When are the two lines x = ay + b, z = cy + d | vedclass

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

Question

(A) The first line is given by $x = ay + b$ and $z = cy + d$. This can be written in symmetric form as $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c

Vedclass · Accepted Answer

$aa' + cc' + 1 = 0$

Vedclass · Answer

(A) The first line is given by $x = ay + b$ and $z = cy + d$. This can be written in symmetric form as $\frac{x - b}{a} = \frac{y}{1} = \frac{z - d}{c}$. The direction ratios of this line are $(a, 1, c)$.The second line is given by $x = a'y + b'$ and $z = c'y + d'$. This can be written in symmetric form as $\frac{x - b'}{a'} = \frac{y}{1} = \frac{z - d'}{c'}$. The direction ratios of this line are $(a', 1, c')$.Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.Applying this condition to our lines: $(a)(a') + (1)(1) + (c)(c') = 0$.Therefore,$aa' + cc' + 1 = 0$.

When are the two lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ perpendicular to each other?

Similar Questions

If the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles,then $p=$

The shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval

If the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}$ intersect,then find the value of $k$.

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module