The direction cosines of the line passing thr | vedclass

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

Question

(C) Let the points be $A(2, 3, -1)$ and $O(0, 0, 0)$.The direction ratios of the line $OA$ are $(2-0, 3-0, -1-0) = (2, 3, -1)$.The distance $OA$ is $\

Vedclass · Accepted Answer

$\frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{1}{\sqrt{14}}$

Vedclass · Answer

(C) Let the points be $A(2, 3, -1)$ and $O(0, 0, 0)$.The direction ratios of the line $OA$ are $(2-0, 3-0, -1-0) = (2, 3, -1)$.The distance $OA$ is $\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.The direction cosines $(l, m, n)$ are given by $\frac{a}{r}, \frac{b}{r}, \frac{c}{r}$, where $(a, b, c)$ are direction ratios and $r$ is the distance.Thus, $l = \frac{2}{\sqrt{14}}, m = \frac{3}{\sqrt{14}}, n = \frac{-1}{\sqrt{14}}$.Alternatively, for the line $AO$, the direction ratios are $(0-2, 0-3, 0-(-1)) = (-2, -3, 1)$.The direction cosines are $\frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{1}{\sqrt{14}}$.Comparing with the options, option $C$ is correct.

The direction cosines of the line passing through $P(2, 3, -1)$ and the origin are

Similar Questions

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:

The acute angle between the lines whose direction cosines are given by the equations $l+m+n=0$ and $2lm+2ln-mn=0$ is

If the angle between lines $AB$ and $CD$ is $\theta$,then the projection of line segment $AB$ on $CD$ is = ..........

If a line in space makes angles $\alpha, \beta$,and $\gamma$ with the coordinate axes,then $\cos 2\alpha + \cos 2\beta + \cos 2\gamma + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ equals:

Find the projection of the line segment joining the points $P(7, -5, 11)$ and $Q(-2, 8, 13)$ on a line $AB$ having direction cosines $\frac{1}{3}, \frac{2}{3}, \frac{2}{3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module