What are the direction ratios of the line rep | vedclass

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

Question

(A) Given the equations of the line are $x = ay + b$ and $z = cy + d$.We can rewrite these equations in terms of $y$ as follows:$x - b = ay \implies \

Vedclass · Accepted Answer

$(a, 1, c)$

Vedclass · Answer

(A) Given the equations of the line are $x = ay + b$ and $z = cy + d$.We can rewrite these equations in terms of $y$ as follows:$x - b = ay \implies \frac{x - b}{a} = y$$z - d = cy \implies \frac{z - d}{c} = y$Equating the expressions for $y$,we get:$\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$This is the symmetric form of the line equation $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$,where $(l, m, n)$ are the direction ratios.Comparing the denominators,the direction ratios are $(a, 1, c)$.

What are the direction ratios of the line represented by the equations $x = ay + b$ and $z = cy + d$?

Similar Questions

The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a + b + c =$

The shortest distance between the lines $r = (3i - 2j - 2k) + t(i)$ and $r = (i - j + 2k) + s(j)$ ($t$ and $s$ being parameters) is

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is . . . . . . .

The angle between two lines $\frac{x+3}{2}=\frac{-y}{3}=\frac{z+5}{-6}$ and $\frac{x-1}{10}=\frac{y+1}{-2}=\frac{z-3}{11}$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module