  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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