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Question: Show that line \[\vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j})\] intersects with the li...

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

Explanation

Solution

Now we know that the two lines r=a1+λb1\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}} and r=a2+λb2\vec{r}={{\vec{a}}_{2}}+\lambda {{\vec{b}}_{2}} are parallel if b1=b2{{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 r=a1i^+a2j^+a3k^+λ(b1i^+b2j^+b3k^)\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 (a1+λb1)i^+(a2+λb2)j^+(a3+λb3)k^\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 μ\mu such that we will get the same equation for a point expressed by both lines.
Hence equating coefficient of i^,j^,k^\hat{i},\hat{j},\hat{k} in the expression for general point on line we will get the conditions on λ and μ\mu

Complete step-by-step answer:
Now we are given with lines r=i^+j^k^+λ(3i^j^)\vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j}) intersects with the line r=4i^k^+μ(2i^+3k^)\vec{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k})
Now we know that the two lines r=a1+λb1\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}} and r=a2+λb2\vec{r}={{\vec{a}}_{2}}+\lambda {{\vec{b}}_{2}} are parallel if b1=b2{{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 r=i^+j^k^+λ(3i^j^)\vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (3\hat{i}-\hat{j}) .

& \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{aligned}$$ Hence any random point on this line will look like $(1+3\lambda ,1-\lambda ,-1).........(1)$. Now consider the line $$\vec{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k})$$ $$\begin{aligned} & \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{aligned}$$ 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. $\begin{aligned} & 4+2\mu =1+3\lambda ............(4) \\\ & 0=1-\lambda ......................(5) \\\ & -1+3\mu =-1...............(6) \\\ \end{aligned}$ Now first consider equation (5) . $\begin{aligned} & 0=1-\lambda \\\ & \Rightarrow \lambda =1 \\\ \end{aligned}$ Hence solving equation (5) we get the value of λ is equal to 1 Now consider equation (6) $\begin{aligned} & -1+3\mu =-1 \\\ & \Rightarrow 3\mu =1-1=0 \\\ & \Rightarrow \mu =0 \\\ \end{aligned}$ Hence we get the value of $\mu $ is equal to 0. Now let us substitute λ = 1 and $\mu $= 0. In equation (4). $\begin{aligned} & LHS=4+2\mu \\\ & =4+2(0) \\\ & =4 \\\ & RHS=1+3\lambda \\\ & =1+3(1) \\\ & =4 \\\ \end{aligned}$ 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