Question

The points of intersection of the line passing through $\overline{i}-2\overline{j}-\overline{k},\overline{2i}+3\overline{j}+\overline{k}$ and the plane passing through $2\overline{i}+\overline{j}-3\overline{k},4\overline{i}-\overline{j}+2\overline{k},3\overline{i}+\overline{k}$

• A. $\frac{5}{3}\overline{i}+\frac{4}{3}\overline{j}+\frac{1}{3}\overline{k}$
• B. $\frac{5}{3}\overline{i}-\frac{4}{3}\overline{j-\frac{1}{3}\overline{k}}$
• C. $\frac{5}{3}\overline{i}+\frac{4}{3}\overline{j-\frac{1}{3}\overline{k}}$
• D. $\frac{5}{3}\overline{i}+\frac{4}{3}\overline{j-\frac{2}{3}\overline{k}}$

Answer 1

The correct option is A $\frac{5}{3}\overline{i}+\frac{4}{3}\overline{j}+\frac{1}{3}\overline{k}$

Equation of line passing three $\overline{i}-\overline{2j}-\overline{k},\overline{2i}+\overline{3j}+\overline{k}$:

$\begin{array}{c}\overline{i}-\overline{2j}-\overline{k},\overline{2i}+\overline{3j}+\overline{k}\\ \frac{x-1}{\left(2-1\right)}=\frac{y+2}{5}=\frac{z+1}{1+1}\\ \frac{x-1}{1}=\frac{y+2}{5}=\frac{z+1}{2}\end{array}$

Equation of plane passing three $2\overline{i}+\overline{j}-3\overline{k},4\overline{i}-\overline{j}+2\overline{k},3\overline{i}+\overline{k}$

$\begin{array}{c}a\left(x-2\right)+b\left(y-1\right)+c\left(z+3\right)=0\\ 2a-2b+5c=0\\ a-b+4c=0\\ \frac{a}{-2+5}=\frac{-b}{8-5}=\frac{c}{-2+2}\\ \frac{a}{-3}=\frac{b}{-3}=\frac{c}{0}\\ -3\left(x-2\right)-3\left(y-1\right)=0\\ x-2+y-1=0\\ x+y-3=0\end{array}$

Ans prove that on the line is $rH,5r-2,2r-1$

AS it lies on plane , no it satisfies the equation $x+y-3=0$

$rH+5r-2-3=0$

$6r=4$

$r=\frac{2}{3}$

$\left(\frac{5}{3},\frac{4}{3},\frac{1}{3}\right)$ is the required prove that.

Then,

Option $A$ is correct answer.