Question

Find the vector and cartesian forms of the equation of the plane passing through the point $(1,2,4)$ and parallel to the lines
$\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$
$\overrightarrow{r}=\hat{i}-3\hat{j}+5\hat{k}+\mu (\hat{i}+\hat{j}-\hat{k})$
Also, the distance of the point $(9,8,10)$ from the plane is $\frac{a}{\sqrt{b}}.$ Find the value of $a+b$

• A. $156$
• B. $160$
• C. $170$
• D. $176$

Answer 1

The correct option is D $176$

$\overrightarrow{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})$
$\overrightarrow{r}=\hat{i}-3\hat{j}+5\hat{k}+\mu (\hat{i}+\hat{j}-\hat{k})$ are equations of given lines
Vector equation of a plane passing through a point $c$ and parallel to vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\overrightarrow{r}=c+t\overrightarrow{a}+s\overrightarrow{b}$
i.e $\overrightarrow{r}=(\hat{i}+2\hat{j}=4\hat{k})+t(2\hat{i}+3\hat{j}+6\hat{k})+s(\hat{i}+\hat{j}-\hat{k})$
Cartesian form of given lines are
$\frac{x-1}{2}$ = $\frac{y-2}{3}$ = $\frac{z+4}{6}$
$\frac{x-1}{1}$ = $\frac{y+3}{1}$ = $\frac{z-5}{-1}$
Equation of plane passing through $(1,2,4)$ and parallel to above lines is $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|$
$\to \left|\begin{array}{ccc}x-1& y-2& z-4\\ 2& 3& 6\\ 1& 1& -1\end{array}\right|$
Required equation of plane is $9x-8y+z+3=0$
Distance of the point $(9,8,10)$ from the plane is $\frac{30}{\sqrt{146}}$

$a=30andb=146$
$\therefore a+b=176$