The equation of acute angle bisector between the lines L1-x-14-y-2-3-z-32- L2-x-13-y-24-z-3-2 is x-1a-y-2-14-z-3b-a-b- Then the value of a-b-

Given : L_1:x 14=y 2 3=z 32 and ⇒ D.R's=(4, 3,2) ⇒ D.C's along L_1=(l_1,m_1,n_1)=(4√(29), 3√(29),2√(29)) L_2:x 13=y 24=z 3 2 nbsp; ⇒ D.R's=(3,4, 2) ⇒ D.C's al ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Mathematics

Bisectors of Angle between Two Lines

The equation ...

Question

The equation of acute angle bisector between the lines $L_{1} : \frac{x - 1}{4} = \frac{y - 2}{- 3} = \frac{z - 3}{2}, L_{2} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - 3}{- 2}$ is $\frac{x - 1}{a} = \frac{y - 2}{- 14} = \frac{z - 3}{b}, (a, b \in R) .$ Then the value of $a + b =$

Open in App

Solution

Given :
$L_{1} : \frac{x - 1}{4} = \frac{y - 2}{- 3} = \frac{z - 3}{2}$ and
$\Rightarrow D . R^{'} s = (4, - 3, 2)$
$\Rightarrow D . C^{'} s$ along $L_{1} = (l_{1}, m_{1}, n_{1}) = (\frac{4}{\sqrt{29}}, \frac{- 3}{\sqrt{29}}, \frac{2}{\sqrt{29}})$
$L_{2} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - 3}{- 2}$
$\Rightarrow D . R^{'} s = (3, 4, - 2)$
$\Rightarrow D . C^{'} s$ along $L_{2} = (l_{2}, m_{2}, n_{2}) = (\frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}, \frac{- 2}{\sqrt{29}})$
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 :
$\frac{x - x_{1}}{l_{1} + l_{2}} = \frac{y - y_{1}}{m_{1} + m_{2}} = \frac{z - z_{1}}{n_{1} + n_{2}}$
$x_{1} = 1, y_{1} = 2, z_{1} = 3$
So, equation is $\frac{x - 1}{1 / \sqrt{29}} = \frac{y - 2}{- 7 / \sqrt{29}} = \frac{z - 3}{4 / \sqrt{29}}$
$\Rightarrow \frac{x - 1}{2} = \frac{y - 2}{- 14} = \frac{z - 3}{8} \Rightarrow a = 2, b = 8$

Suggest Corrections

Similar questions

Two lines $L_{1}$ : $x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2}$ : $x = α, \frac{y}{- 1} = \frac{Z}{2 - α}$ are coplanar. Then $α$ can take value(s)