The lines frac{x - 1}{2} = frac{y + 1}{2} = f | vedclass

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

Question

(B) Let the general point on the first line $L_1$ be $P(r)$.$\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4} = r$$x = 2r + 1, y = 2r - 1, z = 4r +

Vedclass · Accepted Answer

$(-2, -4, -5)$

Vedclass · Answer

(B) Let the general point on the first line $L_1$ be $P(r)$.$\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4} = r$$x = 2r + 1, y = 2r - 1, z = 4r + 1$So,$P = (2r + 1, 2r - 1, 4r + 1)$.Since the lines intersect,this point must satisfy the equation of the second line $L_2$: $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1} = k$.Substituting the coordinates of $P$ into $L_2$:$\frac{2r + 1 - 3}{1} = \frac{2r - 1 - 6}{2} = \frac{4r + 1}{1}$$\frac{2r - 2}{1} = \frac{2r - 7}{2} = 4r + 1$Taking the first and third parts: $2r - 2 = 4r + 1 \Rightarrow -3 = 2r \Rightarrow r = -\frac{3}{2}$.Now,substitute $r = -\frac{3}{2}$ into the coordinates of $P$:$x = 2(-\frac{3}{2}) + 1 = -3 + 1 = -2$$y = 2(-\frac{3}{2}) - 1 = -3 - 1 = -4$$z = 4(-\frac{3}{2}) + 1 = -6 + 1 = -5$Thus,the intersection point is $(-2, -4, -5)$.

The lines $\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1}$ intersect each other at point

Similar Questions

The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is

If the lines $\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1}$ and $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1}$ are perpendicular to each other,then $\lambda=$

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

The foot of the perpendicular drawn from the point $(1, 8, 4)$ on the line joining the points $(0, -11, 4)$ and $(2, -3, 1)$ is

Let $a, b \in R$. If the mirror image of the point $P(a, 6, 9)$ with respect to the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9}$ is $(20, b, -a-9)$,then $|a+b|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module