Question

If a line with direction ratios $2:2:1$ intersects the lines $\frac{x-7}{3}=\frac{y-5}{2}=\frac{z-3}{1}$ and $\frac{x-1}{2}=\frac{y+1}{4}=\frac{x+1}{3}$ at $A$ and $B$, respectively then $AB=$

• A. $\sqrt{2}$
• B. $2$
• C. $\sqrt{3}$
• D. $3$

Answer 1

Answer: (D)

$3$

$Thelinehav\mathrm{in}g\text{direction ratio 2:2:1 intersect the line}\frac{x-7}{3}=\frac{y-5}{2}=\frac{z-3}{1}at\text{point A}$
$\text{Let}\frac{x-7}{3}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda$
$sox=3\lambda +7,y=2\lambda +5,z=\lambda +3are\text{the coordinates of A}$
$\text{also}line\mathrm{int}er\sec t\frac{x-1}{2}=\frac{y+1}{4}=\frac{z+1}{3}atpo\mathrm{int}B$
$solet\frac{x-1}{2}=\frac{y+1}{4}=\frac{z+1}{3}=\mu$
$so\text{coordinates of B are}(2\mu +1,4\mu -1,3\mu -1)$
$now\text{direction ratios of line AB=}(2\mu +1-(3\lambda +7),4\mu -1-(2\lambda +5),3\mu -1-(\lambda +3))$
$=(2\mu -3\lambda -6,4\mu -2\lambda -6,3\mu -\lambda -4)$
$now\text{the point A and B are lying on the line}\text{having direction ratio 2:2:1}$
$\text{so direction ratio of AB will equal to direction ratio of line}$
$\text{so}$
$\frac{2\mu -3\lambda -6}{2}=\frac{4\mu -2\lambda -6}{2}=\frac{3\mu -\lambda -4}{1}$
$\frac{2\mu -3\lambda -6}{2}=\frac{4\mu -2\lambda -6}{2}$
$2\mu -3\lambda -6=4\mu -2\lambda -6$
$\Rightarrow \lambda =-2\mu .....\left(i\right)$
$\frac{2\mu -3\lambda -6}{2}=\frac{3\mu -\lambda -4}{1}$
$2\mu -3\lambda -6=6\mu -2\lambda -8$
$\Rightarrow 4\mu +\lambda -2=0$
$putting\lambda =-2\mu from\left(i\right)weget$
$4\mu -2\mu -2=0$
$\Rightarrow \mu =1$
$sofrom\left(i\right)$
$\lambda =-2$
$HenceCoordinatesofAandBare$
$A(1,1,1),B(3,3,2)$
$Dis\tan cebetweenAB$
$D=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$
$=\sqrt{2^2+2^2+1^2}$
$=\sqrt{9}$
$=3$