Question

A line with direction cosines proportional to 1, -5 and -2 meets line $x=y+5=z+11$ and $x+5=3y=2z$. The coordinates of each of the points of the intersection are given by:

• A. $(2,-3,1)$
• B. $(1,2,3)$
• C. $(0,\frac{5}{3},\frac{5}{2})$
• D. none of these.

Answer 1

Answer: (D)

none of these.

Consider the line

$\frac{x-0}{1}=\frac{y+5}{1}=\frac{z+11}{1}=t$

Any point on the line is
$P=(t,t-5,t-11)$

Similarly consider the line

$\frac{x+5}{6}=\frac{y-0}{2}=\frac{z-0}{3}=mu$

Any point on the line will be

$Q=(6\mu -5,2\mu ,3\mu )$.

Hence, $\overrightarrow{PQ}=(t-6\mu +5)i+(t-2\mu -5)j+(t-3\mu -11)k$

The directional vector of the given line is

$\overrightarrow{n}=i-5j-2k$

Now since $P$ and $Q$ lie on the line

$\overrightarrow{PQ}||\overrightarrow{n}$

Hence, $\overrightarrow{PQ}\times \overrightarrow{n}=0$

$\left(\right(t-6\mu +5)i+(t-2\mu -5)j+(t-3\mu -11\left)k\right)\times (i-5j-2k)=0$
$(3t-11\mu -45)i-(15\mu -3t+21)j+(32\mu -6t-20)k=0$

Hence, $3t-11\mu -45=0$ ...(i)

$15\mu -3t+21=0$...(ii)

$32\mu -6t-20=0$ ...(iii)

Solving (i) and (ii) gives us $t=37$ and $\mu =6$

And solving (ii) and (iii), gives us

$t=162$ and $\mu =31$

Hence, answer is none of the above.