The equation of the line equally inclined wit | vedclass

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

Question

(A) The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{

Vedclass · Accepted Answer

$\frac{x+3}{1}=\frac{y-2}{1}=\frac{z+5}{1}$

Vedclass · Answer

(A) The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.Given that the line passes through $(-3, 2, -5)$,we have $x_1 = -3, y_1 = 2, z_1 = -5$.Since the line is equally inclined to the positive coordinate axes,the direction cosines $(l, m, n)$ are equal,i.e.,$l = m = n$.Thus,the direction ratios can be taken as $a = 1, b = 1, c = 1$.Substituting these values into the equation,we get $\frac{x - (-3)}{1} = \frac{y - 2}{1} = \frac{z - (-5)}{1}$.Therefore,the equation of the line is $\frac{x+3}{1} = \frac{y-2}{1} = \frac{z+5}{1}$.

The equation of the line equally inclined with the positive coordinate axes and passing through $(-3, 2, -5)$ is

Similar Questions

Direction ratios of the line represented by the equations $x = ay + b$ and $z = cy + d$ are:

The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}$ are

The equation of the line passing through the points $(3, 4, -7)$ and $(6, -1, 1)$ is

The equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to each of the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

If the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{-1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module