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Question

$$\displaystyle \vec{r_{1}}=\hat{i}-\hat{j}+3\hat{k}+\lambda\left ( \hat{i}-\hat{j}+\hat{k} \right )$$
$$\displaystyle \vec{r_{2}}=2\hat{i}+4\hat{j}+6\hat{k}+\mu\left ( 2\hat{i}+\hat{j}+3\hat{k} \right )$$


Solution

In this case the two sets of direction ratios are $$1:-1:1$$ and $$2:1:3$$.
They are not equal, so $$\text{these two lines are not parallel.}$$
Now, if the lines intersect it will be at a point where $${ r }_{ 1 }={ r }_{ 2 }$$ i.e., when
$$\left( 1+\alpha  \right) i-\left( 1+\alpha  \right) j+\left( 3+\alpha  \right) k=2\left( 1+\mu  \right) i+\left( 4+\mu  \right) j+\left( 6+3\mu  \right) k$$
Equating the coefficients of $$i$$ and $$j$$, we have
$$1+\alpha =2\left( 1+\mu  \right) $$ and $$-\left( 1+\alpha  \right) =4+\mu $$
Hence, $$\mu =-2$$ and $$\alpha =-3$$
With these values of $$\alpha $$ and $$\mu $$, the coefficients of $$k$$ become
for first line $$3+\alpha =0$$
for second line $$6+3\mu =0$$
Both have equal values. 
So $${ r }_{ 1 }={ r }_{ 2 }$$ when $$\alpha =-3,\mu =-2$$
$$\text{Therefore, the lines intersect at  the point with position vector.}$$
$$\left( 1-3 \right) i-\left( 1-3 \right) j+\left( 3-3 \right) k$$ i.e., $$-2i+2j$$

Physics

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