Question

The vector equation $r=i-2j-k+t(6j-k)$ represents a straight line passing through the points:

• A. $(0,6,-1)$ and $(1,-2,-1)$
• B. $(0,6,-1)$ and $(-1,-4,-2)$
• C. $(1,-2,-1)$ and $(1,4,-2)$
• D. $(1,-2,-1)$ and $(0,-6,1)$

Answer 1

Answer: (C)

$(1,-2,-1)$ and $(1,4,-2)$

Cartesian representation of the given line is,

$\frac{x-1}{0}=\frac{y+2}{6}=\frac{z+1}{-1}=t$

So any point on the given line is of the form $(1,6t-2,-t-1)$

where $t$ can be any real numbers

so for $t=0$ and $1$ the corresponding points are $(1,-2,-1)$ and $(1,4,-2)$

You can check other options does not satisfy above point for any $t$.