Q: If \$ | A \times B | = \sqrt3 A.B , \$ then the value of \$ | A + B | \$ is :

We know that,\$ A \times B = AB \sin \theta \$Also, \$ A.B = AB \cos \theta \$Given : \$ | A \times B | = \sqrt3 A.B \$Using \$ | A \times B | = |A| |B| \sin \theta \$We get \$|A\times B| = AB \sin \theta \$\$ A.B = |A||B| \cos \theta \$\$\therefore\$ \$ AB \sin \theta = \sqrt3 AB \cos \theta \$\$ \tan \theta = \sqrt3 \implies \theta = 60^o \$Now \$ (A+B)^2 = A^2 + B^2 + 2 A.B \$\$ = A^2 + B^2 + 2 AB \cos\theta\$\$ = A^2 + B^2 + 2AB\times \dfrac {1}{2} \$\$ = A^2 + B^2 + AB \$or \$ |A+B| = ( A^2 + B^2 + AB )^{1/2} \$

Q: The distance of the point (1,-2,3) from the plane x-y+z=5, measured parallel to the line \$\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-6}\$

Point is \$(1, -2, 3)\$Equation of plane = \$x-y+z=5\$Equation of parallel line is \$\dfrac{x}{2} = \dfrac{y}{3} =\dfrac{z}{-6}\$\$\dfrac{x-1}{2} = \dfrac{y+2}{3} = \dfrac{z-3}{-6} = \lambda\$\$x=2\lambda + 1 , y=3\lambda-2 , z=-6\lambda+3\$Since,\$x-y+z=5\$\$2\lambda+1-3\lambda+2-6\lambda+3=5\$\$-7\lambda+6=5\$\$\lambda=\dfrac{1}{7}\$Thus,\$x=\dfrac{9}{7} , y=\dfrac{-11}{7} , z=\dfrac{15}{7}\$Therefore,Distance = \$\sqrt{(\dfrac{2}{7})^{2} + (\dfrac{3}{7})^{2} + (\dfrac{-6}{7})^{2}}\$ = \$\sqrt{\dfrac{49}{49}}\$ = \$1\$ unit

Q: A line makes angles \$\alpha,\beta,\gamma,\delta\$ with the four diagonals of a cube then \$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta\$ is equal to

REF.ImageTake 'O' as a corner\$OA,OB,OC\$ are 3 edges through the axesLet \$OA=OB=OC=a\$coordinates of \$O=(o,o,o)\$\$A(a,o,o) B(o,a,o) C(o,o,a)\$\$P(a,a,o) L(o,a,a) M(a,o,a) N(a,a,o)\$The four diagonals \$OP, AL, BM, CN\$Direction cosine of \$OP : a-o,a-o,a-o = a,a,a = 1,1,1\$Direction cosine of \$AL : o-a,a-o,a-o = -a,a,a = -1,1,1\$Direction cosine of \$BM : a-o,o-a,a-o = a,-a,a = 1,-1,1\$Direction cosine of \$CN : a-o,a-o,o-a = a,a,-a = 1,1,-1\$\$\therefore \$ DC's of OP are \$\displaystyle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\$DC's of AL are \$\displaystyle\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\$DC's of BM are \$\displaystyle\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\$DC's of CN are \$\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}}\$Let l,m,n be dc's of line and line makes angle \$\alpha \$with OP :- \$\displaystyle \cos \alpha = l(\frac{1}{\sqrt{3}})+m(\frac{1}{\sqrt{3}})+n(\frac{1}{\sqrt{3}})=\frac{l+m+n}{\sqrt{3}}\$Similarly \$\cos\, \beta =\dfrac{-l+m+n}{\sqrt{3}}\$ \$\cos \delta = \dfrac{l+m-n}{\sqrt{3}}\$\$\cos\gamma =\dfrac{l-m+n}{\sqrt{3}}\$suaring and adding all the fouri.e ; \$\cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta \$\$= \dfrac{1}{3}[(l+m+n)^{2}+(-l+m+n)^{2}+(l-m+n)^{2}+(l+m-n)^{2}]\$\$=\dfrac{1}{3}[4l^{2}+4m^{2}+4n^{2}]=\dfrac{4}{3}(l^{2}+m^{2}+n^{2})\$\$[\because l^{2}+m^{2}+n^{2}=1]=\dfrac{4}{3}\$\$\therefore \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta =\dfrac{4}{3}\$

Q: Find the vector equation of the passing through the point \$2i+3j+k\$ and parallel to the vector \$4i-2j+3k\$.

