Question

The distance of the point $(1,3,-7)$ from the plane passing through the

point $(1,-1,-1)$, having normal perpendicular to both the lines

$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{x-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$, is.

• A. $\frac{20}{\sqrt{74}}$
• B. $\frac{10}{\sqrt{83}}$
• C. $\frac{5}{\sqrt{83}}$
• D. $\frac{10}{\sqrt{74}}$

Answer 1

The correct option is B $\frac{10}{\sqrt{83}}$

$\overline{n}$=$\left[\begin{array}{ccc}i& j& k\\ 1& -2& 3\\ 2& -1& -1\end{array}\right]$

$=5i+7j+3k$

Equation of plane is $\overline{r}.\overline{n}=\overline{n}.\overline{a}$

$\overline{r}.(5i+7j+3k)=(5i+7j+3k)(i-j-k)$

$\overline{r}.(5i+7j+3k)=-5$

Distance =$\frac{5+21-21+5}{\sqrt{5^2+7^2+3^2}}=\frac{10}{\sqrt{83}}$