If the lines frac{x - 1}{2} = frac{y + 1}{3} | vedclass

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

Question

(C) Let the two lines be represented by parameters $\lambda$ and $\mu$ respectively: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \implies x=2\l

Vedclass · Accepted Answer

$\frac{9}{2}$

Vedclass · Answer

(C) Let the two lines be represented by parameters $\lambda$ and $\mu$ respectively: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \implies x=2\lambda+1, y=3\lambda-1, z=4\lambda+1$ $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu \implies x=\mu+3, y=2\mu+k, z=\mu$ If the lines intersect,there must exist a point common to both,so: $2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2 \quad (i)$ $3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1 \quad (ii)$ $4\lambda+1 = \mu \implies 4\lambda - \mu = -1 \quad (iii)$ Subtracting $(i)$ from $(iii)$: $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$ Substituting $\lambda = -\frac{3}{2}$ into $(iii)$: $4(-\frac{3}{2}) + 1 = \mu \implies -6 + 1 = \mu \implies \mu = -5$ Substituting $\lambda = -\frac{3}{2}$ and $\mu = -5$ into $(ii)$: $3(-\frac{3}{2}) - 2(-5) = k+1 \implies -\frac{9}{2} + 10 = k+1 \implies k = 9 - \frac{9}{2} = \frac{9}{2}$

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 $k$ is equal to:

Similar Questions

If $4x + 4y - kz = 0$ is the equation of the plane through the origin that contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4},$ then $k =$

If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$, (where $0 < \alpha < \frac{\pi}{2}$), then $p =$

Let $L$ be a line obtained from the intersection of two planes $x+2y+z=6$ and $y+2z=4$. If point $P(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $(3,2,1)$ on $L$,then the value of $21(\alpha+\beta+\gamma)$ equals ...... .

Let $N$ be the foot of the perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $.......$.

The direction ratios of the line given by the intersection of the planes $x - y + z - 5 = 0$ and $x - 3y - 6 = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module