of intersection of the lines $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}, \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}\mathrm{ }$ is

The angle between the lines whose direction cosines are $\left(\frac{\sqrt{3}}{4},\frac{1}{4},\frac{\sqrt{3}}{2}\right) \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\left(\frac{\sqrt{3}}{4},\frac{1}{4},\frac{-\sqrt{3}}{2}\right)$ is

If direction cosines of two lines are proportional to $\left(\mathrm{2,3},-6\right) \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }(3,-\mathrm{4,5})$ then the acute angle between them is

The cosine of the angle $A$ of the triangle with verities $A\left(1,-\mathrm{1,2}\right), B\left(\mathrm{6,11,2}\right), C(\mathrm{1,2},6)$ is

Let $P(\mathrm{3,2},6)$ be a point in space and $Q$ be a point on the line $\overrightarrow{\mathbf{r}}=\left(\widehat{\dot{\mathbf{i}}}-\widehat{\dot{\mathbf{j}}}+2\widehat{\mathbf{k}}\right)+\mathrm{\mu }\left(-3\widehat{\dot{\mathbf{i}}}+\widehat{\dot{\mathbf{j}}}+5\widehat{\mathbf{k}}\right).$ Then, the value of μ for which the vector $\overrightarrow{\mathbf{P}\mathbf{Q}}$ is parallel to the plane $x-4y+3z=1$ is

The angle between the line

$\frac{x}{1}=\frac{y}{0}=\frac{z}{-1} \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\frac{x}{3}=\frac{y}{4}=\frac{z}{5} \mathrm{i}\mathrm{s}\mathrm{ }\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{t}\mathrm{o}$

The angle between the lines $2x=3y=-z \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }6x=-y=-4z$ is

The angle between the lines $x=1,y=2 \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }y=-1, z=0$ is

The two lines $x=ay+b, z=cy+d \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }x=a^{\prime } y+b^{\prime }, z=c^{\prime } y+d\prime$ are perpendicular to each other, if

The angle between the lines $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}$ and $\frac{x-1}{1}=\frac{2y+3}{3}=\frac{z+5}{2}$, is equal to.