$\text cos\theta = \dfrac{1.0+1.1+0.1}{(√(1^{2}+1^{2}+0^{2}).(√0^{2}+1^{2}+1^{2})}$

Line$:\cfrac { x }{ 2 } =\cfrac { y }{ 3 } =\cfrac { z }{ 4 } $ has directions $(2,3,4)$

Points $A(1,2,3)$ and $B(4,5,6)$ ,  Plane :$2x+y+z=1$

We have,

We have

Direction ratios of given line are $3,4,5$ direction ratios of the perpendicular to the  plane $2x+y-2z=0$ are $2,1,-2$

Consider the problem 

Let $\dfrac{x+3}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{1}=k$

If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$ &  $\vec a\cdot \vec c=n$
Therefore, statement 2 is true.
Since, line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$
Then, $2(1)-3(0)-4(-2)-10=0$
          $\Rightarrow 0=0$
and $(i+2j-k).(2i-3j-4k)=0$
       $\Rightarrow 2-6+4=0$
       $\Rightarrow 0=0$
Therefore, statement 1 is true.

Ans: A

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0$

Let $\vec { OB } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k$
$\vec { AB } =\vec { OB } -\vec { OA } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k-\hat i-2\hat j+3\hat k=3r\hat i-r\hat j+\left( 6+4r \right)\hat k$
Since, $\vec { AB }$ is parallel to $x+y-z=1$
Therefore, $\vec { AB } .\left(\hat i+\hat j-\hat k \right) =0$
$\Rightarrow \left( 3r\hat i-r\hat j+\left( 6+4r \right)\hat k \right) .\left(\hat i+\hat j-\hat k \right)=0 $
$\Rightarrow 3r-r-6-4r=0$
$\Rightarrow r=-3$
Therefore, $\vec { OB } =-8i+5j-9k$

Ans: D

Given plane is $2x−3y+6z−11=0$

Direction ratios of $x=y=z$ are $(1,1,1)$....(1)

Let angle between the above line and $4x -3y+5z=2$ is $ \theta$.

$ \Rightarrow $ angle between the line and normal to the plane is $90^{o} - \theta$...(2)

Direction ratios of normal to the given plane are $(4, -3, 5)$....(3)

From (1), (2) and (3), $ \cos {(90^{o}- \theta)}=\dfrac{1 \times 4 + 1 \times (-3)+ 1 \times 5}{\sqrt{1^2+1^2+1^2}{ \sqrt {4^2+3^2+5^2}}}=\dfrac{6}{\sqrt{3} \times 5 \times \sqrt{2}}$

$ \Rightarrow \sin {\theta}= \dfrac{\sqrt{6}}{5} $

$\Rightarrow \theta = \sin^{-1}{\dfrac{\sqrt{6}}{5}}$

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0,$
$\therefore$ given line is $\bot $ to the normal to the plane i.e., given line is parallel to the given plane.
Also $\left( 1,-1,3 \right) $ lies on the plane $x-2y-z=0$, if $1-2\left( -1 \right) -3=0$ i.e., $1+2-3=0$
which is true 

Given plane is $2x−3y+6z−11=0$

Direction of line $\bar{l}=(0, 1, 0)$
$\bar{n}=$ Normal of plane $=(2, 3, 6)$
$\therefore \alpha =$ Angle between line and plane.
$\therefore \sin\alpha =\left|\dfrac{\bar{l}\cdot \bar{n}}{|\bar{l}||\bar{n}|}\right|$
$=\dfrac{0(2)+1(3)+0(6)}{(1)\sqrt{4+4+36}}$
$=\dfrac{0+3+0}{\sqrt{49}}$
$=\dfrac{3}{7}$
$\therefore$ Measure of an angle between line and plane $\alpha =\sin^{-1}\left(\dfrac{3}{7}\right)$.

Consider the given line equation $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda }$ and plane equation$x+2y+3z=4$.

