Question

The points on the line $\frac{x+1}{1}=\frac{y+3}{3}=\frac{z-2}{-2},$ distance $\sqrt{\left(14\right)}$ from the point in which the line meets the points

• A. $(0,0,0),(2,-4,6)$
• B. $(0,0,0),(3,-4,-5)$
• C. $(0,0,0),(-2,-6,4)$
• D. $(2,6,-4),(3,-4,-5)$

Answer 1

The correct option is C $(0,0,0),(-2,-6,4)$

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

Then,

$x=k-1$

$y=3k-3$

$z=-2k+2$

Now, point in the line$(-1,-3,2)$ ${\left(\sqrt{(k-1-(-1{)}^2+(3k-3+3{)}^2+(-2k+2-2{)}^2}\right)}^2-(\sqrt{14}{)}^2$

(squaring both we get),

$k^2+9k^2+4k^2=14$

$14k^2=14$

$14k^2-14=0$

$14(k^2-1)=0$

$(x+1)(k-1)=0$

$k=1,-1$

Now, i putting the value of $k=1,-1$ in $x$, y and 2 respectively

$x=1-1=0$

$y=3-3=0$

$z=-2+2=\text{0}z=2+2=4$

$\because (0,0,0)$ and $(-2,-6,4)$