Let two lines L1-x-33-y-2-4-z-1 and L2-x-34-y-21-z-3 and x-3a-y-26-zb-a-b- is the acute angle bisector between lines L1 and L2- Then the value of -a-b-

Given lines L_1:x 33=y 2 4=z 1 ⇒ D.R's of L_1=(3, 4, 1) ⇒ D.C's of L_1=(l_1,m_1,n_1)=(3√(26), 4√(26), 1√(26)) and L_2:x 34=y 21=z 3 nbsp; ⇒ D.R's of L_2=(4,1, ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to Be Perpendicular

Let two lines...

Question

Let two lines $L_{1} : \frac{x - 3}{3} = \frac{y - 2}{- 4} = \frac{z}{- 1}$ and $L_{2} : \frac{x - 3}{4} = \frac{y - 2}{1} = \frac{z}{- 3}$ and $\frac{x - 3}{a} = \frac{y - 2}{6} = \frac{z}{b}, (a, b \in R)$ is the acute angle bisector between lines $L_{1}$ and $L_{2} .$ Then the value of $| a + b | =$

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Solution

Given lines $L_{1} : \frac{x - 3}{3} = \frac{y - 2}{- 4} = \frac{z}{- 1}$
$\Rightarrow D . R^{'} s$ of $L_{1} = (3, - 4, - 1)$
$\Rightarrow D . C^{'} s$ of $L_{1} = (l_{1}, m_{1}, n_{1}) = (\frac{3}{\sqrt{26}}, \frac{- 4}{\sqrt{26}}, \frac{- 1}{\sqrt{26}})$
and $L_{2} : \frac{x - 3}{4} = \frac{y - 2}{1} = \frac{z}{- 3}$
$\Rightarrow D . R^{'} s$ of $L_{2} = (4, 1, - 3)$
$\Rightarrow D . C^{'} s$ of $L_{2} = (l_{2}, m_{2}, n_{2}) = (\frac{4}{\sqrt{26}}, \frac{1}{\sqrt{26}}, \frac{- 3}{\sqrt{26}})$
If $l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} > 0$
Acute angle bisector is :
$\frac{x - x_{1}}{l_{1} + l_{2}} = \frac{y - y_{1}}{m_{1} + m_{2}} = \frac{z - z_{1}}{n_{1} + n_{2}}$
Obtuse angle bisector is :
$\frac{x - x_{1}}{l_{1} - l_{2}} = \frac{y - y_{1}}{m_{1} - m_{2}} = \frac{z - z_{1}}{n_{1} - n_{2}}$
Here $(x_{1}, y_{1}, z_{1}) = (3, 2, 0)$
$\Rightarrow \frac{x - 3}{7 / \sqrt{26}} = \frac{y - 2}{- 3 / \sqrt{26}} = \frac{z}{- 4 / \sqrt{26}} \Rightarrow \frac{x - 3}{- 14} = \frac{y - 2}{6} = \frac{z}{8}$
comparing with : $\frac{x - 3}{a} = \frac{y - 2}{6} = \frac{z}{b}$
$\Rightarrow a = - 14, b = 8$

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