The line passes through the point, \$\overrightarrow a = 2\hat i + 3\hat j + \hat k\$ And parallel to the line, \$\overrightarrow b = 4\hat i - 2\hat j + 3\hat k\$ The vector equation of the line passing through a point and parallel to the given line is, \$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b \$ Substituting the values, we get, \$\overrightarrow r = \left( {2\hat i + 3\hat j + \hat k} \right) + \lambda \left( {4\hat i - 2\hat j + 3\hat k} \right)\$ \$\overrightarrow r = \left( {2 + 4\lambda } \right)\hat i + \left( {3 - 2\lambda } \right)\hat j + \left( {1 + 3\lambda } \right)\hat k\$

Q: The equation of the plane containing the two lines \$\displaystyle \frac { x-1 }{ 2 } =\frac { y+1 }{ -1 } =\frac { z }{ 3 } \$ and \$\displaystyle \frac { x }{ 2 } =\frac { y-2 }{ -1 } =\frac { z+1 }{ 3 } \$ is

Equation of any plane through the first line is \$a\left( x-1 \right) +b\left( y+1 \right) +cz=0\$ ...(1)where \$2a-b+3c=0\$ ...(2)It will pass through the second line if \$(0,2,-1),\$ a point on the second line lies on it i.e., if \$-a+3b-c=0\$ ...(3)Solving (2) and (3), we get \$\displaystyle \frac { a }{ -8 } =\frac { b }{ -1 } =\frac { c }{ 5 } \$ or \$\displaystyle \frac { a }{ 8 } =\frac { b }{ 1 } =\frac { c }{ -5 } \$Hence, equation of required plane is \$8\left( x-1 \right) +\left( y+1 \right) -5z=0\$\$\Rightarrow 8x+y-5z-7=0.\$

Q: The image of the point \$(1,3,4)\$ in the plane \$2x -y+z+3=0\$ is

Let \$ P (1,3,4) \$ be a point Let M be the point on the planeEquation of the plane is \$ 2x-y+z+3 = 0 \$Thus equation of the plane is;\$ \dfrac{x-1}{2} = \dfrac{y-3}{-1} = \dfrac{z-4}{1} = k \$Any point on the above line, PM is of the form\$ x = 2k+1;y = -k+3; z = k+4 \$Substituting the above value in the equationof the place we have\$ 2(2k+1)-(-k+3)+(k+4)+3 = 0 \$\$ \Rightarrow 4k+2+k-3+k+4+3 = 0 \$\$ \Rightarrow 6k+6 = 0 \$\$ k = -1 \$Thus coordinates on M are\$ x = 2(-1)+1 = -1 \$\$ y =-(-1)+3 = 4 \$\$ z = (-1)+4 = 3 \$Let \$ Q (x_{1},y_{1},z_{1}) \$ be the image of PThus M is the midpoint of PQ Therefore,\$ -1 = \dfrac{1+x_{1}}{2}; 4 = \dfrac{3+y_{1}}{2}; 3 = \dfrac{4+z_{1}}{2} \$\$ \Rightarrow x_{1} = -3, y_{1} = 5, z_{1} = 2 \$

Q: Find the direction cosines of a line which makes equal angles with the coordinate axes.

Let the direction cosines of the line make an angle \$a\$ with each of the coordinate axes.\$\therefore\$ \$l=\cos {a}, m=\cos {a}, n=\cos {a}\$We know \${l}^{2}+{m}^{2}+{n}^{2}=1\$\$\Rightarrow\$ \$\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \alpha } +\cos ^{ 2 }{ \alpha } =1\$\$\Rightarrow\$ \$3\cos ^{ 2 }{ \alpha } =1\$\$\Rightarrow \$ \$\cos ^{ 2 }{ \alpha } =\cfrac{1}{3}\$\$\Rightarrow\$ \$\cos {\alpha}=\pm \cfrac{1}{\sqrt 3}\$Thus, the direction cosines of the line which is equally inclined to the coordinate axes are \$\pm \cfrac{1}{\sqrt 3}, \pm \cfrac{1}{\sqrt 3}\$ and \$\pm \cfrac{1}{\sqrt 3}\$.

