Question

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 the value of k

• A. $\frac{2}{9}$
• B. $\frac{9}{2}$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

$\frac{9}{2}$

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=u$ where is any constant so for any point on this line has co-ordinates is the form $(2u+1,3u-1,4u+1)$

$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=v$

so for any point on this line has co ordinates in the form $(v+3,2v+k,v)$

Point of intersection of these 2 lines will have co-ordinates of the form

$(2u+1,3u-1,4u-1)\&(v+3,2v+k,v)$

Equating the x, y, z coordinates for both the forms we get 3 equations

$2u+1=v+3$

$2u-v=2$ __(1)

$3u-1=2v+k$

$3u-2v=k+1$ __(2)

$4u+1=v$

$4u-v=1$ __(3)

Subtracting equation (1) from equation (3)

we get,

$2(-\frac{3}{2})-v=2$

$v=-5$

substitute value of v & in equation (2) we get

$3\left(\frac{-5}{2}\right)-2(-5)=k+1$

$k=\frac{9}{2}$