Question

Find the equation of a line passes through $(1,2,3)$ and perpendicular to the vector $(5\hat{i}-3\hat{j}+4\hat{k})$ and $(-7\hat{i}+4\hat{j}+9\hat{k})$

Answer 1

$Let\overrightarrow{r}=x\hat{i}+y\hat{j}+2\hat{k}$

passes through $(1,2,3)$

$\Rightarrow \therefore \overrightarrow{a}=i+2\hat{j}+3\hat{k}$

and direct ratio of $\left(5\hat{i}-3\hat{j}+4k\right)$ and $=\left(-7\hat{i}+4\hat{j}+9\hat{k}\right)$

$\overrightarrow{N}=-12\hat{i}+7\hat{j}+5\hat{k}$

Then, equation of line

$\begin{array}{c}\left(\overrightarrow{r}-\overrightarrow{a}\right).\overrightarrow{N}=0\\ \Rightarrow \left(x\hat{i}+y\hat{j}+2\hat{k}-i-2\hat{j}-3\hat{k}\right)\left(-12\hat{i}+7\hat{j}+5\hat{k}\right)=0\\ \Rightarrow \left\{\left(x-1\right)i+\left(y-2\right)\hat{j}+\left(z-3\right)\hat{k}\right\}.\left(-12\hat{i}+7\hat{j}+5\hat{k}\right)=0\\ \Rightarrow -12\left(x-1\right)+7\left(y-2\right)+5\left(z-3\right)=0\\ \Rightarrow -12x+12+7y-14+5z-15=0\\ -12x+7y+5z-17=0.\end{array}$

$12x-7y-5z+17=0$

Hence, this is the answer.