Q: If a line makes angles \$\alpha, \beta, \gamma\$ with the coordinate axes, then the value of \$\cos 2\alpha + \cos 2\beta + \cos 2\gamma\$ is

\$\cos 2\alpha + \cos 2\beta + \cos 2\gamma\$\$= 2(\cos^{2} \alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3\$\$= 2 - 3 = -1\$\$(\because \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1)\$

Q: Find shortest distance between lines \$\vec r=\hat i+2\hat j+3\hat k+\lambda (2 \hat i+\hat j+4\hat k)\$ and\$\vec r=2\hat i+4 \hat j+5\hat k+\mu( 3\hat i+4\hat j+5\hat k)\$

Given eq of lines \$\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k})\$----(1)\$\vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu(3\hat{i}+4\hat{j}+5\hat{k})\$----(2)position vector and normal vector of line (1)\$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\$\$\vec{n_{1}}=2\hat{i}+3\hat{j}+4\hat{k}\$position vector and normal vector of line (2)\$\vec{c}=2\hat{i}+4\hat{j}+5\hat{k}\$\$\vec{n_{2}}=3\hat{i}+4\hat{j}+5\hat{k}\$shortest between two skews line \$SD=\dfrac{\vec{AC}\cdot(\vec{n_{1}}\times\vec{n_{2}})}{\left | \vec{n_{1}}\times\vec{n_{2}} \right |}\$\$AC=\vec{c}-\vec{a}\$\$AC=2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})\$\$AC=2\hat{i}+4\hat{j}+5\hat{k}-\hat{i}-2\hat{j}-3\hat{k}\$\$AC=\hat{i}+2\hat{j}+2\hat{k}\$\$\vec{n_{1}}\times\vec{n_{2}}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\3&4&5\end{vmatrix}\$\$\vec{n_{1}}\times\vec{n_{2}}=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)\$\$\vec{n_{1}}\times\vec{n_{2}}=-\hat{i}+2\hat{j}-\hat{k}\$\$\left | \vec{n_{1}}\times\vec{n_{2}} \right |=\sqrt{(-1)^2+2^2+(-1)^2}\$\$\left | \vec{n_{1}}\times\vec{n_{2}} \right |=\sqrt{6}\$putting AC \$\vec{n_{1}}\times\vec{n_{2}},\left | \vec{n_{1}}\times\vec{n_{2}} \right |\$ in formula\$SD=\dfrac{\vec{AC}\cdot(\vec{n_{1}}\times\vec{n_{2}})}{\left | \vec{n_{1}}\times\vec{n_{2}} \right |}\$\$SD=\dfrac{(\hat{i}+2\hat{j}+2\hat{k})\cdot(-\hat{i}+2\hat{j}-\hat{k})}{\sqrt{6}}\$\$SD=\dfrac{-1+4-2}{\sqrt{6}}\$\$SD=\dfrac{1}{\sqrt{6}}\$

Question 1

If $ | A \times B | = \sqrt3 A.B , $ then the value of $ | A + B | $ is :

Accepted Answer

We know that,$ A \times B = AB \sin \theta $Also,  $ A.B = AB \cos \theta $Given :  $ | A \times B | = \sqrt3 A.B $Using  $ | A \times B | = |A| |B| \sin \theta $We get  $|A\times B| = AB \sin \theta $$ A.B = |A||B| \cos \theta $$\therefore$  $ AB \sin \theta = \sqrt3 AB \cos \theta $$ \tan \theta = \sqrt3   \implies   \theta = 60^o $Now $ (A+B)^2 = A^2 + B^2 + 2 A.B $$ = A^2 + B^2 + 2 AB \cos\theta$$ = A^2 + B^2 + 2AB\times  \dfrac {1}{2} $$ = A^2 + B^2 + AB $or $ |A+B| = ( A^2 + B^2 + AB )^{1/2} $

Question 2

The distance of the point (1,-2,3) from the plane x-y+z=5, measured parallel to the line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{-6}$

