Question

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{a}& \text{The distance between the line}\overrightarrow{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+4\hat{k})& \text{p}& \frac{25}{3\sqrt{14}}\\ & \text{and plane}\overrightarrow{r}.(\hat{i}+5\hat{j}+\hat{k})=5& & \\ \text{b}& \text{The distance between parallel planes}\overrightarrow{r}.(2\hat{i}-\hat{j}+3\hat{k})=4\text{and}& \text{q}& \frac{13}{7}\\ & \overrightarrow{r}.(6\hat{i}-3\hat{j}+9\hat{k})+13=0\text{is}& & \\ \text{c}& \text{The distance of a point (2, 5, -3) from the plane}\overrightarrow{r}.(6\hat{i}-3\hat{j}+2\hat{k})=4& \text{r}& \frac{10}{3\sqrt{3}}\\ & \text{is}& & \\ \text{d}& \text{The distance of the point (1, 0, -3) from the plane}x-y-z-9=0& \text{s}& 7\\ & \text{measured parallel to line}\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}& & \end{array}$

Then which of the following is correct ?

• A. $a\to r;b\to p;c\to q;d\to s$
• B. $a\to s;b\to p;c\to q;d\to r$
• C. $a\to r;b\to q;c\to p;d\to s$
• D. $a\to s;b\to q;c\to p;d\to r$

Answer 1

The correct option is A $a\to r;b\to p;c\to q;d\to s$

$a.$
The given line and plane are $\overrightarrow{r}=(2\hat{i}-2\hat{j}+3\hat{k})+\lambda (\hat{i}-\hat{j}+4\hat{k})$ and $\overrightarrow{r}.(\hat{i}+5\hat{j}+\hat{k})=5$, respectively.
Since $(\hat{i}-\hat{j}+4\hat{k}).(\hat{i}+5\hat{j}+\hat{k})=0$, line and plane are parallel.
Hence, the required distance is equal to distance of point $(2,-2,3)$ from the plane $x+5y+z-5=0$, which is $\frac{|2-10+3-5|}{\sqrt{1+25+1}}=\frac{10}{3\sqrt{3}}$

$b.$
The distance between two parallel planes $\overrightarrow{r}.(2i-j+3k)=4$ and $\overrightarrow{r}.(6i-3j+9k)+13=0$ is
$d=\frac{|4-\left(\frac{-13}{3}\right)|}{\sqrt{(2{)}^2+(-1{)}^2+(3{)}^2}}=\frac{\left(\frac{25}{3}\right)}{\sqrt{14}}$
$=\frac{25}{3\sqrt{14}}$

$c.$
The perpendicular distance of the point $(2,5,-3)$ from the plane $\overrightarrow{r}.(6i-3j+2k)=4$ or $6x-3y+2z-4=0$ is
$d=\frac{|12-15-6-4|}{\sqrt{36+9+4}}$
$=\frac{13}{7}$

$d.$
The equation of the line $AB$ is,
$\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$
The equation of line passing through $(1,0,-3)$ and parallel to $AB$ is
$\frac{x-1}{2}=\frac{y}{3}=\frac{z+3}{-6}=r$ (say)

The coordinates of any point on line $P(2r+1,3r,-6r-3)$ which lie on the plane
$(2r+1)-\left(3r\right)-(-6r-3)=9$
$\Rightarrow r=1$
Point $P\equiv (3,3,-9)$
Required distance
$PQ=\sqrt{(3-1{)}^2+(3-0{)}^2+(-9+3{)}^2}$
$=\sqrt{4+9+36}=7$