Question

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

Answer 1

Hint:Use the direct formula for angle between two lines when the lines are expressed as the direction ratios. The important point is to note that for two inverse functions in the given domain if the functions are equal then so are the operands. Equate both sides and solve the equations for the variable.

Formula used:
Complete step-by-step answer:
The data given in the problem is,
The given equations of the line are $\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1}$ and $\dfrac{{5 - x}}{{ - 2}} = \dfrac{{7y - 14}}{P} = \dfrac{{z - 3}}{4}$ .
We will use the direct formula for the angle between two line in three dimensions.
The formula states that the angle between the lines $\dfrac{{x - {k_1}}}{{{a_1}}} = \dfrac{{y - {k_2}}}{{{b_1}}} = \dfrac{{z - {k_3}}}{{{c_1}}}$ and $\dfrac{{x - {h_1}}}{{{a_2}}} = \dfrac{{y - {h_2}}}{{{b_2}}} = \dfrac{{z - {h_3}}}{{{c_2}}}$ for all constants ${h_i},{k_i},{a_i},{b_i}$ is given by:
${\cos ^{ - 1}}\left( {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \cdot \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right)$
The first equation is already in its standard form so we need not have to change anything there.

We will write the first equation in standard form as below:
$\dfrac{5 - x}{ - 2} = \dfrac{y - 2}{P/7} = \dfrac{z - 3}{4}$
Now using the formula directly, the angle between these two lines is given as follows:
${\cos ^{ - 1}}\left( {\dfrac{{\left( 2 \right)\left( 2 \right) + \left( 2 \right)\left( {\dfrac{P}{7}} \right) + \left( 1 \right)\left( 4 \right)}}{{\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{\left( {\dfrac{P}{7}} \right)}^2} + {4^2}} }}} \right)$
Simplify the bracket and express the angle as follows:
${\cos ^{ - 1}}\left( {\dfrac{{\left( 2 \right)\left( 2 \right) + \left( 2 \right)\left( {\dfrac{P}{7}} \right) + \left( 1 \right)\left( 4 \right)}}{{\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{\left( {\dfrac{P}{7}} \right)}^2} + {4^2}} }}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{56 + 2P}}{{3\sqrt {{P^2} + 980} }}} \right)$
It is given that the angle between the given lines is ${\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right)$.
Therefore,
${\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{56 + 2P}}{{3\sqrt {{P^2} + 980} }}} \right)$
Taking $\cos $ on both sides we get:
$\dfrac{2}{3} = \dfrac{{56 + 2P}}{{3\sqrt {{P^2} + 980} }}$
Cross multiply to obtain the following:
$\sqrt {{P^2} + 980} = 28 + P$
Square on both the sides,
${P^2} + 980 = {P^2} + 784 + 56P$
Simplify it further,
$56P = 196$
Therefore, $P = \dfrac{7}{2}$.

So, the correct answer is “Option D”.

Note:Here to express the given equations in the standard form is the key point. Make sure that all the variables in the numerator have coefficient equal to $1$. Equate the right-hand side and left-hand side correctly to simplify the problem.