Question

Show that the lines
${{\overrightarrow{r}}_{1}}=3i+2j-4k+\lambda (i+2j+2k),\overrightarrow{{{r}_{2}}}=5i-2j+\mu (3i+2j+6k)$ are intersecting. Hence find their point of intersection.

Answer 1

We will first write both lines in components of i,j,k unit vectors as
${{\overrightarrow{r}}_{1}}=(3+\lambda )i+(2+2\lambda )j+(2\lambda -4)k$ similarly, ${{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k$
Now if they are intersecting it means points are same so we will equate all three components of both lines respectively
$\left( 3+\lambda \right)i=\left( 5+3\mu \right)i$ , $(2+2\lambda )j=(2\mu -2)j$ and $(2\lambda -4)k=6\mu k$
On comparing i component we get $\lambda =2+3\mu$ and j component we get $2\lambda =2\mu -4$
On putting value we get $\lambda =-4$ and $\mu =-2$ , now putting these value back in any of the given lines we get the point of intersection as $-i-6j-12k$

Complete step-by-step answer:
Given two lines as
${{\overrightarrow{r}}_{1}}=3i+2j-4k+\lambda (i+2j+2k),\overrightarrow{{{r}_{2}}}=5i-2j+\mu (3i+2j+6k)$ which are intersecting so we have to find their point of intersection. Here values of $\lambda$ and $\mu$ are unknown.
If the lines are intersecting it means we can equate all three components at a given value of $\lambda$ and $\mu$ from there we will calculate value of $\lambda$ and $\mu$
Splitting in the components we can write ${{\overrightarrow{r}}_{1}}=(3+\lambda )i+(2+2\lambda )j+(2\lambda -4)k$ and ${{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k$ now as they are intersecting , equating each components respectively we get
$\left( 3+\lambda \right)i=\left( 5+3\mu \right)i$, $(2+2\lambda )j=(2\mu -2)j$ and $(2\lambda -4)k=6\mu k$
comparing i component we get $\lambda =2+3\mu .......(1)$ and j component we get $2\lambda =2\mu -4....(2)$
putting equation (1) into the equation (2) we will get expression as $2(2+3\mu )=2\mu -4$
on expanding it looks like $4+6\mu =2\mu -4$ ,on solving we get $4+4\mu =-4$
we get value of $\mu$ equals to -2 , and putting it in equation (1) we get value of $\lambda$ equals
$\lambda =2+3(-2)=2-6=-4$
Now putting back values $\lambda =-4$ and $\mu =-2$ in any of the given line ${{\overrightarrow{r}}_{1}}$ or $\overrightarrow{{{r}_{2}}}$ we will get the intersection point.
${{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k$ putting value of $\mu$ in this equation we will get
$(5+3(-2))i+(2(-2)-2)j+6(-2)k$ which results into $(-1)i+(-6)j+(-12)k$
Hence intersection point is $-i-6j-12k$

Note: In this question we have equated only I and j component but we can consider any two component and equate it, we will get the same result for example we equate j and k component
${{\overrightarrow{r}}_{1}}=(3+\lambda )i+(2+2\lambda )j+(2\lambda -4)k$
, ${{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k$
Equating j component $(2+2\lambda )=(2\mu -2)$ on dividing it by 2 gives $\lambda =\mu -2$
Equating K component $(2\lambda -4)=6\mu$ on dividing it by 2 gives $\lambda =2+3\mu$
Equating both we again got $\lambda =-4$ and $\mu =-2$