Let $\theta $ be the angle between the line and normal to plane converting the given equations into normal form, we have

  $ \overrightarrow{r}=0.\widehat{i}+\widehat{j}+3\widehat{k}+\beta \left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right) $

 $ \overrightarrow{r}=\widehat{i}+2\widehat{j}+3.\widehat{k}=3 $

Now,

  $ \overrightarrow{b}=\left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right) $

 $ \overrightarrow{n}=\widehat{i}+2\widehat{j}+3.\widehat{k}=3 $

We know that,

  $ \cos \theta =\left| \dfrac{\widehat{b}.\widehat{n}}{\left| \widehat{b} \right|\left| \widehat{n} \right|} \right| $

 $ =\left| \dfrac{\left( \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right).\left( \widehat{i}+2\widehat{j}+3.\widehat{k} \right)}{\left| \widehat{i}+2\widehat{j}+\lambda \widehat{k} \right|\left| \widehat{i}+2\widehat{j}+3.\widehat{k} \right|} \right| $

 $ \cos \theta =\left| \dfrac{1+4+3\lambda }{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\lambda }^{2}}}\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}} \right|=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

But given that $\theta ={{\cos }^{-1}}\left( \sqrt{\dfrac{5}{14}} \right)$ ,so

  $ \cos {{\cos }^{-1}}\sqrt{\dfrac{5}{14}}=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

 $ \sqrt{\dfrac{5}{14}}=\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right| $

Taking square both sides ,we get

  $ \dfrac{5}{14}={{\left| \dfrac{5+\lambda }{\sqrt{5+{{\lambda }^{2}}}\sqrt{14}} \right|}^{2}}=\dfrac{{{\left( 5+\lambda  \right)}^{2}}}{\left( 5+{{\lambda }^{2}} \right)\left( 14 \right)} $

 $ \dfrac{{{\left( 5+\lambda  \right)}^{2}}}{\left( 5+{{\lambda }^{2}} \right)}=5 $

 $ 25+{{\lambda }^{2}}+10\lambda =25+5{{\lambda }^{2}} $

 $ 4{{\lambda }^{2}}-10\lambda =0 $

 $ 2\lambda \left( 2\lambda -5 \right)=0 $

 $ \lambda \left( 2\lambda -5 \right)=0 $

 $ \lambda =0,\lambda =\dfrac{5}{2} $

equation of line $\dfrac{x-0}{1}=\dfrac{y-1}{2}=\dfrac{2-3}{\lambda}$ where $a _1=1, a _2=2, a _3=\lambda$

A line is inclined at Φ to a plane. The vector equation of the line is given by 

Angle between the line and plane is same as the angle between the line and normal to the plane
$\displaystyle \therefore \cos \left ( 90^{0}-\theta  \right )=\frac{a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}}{\sqrt{a _{1}^{2}+b _{1}^{2}+c _{1}^{2}}\sqrt{a _{2}^{2}+b _{2}^{2}+c _{2}^{2}}}$
$\displaystyle \therefore\frac{1}{3}=\frac{\left ( 1\times 2+2\times \left ( -1 \right )+2\sqrt{\lambda } \right )}{\sqrt{1^{2}+2^{2}+2^{2}}\sqrt{2^{2}+1^{2}+\lambda }} $

 $\theta$ is angle b/w $\xrightarrow [\gamma]{} =2\hat {  i}+j+k+(i+j+k)t$ and $\rightarrow.(3\hat { i }-4\hat { j }+5k)=q$

Angle b/w line and plane is given by 

$\sin\theta =\dfrac{4 _1a _2+b _1b _2+c _1c _2}{\sqrt{a _1^2+b _1^2+c _1^2}\sqrt{a _2^2+b _2^2+c _2^2}}$   

Where $(a _1,b _1,c _1)$ and $(a _2,b _2,c _2)$ are direction ratios of line and plane Respectively so here 

$a _1,b _1,c _1)=(1,1,1)$ and $(a _2,b _2,c _2)=(3,-4,5)$

So $\sin \theta=\dfrac{3-4+5}{\sqrt{1+1+1}\sqrt{9+16+25}}$

$\dfrac{4}{\sqrt{3}\sqrt{50}}=\dfrac{4}{\sqrt{3}5\sqrt{2}}=\dfrac{4}{\sqrt{6.5}}\times \dfrac{\sqrt{6}}{\sqrt{6}}=\dfrac{2\sqrt{6}}{5.3}=\dfrac{2\sqrt{6}}{15}$

so here $\sin\theta =\dfrac{2\sqrt{6}}{15} \Rightarrow \theta =\sin\dfrac{2\sqrt{6}}{15}$