Question

If the angle between the lines, $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ and $\frac{5-x}{-2}=\frac{7y-14}{p}=\frac{z-4}{4}$ is ${\cos }^{-1}\left(\frac{2}{3}\right)$, then $p$ is equal to:

• A. $\frac{2}{7}$
• B. $\frac{7}{2}$
• C. $\frac{-4}{7}$
• D. $\frac{-7}{4}$

Answer 1

The correct option is B $\frac{7}{2}$

$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\text{and}\frac{x-5}{2}=\frac{y-2}{\frac{p}{7}}=\frac{z-4}{4}$
$\text{angle between both lines is}{\cos }^{-1}\left(\frac{2}{3}\right)={\cos }^{-1}(\frac{4+\frac{2p}{7}+4}{3\sqrt{4+\frac{p^2}{49}+16}})$
$\Rightarrow \frac{2}{3}=(\frac{4+\frac{2p}{7}+4}{3\sqrt{4+\frac{p^2}{49}+16}})$
$\Rightarrow \frac{2}{3}=\frac{2p+56}{3\sqrt{p^2+980}}$
$\Rightarrow \sqrt{p^2+980}=p+28$
$\Rightarrow p^2+980=p^2+56p+784$
$\Rightarrow 56p=196$
$\Rightarrow p=\frac{7}{2}$