The angle between the lines whose direction c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We are given the equations for direction cosines $l, m, n$ as $l + m + n = 0$ and $l^2 = m^2 + n^2$.From the first equation,$n = -(l + m)$.Substit

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) We are given the equations for direction cosines $l, m, n$ as $l + m + n = 0$ and $l^2 = m^2 + n^2$.From the first equation,$n = -(l + m)$.Substituting this into the second equation: $l^2 = m^2 + (-(l + m))^2$.$l^2 = m^2 + l^2 + m^2 + 2lm$.$0 = 2m^2 + 2lm$.$2m(m + l) = 0$.This gives two cases: $m = 0$ or $m = -l$.Case $1$: If $m = 0$,then $l + 0 + n = 0 \Rightarrow n = -l$. The direction ratios are $(l, 0, -l)$,which is proportional to $(1, 0, -1)$. Let $\vec{a} = \hat{i} - \hat{k}$.Case $2$: If $m = -l$,then $l + (-l) + n = 0 \Rightarrow n = 0$. The direction ratios are $(l, -l, 0)$,which is proportional to $(1, -1, 0)$. Let $\vec{b} = \hat{i} - \hat{j}$.The angle $	heta$ between the lines is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.$\vec{a} \cdot \vec{b} = (1)(1) + (0)(-1) + (-1)(0) = 1$.$|\vec{a}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$.$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$.$\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.Therefore,$	heta = \frac{\pi}{3}$.

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^2 = m^2 + n^2$ is

Similar Questions

$A$ ray of light is incident upon an air/water interface (it passes from air into water) at an angle of $45^o$. Which of the following quantities change as the light enters the water?
$(I)$ wavelength
$(II)$ frequency
$(III)$ speed of propagation
$(IV)$ direction of propagation

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to

$A$ coil of wire of radius $r$ has $600$ turns and a self-inductance of $108 \ mH$. The self-inductance of a coil with the same radius and $500$ turns is: (in $mH$)

Heroin is chemically known as $.........$.

The solution of the differential equation $\frac{dy}{dx} + \frac{3x^2}{1+x^3}y = \frac{\sin^2 x}{1+x^3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module