Accepted Answer

Point is $(1, -2, 3)$Equation of plane = $x-y+z=5$Equation of parallel line is $\dfrac{x}{2} = \dfrac{y}{3} =\dfrac{z}{-6}$$\dfrac{x-1}{2} = \dfrac{y+2}{3} = \dfrac{z-3}{-6} = \lambda$$x=2\lambda + 1 , y=3\lambda-2 , z=-6\lambda+3$Since,$x-y+z=5$$2\lambda+1-3\lambda+2-6\lambda+3=5$$-7\lambda+6=5$$\lambda=\dfrac{1}{7}$Thus,$x=\dfrac{9}{7} , y=\dfrac{-11}{7} , z=\dfrac{15}{7}$Therefore,Distance = $\sqrt{(\dfrac{2}{7})^{2} + (\dfrac{3}{7})^{2} + (\dfrac{-6}{7})^{2}}$ = $\sqrt{\dfrac{49}{49}}$ = $1$ unit

Question 3

A line makes angles $\alpha,\beta,\gamma,\delta$ with the four diagonals of a cube then $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma+\cos^{2}\delta$ is equal to

Accepted Answer

REF.ImageTake 'O' as a corner$OA,OB,OC$ are 3 edges through the axesLet $OA=OB=OC=a$coordinates of $O=(o,o,o)$$A(a,o,o) B(o,a,o) C(o,o,a)$$P(a,a,o) L(o,a,a) M(a,o,a) N(a,a,o)$The four diagonals $OP, AL, BM, CN$Direction cosine of $OP : a-o,a-o,a-o = a,a,a = 1,1,1$Direction cosine of $AL : o-a,a-o,a-o = -a,a,a = -1,1,1$Direction cosine of $BM : a-o,o-a,a-o = a,-a,a = 1,-1,1$Direction cosine of $CN : a-o,a-o,o-a = a,a,-a = 1,1,-1$$\therefore $ DC's of OP are $\displaystyle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$DC's of AL are $\displaystyle\dfrac{-1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$DC's of BM are $\displaystyle\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$DC's of CN are $\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{-1}{\sqrt{3}}$Let l,m,n be dc's of line and line makes angle $\alpha $with OP :- $\displaystyle \cos \alpha = l(\frac{1}{\sqrt{3}})+m(\frac{1}{\sqrt{3}})+n(\frac{1}{\sqrt{3}})=\frac{l+m+n}{\sqrt{3}}$Similarly $\cos\, \beta =\dfrac{-l+m+n}{\sqrt{3}}$ $\cos \delta  = \dfrac{l+m-n}{\sqrt{3}}$$\cos\gamma =\dfrac{l-m+n}{\sqrt{3}}$suaring and adding all the fouri.e ; $\cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta $$= \dfrac{1}{3}[(l+m+n)^{2}+(-l+m+n)^{2}+(l-m+n)^{2}+(l+m-n)^{2}]$$=\dfrac{1}{3}[4l^{2}+4m^{2}+4n^{2}]=\dfrac{4}{3}(l^{2}+m^{2}+n^{2})$$[\because l^{2}+m^{2}+n^{2}=1]=\dfrac{4}{3}$$\therefore \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta =\dfrac{4}{3}$

Question 4

Find the vector equation of the passing through the point $2i+3j+k$ and parallel to the vector $4i-2j+3k$.

Accepted Answer

The line passes through the point,&#10;&#10;$\overrightarrow a &#10;= 2\hat i + 3\hat j + \hat k$&#10;&#10;And parallel to the line,&#10;&#10;$\overrightarrow b &#10;= 4\hat i - 2\hat j + 3\hat k$&#10;&#10;The vector equation of the line passing through a&#10;point and parallel to the given line is,&#10;&#10;$\overrightarrow r &#10;= \overrightarrow a  + \lambda&#10;\overrightarrow b $&#10;&#10;Substituting the values, we get,&#10;&#10;$\overrightarrow r &#10;= \left( {2\hat i + 3\hat j + \hat k} \right) + \lambda \left( {4\hat i&#10;- 2\hat j + 3\hat k} \right)$&#10;&#10;$\overrightarrow r &#10;= \left( {2 + 4\lambda } \right)\hat i + \left( {3 - 2\lambda }&#10;\right)\hat j + \left( {1 + 3\lambda } \right)\hat k$

