A straight line is equally inclined to all th | vedclass

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

Question

(B) Let the direction cosines of the line be $(l, m, n)$.Since the line is equally inclined to the coordinate axes,the angles $\alpha, \beta, \gamma$

Vedclass · Accepted Answer

$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Vedclass · Answer

(B) Let the direction cosines of the line be $(l, m, n)$.Since the line is equally inclined to the coordinate axes,the angles $\alpha, \beta, \gamma$ made by the line with the $x, y, z$ axes respectively are equal,i.e.,$\alpha = \beta = \gamma$.Therefore,the direction cosines are equal: $l = m = n$.We know that the sum of the squares of the direction cosines is $l^2 + m^2 + n^2 = 1$.Substituting $l = m = n$,we get $l^2 + l^2 + l^2 = 1$,which implies $3l^2 = 1$.Thus,$l^2 = \frac{1}{3}$,so $l = \pm \frac{1}{\sqrt{3}}$.The angle $\beta$ made by the line with the $y$-axis is given by $\cos \beta = m = \pm \frac{1}{\sqrt{3}}$.Taking the principal value,$\beta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

$A$ straight line is equally inclined to all the three coordinate axes. Then,the angle made by the line with the $y$-axis is

Similar Questions

If the direction cosines of a line satisfy the relations $l-m+n=0$ and $lm+mn-4nl=0$,then the direction cosines of the line are

The direction cosines of the line $\frac{3x + 1}{-3} = \frac{3y + 2}{6} = \frac{z}{-1}$ are

The direction cosines of the supporting line of the vector $\hat{i}+\hat{j}-2 \hat{k}$ are

If the direction cosines of two lines are such that $l+m+n=0$ and $l^2+m^2-n^2=0$, then the angle between them is:

$A$ line makes angles $\alpha, \beta, \gamma$ with the coordinate axes. If $\alpha + \beta = 90^\circ$,then $\gamma = \dots \dots ^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module