Browse Questions

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)

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)]$
$\Rightarrow\bigg [\large\frac{56}{17},\frac{43}{17},\frac{111}{17}\bigg]$