Straight Lines and Angles

This is a standard vector geometry problem involving the angle between a vector and a plane defined by two other vectors. The formula derived for the angle between a unit vector and the plane of two other unit vectors equally inclined at theta is indeed cos^-1(cos(theta) / cos(theta/2)).

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

The sine of the angle between a line with direction vector v and a plane with normal vector n is given by |v.n| / (|v||n|). Here v = (1, -1, 1) and n = (2, -1, 1). The dot product is 2 + 1 + 1 = 4, and the magnitudes are sqrt(3) and sqrt(6). Thus sin(theta) = 4 / (sqrt(3)*sqrt(6)) = 4 / (3*sqrt(2)) = 2*sqrt(2)/3. Using cos(theta) = sqrt(1 - sin^2(theta)), we calculate the expression.

The angle theta between a line with direction (1, 2, lambda) and a plane with normal (1, 2, 3) satisfies sin(theta) = |(1, 2, lambda).(1, 2, 3)| / (sqrt(1+4+lambda^2) * sqrt(1+4+9)). Given cos(theta) = sqrt(5/14), then sin(theta) = sqrt(1 - 5/14) = 3/sqrt(14). Solving |5 + 3*lambda| / (sqrt(5+lambda^2) * sqrt(14)) = 3/sqrt(14) leads to lambda = 3/2.

This is a variation of the previous question with a corrected direction vector for the line. The steps involve finding the plane equation using the point and line, then calculating the angle between the plane's normal and the x-axis vector.

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 $a _1=1, a _2=2, a _3=\lambda$

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

According to the question:

The line is x/2 = (y-2)/3 = (z-3)/4 = k. Point P = (2k, 3k+2, 4k+3). Vector from P to (3, -1, 11) is (3-2k, -3-3k, 8-4k). This must be perpendicular to the line direction (2, 3, 4). 2(3-2k) + 3(-3-3k) + 4(8-4k) = 0 => 6-4k-9-9k+32-16k = 0 => 29-29k=0 => k=1. Point P = (2, 5, 7). Distance = sqrt((3-2)^2 + (-1-5)^2 + (11-7)^2) = sqrt(1 + 36 + 16) = sqrt(53).