Question
Find the angle between the lines whose direction cosines are given by the equations $l + 2m + 3n = 0$ and $3lm - 4ln + mn = 0$
✓
Answer
Given that,
$l + 2m + 3n = 0 .....(1)$
$3lm - 4ln + mn = 0 .....(2)$
From equation $(1),$
$l + 2m + 3n = 0$
$l = -2m - 3n$
Put the value of l in equation $(2),$
$3lm - 4ln + mn = 0$
$3(-2m - 3n) m - 4(-2m - 3n) n + mn = 0$
$-6m^{2 }- 9nm + 8mn + 12n^2 + mn = 0$
$-6m^2 + 12n^2 = 0$
$m^2 = 2n^2$
$\text{m}=\pm\sqrt{2\text{n}^2}$
$\text{m}=\text{n}\sqrt{2}$ or $\text{m}=-\text{n}\sqrt{2}$
Put $\text{m}=\text{n}\sqrt{2}$ in equation $(1)$
$l + 2m + 3n = 0$
$\text{l}+2\big(\text{n}\sqrt{2}\big)+3\text{n}=0$
$\text{l}+\text{n}\big(2\sqrt{2}+3\big)=0$
$\text{l}+-\big(2\sqrt{2}+3\big)\text{n}$
Again, $\text{m}=-\sqrt{2\text{n}}$ in equation $(1)$
$l + 2m + 3n = 0$
$\text{l}+2\big(-\sqrt{2\text{n}}\big)+3\text{n}=0$
$\text{l}-2\sqrt{2\text{n}}+3\text{n}=0$
$\text{l}+\text{n}\big(-2\sqrt{2\text{n}}+3\big)=0$
$\text{l}=\big(2\sqrt{2\text{n}}-3\big)\text{n}$
Thus, direction cosines of the lines are given by,
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2\text{n}},\text{n}$ or $\big(2\sqrt{2}-3\big)\text{n},\sqrt{2\text{n}},\text{n}$
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2},1$ or $\big(2\sqrt{2}-3\big)\text{n},-\sqrt{2},1$
So, vectors parallel to these lines are
$\vec{\text{a}}=-\Big(2\sqrt{2}+3\Big)\hat{\text{i}}+\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\Big(2\sqrt{2}-3\Big)\hat{\text{i}}-\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\times\big|\vec{\text{b}}\big|}$
$=\frac{-\big(2\sqrt{2}+3\big)\times{\big(2\sqrt{2}-3\big)+\big(\sqrt{2}\big)}\times\big(-\sqrt{2}\big)+(1)(1)}{\sqrt{\big(2\sqrt{2}+3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}\sqrt{\big(2\sqrt{2}-3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}}$
$=\frac{-(8-9)-2+1}{\sqrt{8+9+12\sqrt{2}+2+1\sqrt{8+9-12\sqrt{2}+2+1}}}$
$=\frac{-(-1)-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$=\frac{1-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$.
$l + 2m + 3n = 0 .....(1)$
$3lm - 4ln + mn = 0 .....(2)$
From equation $(1),$
$l + 2m + 3n = 0$
$l = -2m - 3n$
Put the value of l in equation $(2),$
$3lm - 4ln + mn = 0$
$3(-2m - 3n) m - 4(-2m - 3n) n + mn = 0$
$-6m^{2 }- 9nm + 8mn + 12n^2 + mn = 0$
$-6m^2 + 12n^2 = 0$
$m^2 = 2n^2$
$\text{m}=\pm\sqrt{2\text{n}^2}$
$\text{m}=\text{n}\sqrt{2}$ or $\text{m}=-\text{n}\sqrt{2}$
Put $\text{m}=\text{n}\sqrt{2}$ in equation $(1)$
$l + 2m + 3n = 0$
$\text{l}+2\big(\text{n}\sqrt{2}\big)+3\text{n}=0$
$\text{l}+\text{n}\big(2\sqrt{2}+3\big)=0$
$\text{l}+-\big(2\sqrt{2}+3\big)\text{n}$
Again, $\text{m}=-\sqrt{2\text{n}}$ in equation $(1)$
$l + 2m + 3n = 0$
$\text{l}+2\big(-\sqrt{2\text{n}}\big)+3\text{n}=0$
$\text{l}-2\sqrt{2\text{n}}+3\text{n}=0$
$\text{l}+\text{n}\big(-2\sqrt{2\text{n}}+3\big)=0$
$\text{l}=\big(2\sqrt{2\text{n}}-3\big)\text{n}$
Thus, direction cosines of the lines are given by,
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2\text{n}},\text{n}$ or $\big(2\sqrt{2}-3\big)\text{n},\sqrt{2\text{n}},\text{n}$
$-\big(2\sqrt{2}+3\big)\text{n},\sqrt{2},1$ or $\big(2\sqrt{2}-3\big)\text{n},-\sqrt{2},1$
So, vectors parallel to these lines are
$\vec{\text{a}}=-\Big(2\sqrt{2}+3\Big)\hat{\text{i}}+\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\Big(2\sqrt{2}-3\Big)\hat{\text{i}}-\sqrt{2}\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\times\big|\vec{\text{b}}\big|}$
$=\frac{-\big(2\sqrt{2}+3\big)\times{\big(2\sqrt{2}-3\big)+\big(\sqrt{2}\big)}\times\big(-\sqrt{2}\big)+(1)(1)}{\sqrt{\big(2\sqrt{2}+3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}\sqrt{\big(2\sqrt{2}-3\big)^2+\big(-\sqrt{2}\big)^2+(1)^2}}$
$=\frac{-(8-9)-2+1}{\sqrt{8+9+12\sqrt{2}+2+1\sqrt{8+9-12\sqrt{2}+2+1}}}$
$=\frac{-(-1)-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$=\frac{1-2+1}{\sqrt{20+12\sqrt{2}}\sqrt{20-12\sqrt{2}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate the following integrals:$\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$
→Prove that:
$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
→$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
Using properties of determinants, show the following: $\begin{vmatrix} (\text{b}+\text{c})^2& \text{ab} & \text{ca} \\ \text{ab} & (\text{b}+\text{c})^2 & \text{bc} \\ \text{ac} & \text{bc} & \text{(a+b)}^2 \end{vmatrix} =2\text{abc}\ (\text{a+b+c})^3\dot{}$
→Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
→$\int\frac{\text{x}^2+\text{x}+1}{(\text{x}+1)^2(\text{x}+2)}\ \text{dx}$
If four points A, B, C and D with position vectors $4\hat{\text{i}}+3\hat{\text{j}},5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}$ respectively are coplanar, then find the value of x.
→Evaluate the following intregals:
$\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
→$\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
Evaluate the following integrals:$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$
→Solve the following initial value problems:
$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0,\text{ y}(0)=0$
→$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0,\text{ y}(0)=0$
An oil company has two depots$, A$ and $B,$ with capacities of $7000$ litres and $4000$ litres respectively. The company is to supply oil to three petrol pumps, $\text{D, E, F}$ whose requirements are $4500, 3000$ and $3500$ litres respectively. The distance $($in $\ km)$ between the depots and petrol pumps is given in the following table:
Assuming that the transportation cost per km is $Rs. 1.00$ per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
→Assuming that the transportation cost per km is $Rs. 1.00$ per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
Evaluate the following intregals:
$\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$
→$\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$