Question

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

then p is equal to?

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

Answer 1

Answer: (D)

$\frac{7}{2}$

Above formula is used to find angle $\theta$ between two lines having direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$
Now let us find the direction cosines of the given lines
$\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$ direction cosines are $2,2,1$
next line can also be written as
$\frac{x-5}{2}=\frac{y-2}{\frac{P}{7}}=\frac{z-3}{4}$
hence direction cosines are $2,\frac{P}{7},4$
and we know that $cos\theta =\frac{2}{3}$
Keeping the value of these cosines in above formula we get
$\frac{2}{3}=\left|\frac{2\times 2+2\times \frac{P}{7}+1\times 4}{\sqrt{2^2+2^2+1^2}\sqrt{2^2+\frac{P^2}{49}+4^2}}\right|=\frac{8+\frac{2P}{7}+4}{3\times \sqrt{2^2+\frac{P^2}{49}+4^2}}$
after cross multiplication we get
$\Rightarrow {(4+\frac{P}{7})}^2=20+\frac{P^2}{49}$
$\Rightarrow 16+\frac{8P}{7}+\frac{P^2}{49}=20+\frac{P^2}{49}$
$\Rightarrow \frac{8P}{7}=4$
$\Rightarrow P=\frac{7}{2}$
Therefore Answer is $D$