Name: Angle Between Two Lines
Uploaded: 2019-07-19
Duration: 2 min 56 s
Description: The measure of the angle formed by the lines of two vectors can be found in the three-dimensional plane. Watch this video to understand the concept.

Question

Find the angle between the lines whose direction cosines are given by the equations

l + 2 m + 3 n = 0

and

3 l m - 4 l n + m n = 0

.

Accepted Answer

The equations given in the questions are

$l + 2 m + 3 n = 0 \dots (1)$

$3 l m - 4 l n + m n = 0 \dots (2)$

From equation $(1)$ , we can write $n = - (\frac{l + 2 m}{3})$ .

Now substituting it in equation $(2)$ ,

$3 l m + 4 l (\frac{l + 2 m}{3}) - m (\frac{l + 2 m}{3}) = 0$

$\Rightarrow 3 l m + \frac{4}{3} l^{2} + \frac{8}{3} l m - \frac{1}{3} l m - \frac{2}{3} m^{2} = 0$

$\Rightarrow \frac{4}{3} l^{2} + 3 l m + \frac{7}{3} l m - \frac{2}{3} m^{2} = 0$

Multiplying the equation with $3$ , we get,

$4 l^{2} + 16 l m - 2 m^{2} = 0 \Rightarrow 2 l^{2} + 8 l m - 1 m^{2} = 0$

Dividing the equation by $m^{2}$ ,

$2 {(\frac{l}{m})}^{2} + 8 (\frac{l}{m}) - 1 = 0$

Using the formula to find the roots of the above quadratic equation,

$\frac{l}{m} = \frac{(- 8) \pm \sqrt{64 + 8}}{4} = \frac{(- 8) \pm \sqrt{72}}{4}$

$= \frac{(- 8) \pm 6 \sqrt{2}}{4} = \frac{(- 4) \pm 3 \sqrt{2}}{2}$

$∴ \frac{l}{m} = \frac{(- 4) + 3 \sqrt{2}}{2}, \frac{(- 4) - 3 \sqrt{2}}{2}$

The first root can be used for the first line and the second root can be used for the second line, i.e.,

$\frac{l_{1}}{m_{1}} = \frac{(- 4) + 3 \sqrt{2}}{2}, \frac{l_{2}}{m_{2}} = \frac{(- 4) - 3 \sqrt{2}}{2}$

For the first line, $l_{1} = \frac{(- 4) + 3 \sqrt{2}}{2} m_{1}$ .

Substitute it in equation $(1)$ ,

$\frac{(- 4) + 3 \sqrt{2}}{2} m_{1} + 2 m_{1} + 3 n_{1} = 0$

$\Rightarrow \frac{3 \sqrt{2}}{2} m_{1} + 3 n_{1} = 0$

$\Rightarrow n_{1} = - \frac{\sqrt{2}}{2} m_{1}$

So, for the first line, $l_{1} : m_{1} : n_{1} \Leftrightarrow \frac{(- 4) + 3 \sqrt{2}}{2} m_{1} : m_{1} : - \frac{\sqrt{2}}{2} m_{1}$ .

The direction ratios are $\frac{(- 4) + 3 \sqrt{2}}{2} : 1 : - \frac{\sqrt{2}}{2}$ .

The direction cosines can be found as $\frac{\frac{(- 4) + 3 \sqrt{2}}{2}}{\sqrt{{(\frac{(- 4) + 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(- \frac{\sqrt{2}}{2})}^{2}}}, \frac{1}{\sqrt{{(\frac{(- 4) + 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(- \frac{\sqrt{2}}{2})}^{2}}}, \frac{- \frac{\sqrt{2}}{2}}{\sqrt{{(\frac{(- 4) + 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(- \frac{\sqrt{2}}{2})}^{2}}}$ . For ease of representation, consider $\sqrt{{(\frac{(- 4) + 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(- \frac{\sqrt{2}}{2})}^{2}} = a$ .

Therefore, the direction ratios are $\frac{\frac{(- 4) + 3 \sqrt{2}}{2}}{a}, \frac{1}{a}, \frac{- \frac{\sqrt{2}}{2}}{a}$ .