Question 5

The equation of the plane containing the two lines $\displaystyle \frac { x-1 }{ 2 } =\frac { y+1 }{ -1 } =\frac { z }{ 3 } $ and $\displaystyle \frac { x }{ 2 } =\frac { y-2 }{ -1 } =\frac { z+1 }{ 3 } $ is

Accepted Answer

Equation of any plane through the first line is $a\left( x-1 \right) +b\left( y+1 \right) +cz=0$    ...(1)where $2a-b+3c=0$    ...(2)It will pass through the second line if $(0,2,-1),$ a point on the second line lies on it i.e., if $-a+3b-c=0$     ...(3)Solving (2) and (3), we get $\displaystyle \frac { a }{ -8 } =\frac { b }{ -1 } =\frac { c }{ 5 } $ or $\displaystyle \frac { a }{ 8 } =\frac { b }{ 1 } =\frac { c }{ -5 } $Hence, equation of required plane is $8\left( x-1 \right) +\left( y+1 \right) -5z=0$$\Rightarrow 8x+y-5z-7=0.$

Question 6

The image of the point $(1,3,4)$ in the plane $2x -y+z+3=0$ is

Accepted Answer

Let $ P (1,3,4) $ be a point Let M be the point on the planeEquation of the plane is $ 2x-y+z+3 = 0 $Thus equation of the plane is;$ \dfrac{x-1}{2} = \dfrac{y-3}{-1} = \dfrac{z-4}{1} = k $Any point on the above line, PM is of the form$ x = 2k+1;y = -k+3; z = k+4 $Substituting the above value in the equationof the place we have$ 2(2k+1)-(-k+3)+(k+4)+3 = 0 $$ \Rightarrow 4k+2+k-3+k+4+3 = 0 $$ \Rightarrow 6k+6 = 0 $$ k = -1 $Thus coordinates on M are$ x = 2(-1)+1 = -1 $$ y =-(-1)+3 = 4 $$ z = (-1)+4 = 3 $Let $ Q (x_{1},y_{1},z_{1}) $ be the image of PThus M is the midpoint of PQ Therefore,$ -1 = \dfrac{1+x_{1}}{2}; 4 = \dfrac{3+y_{1}}{2}; 3 = \dfrac{4+z_{1}}{2} $$ \Rightarrow x_{1} = -3, y_{1} = 5, z_{1} = 2 $

Question 7

Find the equation of the plane passing through the following points:$(2,3,4), (-3, 5, 1)$ and $(4,-1,2)$

Accepted Answer

Vector equation of a plane passing through three points with position vectors $\vec{a},\vec{b},\vec{c}$ is $\left(\vec{r}-\vec{a}\right).\left[\left(\vec{b}-\vec{a}\right)\times\left(\vec{c}-\vec{a}\right)\right]=0$Now, the plane passes through the points$A\left(2,3,4\right),\,B\left(-3,5,1\right)$ and $C\left(4,-1,2\right)$$\vec{b}-\vec{a}=-3\hat{i}+5\hat{j}+\hat{k}-2\hat{i}-3\hat{j}-4\hat{k}=-5\hat{i}+2\hat{j}-3\hat{k}$$\vec{c}-\vec{a}=4\hat{i}-\hat{j}+2\hat{k}-2\hat{i}-3\hat{j}-4\hat{k}=2\hat{i}-4\hat{j}-2\hat{k}$$\left(\vec{b}-\vec{a}\right)\times\left(\vec{c}-\vec{a}\right)=\left| \begin{matrix} \hat{i} & \hat{j}  & \hat{k} \ -5 & 2 & -3 \ 2 & -4  & -2 \end{matrix} \right|$$=\left(-4-12\right)\hat{i}-\left(10+6\right)\hat{j}+\left(20-4\right)\hat{k}$$=-16\hat{i}-16\hat{j}+16\hat{k}$$\therefore\,$vector equation of plane is $\left(x\hat{i}+y\hat{j}+z\hat{k}-2\hat{i}-3\hat{j}-4\hat{k}\right)\left(8\hat{i}-16\hat{j}-16\hat{k}\right)=0$$\Rightarrow\,\left[\left(x-2\right)\hat{i}+\left(y-3\right)\hat{j}+\left(z-4\right)\hat{k}\right]\left(8\hat{i}-16\hat{j}-16\hat{k}\right)=0$$\Rightarrow\,8x-16-16y+48-16z+64=0$$\Rightarrow\,8x-16y-16z+96=0$or $x-2y-2z+12=0$ is the required equation of the plane.

