How to find the distance from the point A(3,-5,5) to the line x = 2 + 3t , y= 1-2t, z= -1 + t ?

Jul 7, 2016

$\frac{5}{\sqrt{6}}$

Explanation:

there is a equation
$x + 2 y + z - 3 = 0$
use distance formula
=$\frac{\left(1 \cdot 3 - 5 \cdot 2 + 5 \cdot 1\right) - 3}{\sqrt{{1}^{2} + {2}^{2} + {1}^{2}}}$
=$- \frac{5}{\sqrt{6}}$

$\left\mid - \frac{5}{\sqrt{6}} \right\mid$
=$\frac{5}{\sqrt{6}}$

Jul 7, 2016

$\sqrt{\frac{83}{2}}$

Explanation:

Defining

${p}_{0} = \left\{2 , 1 , - 1\right\}$
$\vec{v} = \left\{3 , - 2 , 1\right\}$
${p}_{A} = \left\{3 , - 5 , 5\right\}$

we have to determine the distance between the line
$r \to {p}_{0} + t \vec{v}$ and the point ${p}_{A}$

Using Pitagoras we have

$a = \left\lVert {p}_{a} - {p}_{0} \right\rVert$

$b = \left\mid \left\langle{p}_{A} - {p}_{0} , \frac{\vec{v}}{\left\lVert \vec{v} \right\rVert}\right\rangle \right\mid$

$d = \sqrt{{a}^{2} - {b}^{2}}$ which is the sought distance

a = sqrt((3-2)^2+(-5-1)^2+(5+1)^2

$\frac{\vec{v}}{\left\lVert \vec{v} \right\rVert} = \frac{\left\{3 , - 2 , 1\right\}}{\sqrt{{3}^{2} + {2}^{2} + 1}}$

$b = \left\mid \frac{\left(3 - 2\right) \cdot 3 + \left(5 + 1\right) \cdot 2 + \left(5 + 1\right) \cdot 1}{\sqrt{{3}^{2} + {2}^{2} + 1}} \right\mid$

Finally

$d = \sqrt{\frac{83}{2}}$

Jul 7, 2016

$\sqrt{\frac{83}{2}} .$

Explanation:

We find the co-ords. of the foot $M$ of the perp. from $A \left(3 , - 5 , 5\right)$ on the given line $L : x = 2 + 3 t , y = 1 - 2 t , z = - 1 + t , t \in \mathbb{R} .$

We take a note that since $M \in L , M \left(2 + 3 t , 1 - 2 t , - 1 + t\right)$ for some $t \in \mathbb{R} .$

Also $A \left(3 , - 5 , 5\right) \Rightarrow \vec{A M} = \left(2 + 3 t - 3 , 1 - 2 t + 5 , - 1 + t - 5\right) = \left(3 t - 1 , 6 - 2 t , t - 6\right)$

The Direction Vector $\vec{l}$ of line $L$ is $\vec{l} = \left(3 , - 2 , 1\right)$

Knowing that $\vec{A M}$ is perp. to $\vec{l}$, we have, $\vec{A M} . \vec{l} = 0 \Rightarrow \left(3 t - 1 , 6 - 2 t , t - 6\right) . \left(3 , - 2 , 1\right) = 0$
$\therefore 3 \left(3 t - 1\right) - 2 \left(6 - 2 t\right) + \left(t - 6\right) = 0$
$\therefore 9 t - 3 - 12 + 4 t + t - 6 = 0$
$\therefore 14 t = 21 \Rightarrow t = \frac{3}{2} \Rightarrow \vec{A M} = \left(\frac{9}{2} - 1 , 6 - 3 , \frac{3}{2} - 6\right) = \left(\frac{7}{2} , 3 , - \frac{9}{2}\right)$

Hence the Dist. $A M = | | \vec{A M} | | = \sqrt{\frac{49}{4} + 9 + \frac{81}{4}} = \sqrt{\frac{166}{4}} = \sqrt{\frac{83}{2}} ,$ as derived by Cesareo R. Sir!
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