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

Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals...

The distance of the point $P(1, 2, 1)$ from the plane $2x + y - z = 10$ measured along the line $\frac{x - 5}{1} = \frac{2y - 3}{2} = \frac{z - \frac{5}{2}}{1}$ is

If two distinct points $Q$ and $R$ lie on the line of intersection of the planes $-x + 2y - z = 0$ and $3x - 5y + 2z = 0$,and $PQ = PR = \sqrt{18}$,where the point $P$ is $(1, -2, 3)$,then the area of the triangle $PQR$ is equal to

If the plane $4x + 4y - kz = 0$ is the equation of the plane passing through the origin and containing the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$,find the value of $k$.

The distance of the point $A(3, -4, 5)$ from the plane $2x + 5y - 6z = 16$ measured along the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module