question_answer

Find the angle between the line

\[\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\] and the plane \[10x+2y-11z=3.\]

OR

Find the equation of the plane which contains the line of intersection of

planes   \[\vec{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})-4=0,\]

\[\vec{r}\cdot (2\hat{i}+\hat{j}+\hat{k})-15=0\] and is perpendicular to the plane

\[\vec{r}\cdot (5\hat{i}+3\hat{j}-6\hat{k})+8=0.\]

Answer:

The angle between a line\[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\]

and the normal to the plane\[Ax+By+Cz=D\]is given by

\[\cos \,(90{}^\circ -\phi )=\left| \frac{Aa+Bb+Cc}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}} \right|\]

So, the angle between line and plane is given by

\[\sin \phi =\left| \frac{Aa+Bb+Cc}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}} \right|\]

Given line is

\[\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\Rightarrow \frac{x-(-\,1)}{2}=\frac{y-0}{3}=\frac{z-3}{6}\]

Comparing it with\[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c},\]we get

\[a=2,\]\[b=3\]and \[c=6\]

The plane is\[10x+2y-11z=3\]

Comparing it with\[Ax+By+Cz=D,\]we get

\[A=10,\]\[B=2,\]and \[C=-\,11\]

\[\therefore \]\[\sin \phi =\left| \frac{(10\times 2)+(2\times 3)+(-\,11\times 6)}{\sqrt{{{2}^{3}}+{{3}^{2}}+{{6}^{2}}}\sqrt{{{10}^{2}}+{{2}^{2}}+{{(-\,11)}^{2}}}} \right|\]

\[=\left| \frac{20+6-66}{\sqrt{4+9+36}\sqrt{10+4+121}} \right|\]

\[=\left| \frac{-\,40}{7\times 15} \right|\,\,=\frac{8}{21}\]

\[\Rightarrow \]   \[\phi ={{\sin }^{-\,1}}\left( \frac{8}{21} \right)\]

\[\therefore \]The angle between the given line and plane is\[{{\sin }^{-\,1}}\left( \frac{8}{21} \right)\].

OR

The equation of any plane through the line of intersection of the given planes is

\[[\overrightarrow{r}\cdot (\hat{i}+2\hat{j}+3\hat{k})-4]+\lambda \,[\overrightarrow{r}\cdot (2\,\hat{i}+\hat{j}+\hat{k})-15]=0\]

as \[\overrightarrow{r}\cdot [(1+2\lambda )\,\hat{i}+(2+\lambda )\hat{j}+(3+\lambda )\hat{k}]=4+15\lambda \]

?(i)

If plane (i) is perpendicular to \[\overrightarrow{r}\cdot (5\,\hat{i}+3\hat{j}-6\hat{k})+8=0,\]

then,

\[[(1+2\lambda )\,\hat{i}+(2+\lambda )\hat{j}+(3+\lambda )\hat{k}]\cdot (5\,\hat{i}+3\hat{j}-6\hat{k})=0\]

\[\Rightarrow \]\[5\,(1+2\lambda )+3\,(2+\lambda )+(3+\lambda )(-\,6)=0\]

\[\Rightarrow \]\[5+10\lambda +6+3\lambda -18-6\lambda =0\]

\[\Rightarrow \]\[7\lambda -7=0\Rightarrow \lambda =1\]

Putting \[\lambda =1\]in Eq. (i), we obtain the equation of the required plane as

\[\overrightarrow{r}\cdot \{(1+2)\,\hat{i}+(2+1)\hat{j}+(3+1)\hat{k}=4+15\}\]

\[\Rightarrow \]   \[\overrightarrow{r}\cdot (3\,\hat{i}+3\hat{j}+4\hat{k})=19\]