Now for the second line, $l_{2} = \frac{(- 4) - 3 \sqrt{2}}{2} m_{2}$ .

Substitute it in equation $(1)$ ,

$\frac{(- 4) - 3 \sqrt{2}}{2} m_{2} + 2 m_{2} + 3 n_{2} = 0$

$\Rightarrow \frac{- 3 \sqrt{2}}{2} m_{2} + 3 n_{2} = 0$

$\Rightarrow n_{2} = \frac{\sqrt{2}}{2} m_{2}$

So, for the second line, $l_{2} : m_{2} : n_{2} \Leftrightarrow \frac{(- 4) - 3 \sqrt{2}}{2} m_{2} : m_{2} : \frac{\sqrt{2}}{2} m_{2}$ .

The direction ratios are $\frac{(- 4) - 3 \sqrt{2}}{2} : 1 : \frac{\sqrt{2}}{2}$ .

The direction cosines can be found as $\frac{\frac{(- 4) - 3 \sqrt{2}}{2}}{\sqrt{{(\frac{(- 4) - 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(\frac{\sqrt{2}}{2})}^{2}}}, \frac{1}{\sqrt{{(\frac{(- 4) - 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(\frac{\sqrt{2}}{2})}^{2}}}, \frac{\frac{\sqrt{2}}{2}}{\sqrt{{(\frac{(- 4) - 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(\frac{\sqrt{2}}{2})}^{2}}}$ . For ease of representation, consider $\sqrt{{(\frac{(- 4) - 3 \sqrt{2}}{2})}^{2} + 1^{2} + {(\frac{\sqrt{2}}{2})}^{2}} = b$

Therefore, the direction ratios are $\frac{\frac{(- 4) - 3 \sqrt{2}}{2}}{b}, \frac{1}{b}, \frac{\frac{\sqrt{2}}{2}}{b}$ .

The angle between two lines whose direction cosines are known is $\cos (θ) = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$ .

Therefore,

$\cos (θ) = (\frac{\frac{(- 4) + 3 \sqrt{2}}{2}}{a} \times \frac{\frac{(- 4) - 3 \sqrt{2}}{2}}{b}) + (\frac{1}{a} \times \frac{1}{b}) + (\frac{- \frac{\sqrt{2}}{2}}{a} \times \frac{\frac{\sqrt{2}}{2}}{b})$

$\cos (θ) = \frac{((- 4) + 3 \sqrt{2}) \times ((- 4) - 3 \sqrt{2})}{a b} + \frac{1}{a b} - \frac{\frac{2}{4}}{a b}$

$\cos (θ) = \frac{\frac{{(- 4)}^{2} - {(3 \sqrt{2})}^{2}}{4}}{a b} + \frac{1}{a b} - \frac{1}{2 a b}$

$\cos (θ) = \frac{\frac{16 - 18}{4}}{a b} + \frac{1}{a b} - \frac{1}{2 a b}$

$\cos (θ) = \frac{\frac{- 2}{4}}{a b} + \frac{1}{a b} - \frac{1}{2 a b}$

$\cos (θ) = \frac{- 2}{4 a b} + \frac{1}{a b} - \frac{1}{2 a b}$

$\cos (θ) = \frac{(- 2) + 4 - 2}{4 a b} = 0$

$∴ θ = \cos^{- 1} (0) = \frac{π}{2}$ .

Find the angle between the lines whose direction cosines are given by the equations $l + 2 m + 3 n = 0$ and $3 l m - 4 l n + m n = 0$ .

Important Questions on Angle Between Two Lines

Learn and Prepare for any exam you want !

Find the angle between the lines whose direction cosines are given by the equations l+2m+3n=0 and 3lm-4ln+mn=0.

Important Questions on Angle Between Two Lines

Learn and Prepare for any exam you want !

Find the angle between the lines whose direction cosines are given by the equations $l + 2 m + 3 n = 0$ and $3 l m - 4 l n + m n = 0$ .