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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the point on the line \( \large\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2} \) at a distance \( 3\sqrt 2\) from the point (1,2,3)

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Toolbox:
  • Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
The coordinate of any point on the line $\large\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$
This is given by
$\large\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$$=\lambda$
$\Rightarrow x+2=3\lambda,y+1=2\lambda,z-3=2\lambda$
$\Rightarrow x=3\lambda-2,y=2\lambda-1,z=2\lambda +3$
So,let the coordinates of the desired points are $(3\lambda-2,2\lambda-1,2\lambda+3)$.
Step 2:
The distance between this point and (1,2,3) is given as $3\sqrt 2$
$\therefore\sqrt{3\lambda-2-1)^2+(2\lambda-1-2)^2+(2\lambda+3-3)^2}=3\sqrt 2$
Squaring on both sides we get,
$9(\lambda-1)^2+(2\lambda-3)^2+4\lambda^2=18$
$17\lambda^2-30\lambda=0$
$\lambda=0,\lambda=\large\frac{30}{17}$
Step 3:
Now substituting the values of $\lambda$ we get,
$(3(0)-2,2(0)-1,2(0)+3)$
$\Rightarrow (-2,-1,3)$
$[3(\large\frac{30}{17}\big),$$2\big(\large\frac{30}{17}$$-1),2(\large\frac{30}{17}$$+3)]$
$\Rightarrow\bigg [\large\frac{56}{17},\frac{43}{17},\frac{111}{17}\bigg]$
answered Sep 27, 2013 by sreemathi.v
 

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