Question

If distance between two lines $\frac{x-5}{7}=\frac{y-7}{1}=\frac{z+3}{c}$ and $\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{c}$ is $65.5$, find $c$.

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

Answer: (C)

$3$

Given lines

$\frac{x-5}{7}=\frac{y-7}{1}=\frac{z+3}{c}$

position vector

$\overrightarrow{a}=5\hat{i}+7\hat{j}-3\hat{k}$

normal vector

$\overrightarrow{b}=7\hat{i}+\hat{j}+c\hat{k}$

and

$\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{c}$

position vector

$\overrightarrow{c}=8\hat{i}+7\hat{j}+5\hat{k}$

$\overrightarrow{c}-\overrightarrow{a}=8\hat{i}+7\hat{j}+5\hat{k}-(5\hat{i}+7\hat{j}-3\hat{k})$

$\overrightarrow{c}-\overrightarrow{a}=3\hat{i}+8\hat{k}$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 0& 8\\ 7& 1& c\end{array}\right|$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=\hat{i}(-8)-\hat{j}(3c-56)+\hat{k}\left(3\right)$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=-8\hat{i}-\hat{j}(3c-56)+3\hat{k}$

$distance=65.5$

$\frac{|-8\hat{i}-\hat{j}(3c-56)+3\hat{k}|}{\sqrt{7^2+1^2+c^2}}=65.5$

$\frac{\sqrt{(-8{)}^2+(-(3c-56){)}^2+3^2}}{\sqrt{49+1+c^2}}=65.5$

$\frac{\sqrt{64+(3c-56{)}^2+9}}{\sqrt{50+c^2}}=65.5$

$\frac{\sqrt{75+(3c-56{)}^2}}{\sqrt{50+c^2}}=65.5$

$\frac{75+(3c-56{)}^2}{50+c^2}={65.5}^2$

$75+9c^2+{56}^2-336c={65.5}^2(50+c^2)$

$75+9c^2+3136-336c=214512.5+4290.25c^2$

$4281.25c^2+336c-211301.5=0$

$c=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$c=\frac{-336\pm \sqrt{{336}^2-4\left(4281.25\right)(-211301.5)}}{2\left(4281.25\right)}$

$c=\frac{-336\pm \sqrt{112896+3618538187.5}}{8562.5}$

$c=\frac{-336\pm \sqrt{3615951083.5}}{8562.5}$

$c=\frac{-336\pm 60132.77}{8562.5}$

On taking (+) sign

$c=\frac{59796.77}{8562.5}$

$c=6.98$

on taking (-) sign

$c=\frac{-336-60132.77}{8562.5}$

$c=\frac{60468.77}{8562.5}$

$c=7.06$

So $c=6$ accept

and it is direction ratio so it may be

$c=3$ by dividing by 2

Hence option C is correct