Question

Show that line \[\vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j})\] intersects with the line \[\vec{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k})\] and also find the point of intersection.

Answer 1

Hint: Now we know that the two lines $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{2}}+\lambda {{\vec{b}}_{2}}$ are parallel if ${{b}_{1}}={{b}_{2}}$ .
Hence we know that the lines are not parallel. Now there are two possibilities that the lines are intersecting or skew lines. Now first we know that for \[\vec{r}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}+\lambda \left( {{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k} \right)\] ant point is expressed as $\left( {{a}_{1}}+\lambda {{b}_{1}} \right)\hat{i}+\left( {{a}_{2}}+\lambda {{b}_{2}} \right)\hat{j}+\left( {{a}_{3}}+\lambda {{b}_{3}} \right)\hat{k}$ \[\].
Now we know that lines intersect then there is a common point on both lines and that point is called point of intersection. Now if there is a common point then there will exist one value of λ and $\mu $ such that we will get the same equation for a point expressed by both lines.
Hence equating coefficient of $\hat{i},\hat{j},\hat{k}$ in the expression for general point on line we will get the conditions on λ and $\mu $

Complete step-by-step answer:
Now we are given with lines \[\vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j})\] intersects with the line \[\vec{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k})\]
Now we know that the two lines $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{2}}+\lambda {{\vec{b}}_{2}}$ are parallel if ${{b}_{1}}={{b}_{2}}$ .
Hence we know that the lines are not parallel. Now there are two possibilities that the lines are intersecting or skew lines.
Now first consider the line \[\vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j})\] .
\[\begin{align}
  & \vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j}) \\
 & =\hat{i}+\hat{j}-\hat{k}+3\lambda \hat{i}-\lambda \hat{j} \\
 & \vec{r}=(1+3\lambda )\hat{i}+(1-\lambda )\hat{j}-\hat{k} \\
\end{align}\]
Hence any random point on this line will look like $(1+3\lambda ,1-\lambda ,-1).........(1)$.
Now consider the line \[\vec{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k})\]
\[\begin{align}
  & \vec{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k}) \\
 & =4\hat{i}-\hat{k}+2\mu \hat{i}+3\mu \hat{k} \\
 & =\left( 4+2\mu \right)i+0j+\left( -1+3\mu \right)k \\
\end{align}\]
Hence any random point on the line $\left( (4+2\mu ),0,(-1+3\mu ) \right)............(2)$ .
Now Let us say that the two lines intersect.
Now we know that lines intersect then there is a common point on both lines and that point is called point of intersection. Now if there is a common point then there will exist one value of λ and $\mu $ such that $\left( (4+2\mu ),0,(-1+3\mu ) \right)=(1+3\lambda ,1-\lambda ,-1))$
Now equating this we get three equations.
$\begin{align}
  & 4+2\mu =1+3\lambda ............(4) \\
 & 0=1-\lambda ......................(5) \\
 & -1+3\mu =-1...............(6) \\
\end{align}$
Now first consider equation (5) .
$\begin{align}
  & 0=1-\lambda \\
 & \Rightarrow \lambda =1 \\
\end{align}$
Hence solving equation (5) we get the value of λ is equal to 1
Now consider equation (6)
$\begin{align}
  & -1+3\mu =-1 \\
 & \Rightarrow 3\mu =1-1=0 \\
 & \Rightarrow \mu =0 \\
\end{align}$
Hence we get the value of $\mu $ is equal to 0.
Now let us substitute λ = 1 and $\mu $= 0. In equation (4).
$\begin{align}
  & LHS=4+2\mu \\
 & =4+2(0) \\
 & =4 \\
 & RHS=1+3\lambda \\
 & =1+3(1) \\
 & =4 \\
\end{align}$
Now LHS = RHS . This means the value λ = 1 and $\mu $= 0 satisfies the equation (4).
Hence the lines are not skew lines.
Hence we have that the lines are intersecting
Now from equation (2) we have the intersecting point as $\left( (4+2\mu ),0,(-1+3\mu ) \right)$ Now substituting $\mu $= 0 in the equation we get the point of intersection as $\left( 4,0,-2 \right)$

Note: Now in 2d we have that if the lines are not parallel they are intersecting. But in 3d there is a possibility that the lines are skew lines. Hence once we find the value of λ and $\mu $ we check if it also satisfies the third equation which is not used to find the value of λ and $\mu $. If it satisfies then the lines are intersecting or else they are skew lines