The equation of the line passing through the point $(1,2,3)$ and perpendicular to the lines…

The equation of the line passing through the point $(1,2,3)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\bar{r}=\lambda(-3 \hat{\imath}+2 \hat{\jmath}+5 \hat{k})$ is
  1. $\bar{r}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+7 \hat{\jmath}-4 \hat{k})$
  2. $\bar{r}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+7 \hat{\jmath}+4 \hat{k})$
  3. $\bar{r}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}-7 \hat{\jmath}-4 \hat{k})$
  4. $\bar{r}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}-7 \hat{\jmath}+4 \hat{k})$

Solution

The vector perpendicular to both the given lines is given by $\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1\end{array}\right|=\hat{i}(-6)-\hat{j}(-7)+\hat{k}(-2)=-6 \hat{i}+7 \hat{j}-2 \hat{k}$ Hence d.r.s. of required line are $6,-7,2$. Thus eq. of required line is $\frac{x-1}{6}=\frac{y-2}{-7}=\frac{2-3}{2} \text { i.e. } \frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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