Angle between Two Line Segments

Q. $L_1$ and $L_2$ are two lines whose vector equations are
$L_1:\overrightarrow{r}=\lambda (\cos \theta +\sqrt{3})\hat{i}+\left(\sqrt{2}\sin \theta \right)\hat{j}+(\cos \theta -\sqrt{3})\hat{k}$
$L_2:\overrightarrow{r}=\mu (a\hat{i}+b\hat{j}+c\hat{k}),$ where $\lambda$ and $\mu$ are scalars and $\alpha$ is the acute angle between $L_1$ and $L_2.$
If the angle $\alpha$ is independent of $\theta ,$ then the value of $\alpha$ is

1. $\frac{\pi }{6}$
2. $\frac{\pi }{4}$
3. $\frac{\pi }{3}$
4. $\frac{\pi }{2}$

Q.

A variable line $L$ is drawn through $O(0,0)$ to meet the lines$L_1:Y-X-10=0$ And $L_2:Y-X-20=0$ . At points $A$ And $B$ Respectively. A Point $P$ is taken on $L$ such that $\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}$ And $P,A,B$ lies on the same side of origin$O$. The locus of P is

Q. A line makes angles $a,b,c,d$ with the four diagonals of a cube, then ${\cos }^{2}a+{\cos }^{2}b+{\cos }^{2}c+{\cos }^{2}d=$

1. $\frac{1}{3}$
2. $\frac{4}{3}$
3. $\frac{2}{3}$
4. $\frac{5}{3}$

Q. The lines $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{-1},\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{4}$ are

1. parallel lines
2. intersecting lines
3. perpendicular lines
4. skew lines

Q.

The angle between any two diagonals of a cube is:

1. $co{s}^{-1}\left(\frac{1}{2}\right)$

2. $co{s}^{-1}\left(\frac{1}{3}\right)$

3. $co{s}^{-1}\left(\frac{1}{4}\right)$

4. $\frac{\pi }{2}$

Q. The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right-angled triangle whose opposite vertex is $(7,2,4)$. Then which of the following is not the side of the triangle?

1. $\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$
2. $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$
3. $\frac{x-7}{3}=\frac{y-2}{5}=\frac{z-4}{-1}$
4. none of these

Q. If the angle between two intersecting lines having direction ratios $(5,7,3)$ & $(3,4,5)$ respectively can be given by
${\cos }^{-1}\left(\frac{58}{\sqrt{b}}\right)$,
then what will be the value of $b$ ?

1. 4150
2. 4052
3. 3971
4. 4167

Q. The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right-angled triangle whose opposite vertex is $(7,2,4)$. Then which of the following is not the side of the triangle?

1. $\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$
2. $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$
3. $\frac{x-7}{3}=\frac{y-2}{5}=\frac{z-4}{-1}$
4. none of these

Q. If $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2,)$ are d.c.'s of $\overline{OA,}$ $\overline{OB}$ such that $\angle AOB=\theta$ where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle $\angle AOB$ are

1. $\frac{l_1+l_2}{2sin\frac{\theta }{2}},\frac{m_1+m_2}{2sin\frac{\theta }{2}},\frac{n_1+n_2}{2sin\frac{\theta }{2}}$
2. $\frac{l_1+l_2}{2cos\frac{\theta }{2}},\frac{m_1+m_2}{2cos\frac{\theta }{2}},\frac{n_1+n_2}{2cos\frac{\theta }{2}}$
3. $\frac{l_1-l_2}{2sin\frac{\theta }{2}},\frac{m_1-m_2}{2sin\frac{\theta }{2}},\frac{n_1-n_2}{2sin\frac{\theta }{2}}$
4. $\frac{l_1-l_2}{2cos\frac{\theta }{2}},\frac{m_1-m_2}{2cos\frac{\theta }{2}},\frac{n_1-n_2}{2cos\frac{\theta }{2}}$

Q. The angle between the lines $x-3y-4=0,$ $4y-z+5=0$ and $x+3y-11=0,$ $2y-z+6=0$ is

1. ${\cos }^{-1}\left(\sqrt{\frac{2}{13}}\right)$
2. ${\cos }^{-1}\left(\sqrt{\frac{1}{14}}\right)$
3. $\frac{\pi }{2}$
4. $0$