Question 8

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Accepted Answer

Let the direction cosines of the line make an angle $a$ with each of the coordinate axes.$\therefore$ $l=\cos {a}, m=\cos {a}, n=\cos {a}$We know ${l}^{2}+{m}^{2}+{n}^{2}=1$$\Rightarrow$ $\cos ^{ 2 }{ \alpha  } +\cos ^{ 2 }{ \alpha  } +\cos ^{ 2 }{ \alpha  } =1$$\Rightarrow$ $3\cos ^{ 2 }{ \alpha  } =1$$\Rightarrow $ $\cos ^{ 2 }{ \alpha  } =\cfrac{1}{3}$$\Rightarrow$ $\cos {\alpha}=\pm \cfrac{1}{\sqrt 3}$Thus, the direction cosines of the line which is equally inclined to the coordinate axes are $\pm \cfrac{1}{\sqrt 3}, \pm \cfrac{1}{\sqrt 3}$  and $\pm \cfrac{1}{\sqrt 3}$.

Question 9

If a line makes angles $\alpha, \beta, \gamma$ with the coordinate axes, then the value of $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ is

Accepted Answer

$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$$= 2(\cos^{2} \alpha + \cos^{2}\beta + \cos^{2}\gamma) - 3$$= 2 - 3 = -1$$(\because \cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1)$

Question 10

Find shortest distance between lines $\vec r=\hat i+2\hat j+3\hat k+\lambda (2 \hat i+\hat j+4\hat k)$ and$\vec r=2\hat i+4 \hat j+5\hat k+\mu( 3\hat i+4\hat j+5\hat k)$

Accepted Answer

Given eq of lines $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$----(1)$\vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu(3\hat{i}+4\hat{j}+5\hat{k})$----(2)position vector and normal vector of line (1)$\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$$\vec{n_{1}}=2\hat{i}+3\hat{j}+4\hat{k}$position vector and normal vector of line (2)$\vec{c}=2\hat{i}+4\hat{j}+5\hat{k}$$\vec{n_{2}}=3\hat{i}+4\hat{j}+5\hat{k}$shortest between two skews line $SD=\dfrac{\vec{AC}\cdot(\vec{n_{1}}\times\vec{n_{2}})}{\left | \vec{n_{1}}\times\vec{n_{2}} \right |}$$AC=\vec{c}-\vec{a}$$AC=2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})$$AC=2\hat{i}+4\hat{j}+5\hat{k}-\hat{i}-2\hat{j}-3\hat{k}$$AC=\hat{i}+2\hat{j}+2\hat{k}$$\vec{n_{1}}\times\vec{n_{2}}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\2&3&4\3&4&5\end{vmatrix}$$\vec{n_{1}}\times\vec{n_{2}}=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)$$\vec{n_{1}}\times\vec{n_{2}}=-\hat{i}+2\hat{j}-\hat{k}$$\left | \vec{n_{1}}\times\vec{n_{2}} \right |=\sqrt{(-1)^2+2^2+(-1)^2}$$\left | \vec{n_{1}}\times\vec{n_{2}} \right |=\sqrt{6}$putting AC $\vec{n_{1}}\times\vec{n_{2}},\left | \vec{n_{1}}\times\vec{n_{2}} \right |$ in formula$SD=\dfrac{\vec{AC}\cdot(\vec{n_{1}}\times\vec{n_{2}})}{\left | \vec{n_{1}}\times\vec{n_{2}} \right |}$$SD=\dfrac{(\hat{i}+2\hat{j}+2\hat{k})\cdot(-\hat{i}+2\hat{j}-\hat{k})}{\sqrt{6}}$$SD=\dfrac{-1+4-2}{\sqrt{6}}$$SD=\dfrac{1}{\sqrt{6}}$