Question

find the point on the line x+2/3=y+1/2=z-3/2 at a distance 3root2 from the point (1,2,3)

Answer 1

(-2,-1,3) and $(\dfrac{56}{17},\dfrac{43}{17},\dfrac{111}{17})$

Step-by-step explanation:

Equation of line: -

$\dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2}$

This is Cartesian equation of line in 3-D.

$\dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2}=t$

Parametric equation of line: -

$x(t)=3t-2$

$y(t)=2t-1$

$z(t)=2t+3$

Distance from point (3t-2,2t-1,2t+3) and point (1,2,3) is $3\sqrt{2}$

Using distance formula:-

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)$

$3\sqrt{2}=\sqrt{(3t-3)^2+(2t-3)^2+(2t)^2}$

Taking square both sides

$18=9t^2+9-18t+4t^2+9-12t+4t^2$

$17t^2-30t=0$

$t=0,\dfrac{30}{17}$

For t=0,

• Point on line: (-2,-1,3)

For $t=\dfrac{30}{17}$,

• Point on line: $(\dfrac{56}{17},\dfrac{43}{17},\dfrac{111}{17})$

# Learn more:

Equation of line:

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