Question

Find the value of p so that the lines ${{l}_{1}}$ : $\dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2}$ and ${{l}_{2}}$: $\dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5}$are at right angle. Also, find the equation of a line passing through the point (3,2,-4) and parallel to ${{l}_{1}}$.

Answer 1

Hint: To solve this question we will first write the equation of lines ${{l}_{1}}$ and ${{l}_{2}}$ ad will get direction ratios of both lines ${{l}_{1}}$ and ${{l}_{2}}$. Then we will find value of p using condition of intersection of two lines at right angle and then we will put value of p in direction ratios of ${{l}_{1}}$ and hence find the equation of line parallel to line ${{l}_{1}}$ and passing through point ( 3, 2, -4 ).

Complete step by step answer:
In question it is given that lines ${{l}_{1}}$ : $\dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2}$ and ${{l}_{2}}$: $\dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5}$are at right angle.
So, firstly writing, ${{l}_{1}}$ and ${{l}_{2}}$in standard form of equation of lines, we get
${{l}_{1}}$ : $\dfrac{x-1}{-3}=\dfrac{y-2}{\dfrac{2p}{7}}=\dfrac{z-3}{2}$ and ${{l}_{2}}$: $\dfrac{x-1}{-\dfrac{3p}{7}}=\dfrac{y-5}{1}=\dfrac{z-6}{-5}$.
Now, comparing lines ${{l}_{1}}$ and ${{l}_{2}}$with standard form of equation $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line, we get
direction ratios of line ${{l}_{1}}$ will be -3, $\dfrac{2p}{7}$and 2, also direction cosines of line ${{l}_{2}}$ will be $-\dfrac{3p}{7}$, 1, -5.
Now, we know that if line ${{l}_{1}}$ have direction cosines as a, b, c and line ${{l}_{2}}$ have direction cosines as d, e,f then lines ${{l}_{1}}$and ${{l}_{2}}$ will be at right angle if ad + be + cf = 0.
So, lines ${{l}_{1}}$and ${{l}_{2}}$will be at right angle if \[(-3)\left( \dfrac{-3p}{7} \right)+\left( \dfrac{2p}{7} \right)\cdot 1+2\cdot (-5)=0\]
On solving, we get
\[\left( \dfrac{9p}{7} \right)+\left( \dfrac{2p}{7} \right)-10=0\]
On simplifying we get
$\dfrac{11p}{7}=10$
$p=\dfrac{70}{11}$
Now we have to find the equation of another line which passes through point ( 3, 2, -4 ) and is parallel to line ${{l}_{1}}$.
Now we know that direction ratios of line ${{l}_{1}}$ will be -3, $\dfrac{2p}{7}$and 2.
Putting $p=\dfrac{70}{11}$ in direction ratios of line ${{l}_{1}}$, we get
Direction ratios of line ${{l}_{1}}$ as - 3, $\dfrac{20}{11}$and 2.
So, direction ratios of line parallel to line ${{l}_{1}}$ will be the same as direction ratios of line ${{l}_{1}}$.
Now it is given that equation of line passes through point ( 3, 2, -4 )
So, equation of line which passes through point ( 3, 2, -4 ) and parallel to line ${{l}_{1}}$, will be
$\dfrac{x-3}{-3}=\dfrac{y-2}{\dfrac{20}{11}}=\dfrac{z+4}{2}$


Note:
Always convert the equation of line firstly to standard line equation which is $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line. Always remember that the linear combination of direction ratios of two lines meeting at right angles is zero. Avoid calculation error while solving the question.