# Question 64850

Aug 25, 2016

$P = \left\{3 , 4 , 2\right\}$

For $Q$ there are two feasible solutions

${Q}_{1} = \left\{2 , 2 , 0\right\}$
${Q}_{2} = \left\{4 , 6 , 4\right\}$

#### Explanation:

Let $L \to {p}_{0} + \lambda \vec{v}$ where

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

If the segment $\overline{A P}$ is orthogonal to $L$ then $\left\lVert A - P \right\rVert$ represents the minimum distance between $A$ and $L$. The point $P$ of minimum distance is obtained by minimizing

$d = {\min}_{\lambda} \left\lVert A - L \left(\lambda\right) \right\rVert = {\min}_{\lambda} \left\lVert A - {p}_{0} - \lambda \vec{v} \right\rVert$

but

${d}^{2} \left(\lambda\right) = {\left\lVert A - {p}_{0} \right\rVert}^{2} - 2 \lambda \left\langleA - {p}_{0} , \vec{v}\right\rangle + {\lambda}^{2} {\left\lVert \vec{v} \right\rVert}^{2}$

Its minimum is located at a $\lambda$ such that

$\frac{d}{d \lambda} {d}^{2} \left(\lambda\right) = - 2 \left\langleA - {p}_{0} , \vec{v}\right\rangle + 2 \lambda {\left\lVert \vec{v} \right\rVert}^{2} = 0$ for

${\lambda}_{P} = \frac{\left\langleA - {p}_{0} , \vec{v}\right\rangle}{\left\lVert \vec{v} \right\rVert} ^ 2 = 2$ so $P$ is located at

$P = {p}_{0} + {\lambda}_{P} \vec{v} = \left\{3 , 4 , 2\right\}$.

Once obtained $P$ the next point, $Q$ obeys the relationship

$\left\lVert A - P \right\rVert = \left\lVert P - Q \right\rVert$ but $Q \in L$ so

$\left\lVert A - P \right\rVert = \left\lVert P - {p}_{0} - \lambda \vec{v} \right\rVert$

Squaring both sides

${\left\lVert A - P \right\rVert}^{2} = {\left\lVert P - {p}_{0} \right\rVert}^{2} - 2 \lambda \left\langleP - {p}_{0} , \vec{v}\right\rangle + {\lambda}^{2} {\left\lVert \vec{v} \right\rVert}^{2}$.

Solving for $\lambda$ we have two outcomes:

${\lambda}_{Q} = \left\{1 , 3\right\}$

So

${Q}_{1} = {p}_{0} + {\lambda}_{{Q}_{1}} \vec{v} = \left\{2 , 2 , 0\right\}$
${Q}_{2} = {p}_{0} + {\lambda}_{{Q}_{2}} \vec{v} = \left\{4 , 6 , 4\right\}$

Aug 25, 2016

The Reqd. Pts. are $P \left(3 , 4 , 2\right) , {Q}_{1} \left(4 , 6 , 4\right) , \mathmr{and} , {Q}_{2} \left(2 , 2 , 0\right)$.

#### Explanation:

Reqd. pt. $P \in L = \left\{\left(1 , 0 , - 2\right) + \lambda \left(1 , 2 , 2\right) | \lambda \in \mathbb{R}\right\}$

$\Rightarrow \exists t \in \mathbb{R} , s . t . , P = P \left(1 + t , 2 t , 2 t - 2\right)$

Now, $A \left(1 , 3 , 4\right) , P \left(1 + t , 2 t , 2 t - 2\right) \Rightarrow \vec{A P} = \left(t , 2 t - 3 , 2 t - 6\right)$.

Since, $\vec{A P} \bot L \Rightarrow \vec{A P} \bot \text{the direction vector of} L$

$\Rightarrow \vec{A P} \bot \left(1 , 2 , 2\right) \Rightarrow \vec{A P} . \left(1 , 2 , 2\right) = 0$

$\therefore \left(t , 2 t - 3 , 2 t - 6\right) . \left(1 , 2 , 2\right) = 0. \therefore t + 4 t - 6 + 4 t - 12 = 0$

$\therefore t = 2 \Rightarrow P \left(1 + t , 2 t , 2 t - 2\right) = P \left(3 , 4 , 2\right)$.

As regards $Q \in L$, like $P$, $Q$ has to be $Q \left(1 + s , 2 s , 2 s - 2\right)$ for

some $s \in \mathbb{R}$

Because in right-isosceles $\Delta A P Q , m \angle A P Q = {90}^{\circ} , A P = P Q$.

With $A \left(1 , 3 , 4\right) , P \left(3 , 4 , 2\right) , Q \left(1 + s , 2 s , 2 s - 2\right) , \mathmr{and} , A {P}^{2} = P {Q}^{2}$, we have,

${\left(1 - 3\right)}^{2} + {\left(3 - 4\right)}^{2} + {\left(4 - 2\right)}^{2} = {\left(1 + s - 3\right)}^{2} + {\left(2 s - 4\right)}^{2} + {\left(2 s - 2 - 2\right)}^{2}$

$\therefore 4 + 1 + 4 = {\left(s - 2\right)}^{2} + 4 {\left(s - 2\right)}^{2} + 4 {\left(s - 2\right)}^{2}$.

$\therefore 9 {\left(s - 2\right)}^{2} = 9 \Rightarrow s - 2 = \pm 1 \Rightarrow s = 3 , \mathmr{and} , s = 1$.

Accordingly, we have $2$ pts. ${Q}_{1} , {Q}_{2} \in L$ satisfying the given cond., namely,

for s=3, Q_1(1+s,2s,2s-2)=Q_1(4,6,4), &, "for" s=1, Q_2(2,2,0)#.

Thus, the Reqd. Pts. are $P \left(3 , 4 , 2\right) , {Q}_{1} \left(4 , 6 , 4\right) , \mathmr{and} , {Q}_{2} \left(2 , 2 , 0\right)$, as

obtained by Respected Cesareo R.

Enjoy Maths.!.