Name: Reflection of a Point Over the Coordinate Axes
Uploaded: 2019-07-19
Duration: 2 min 41 s
Description: What happens to the location of the point (3,2) if Y-coordinate is changed to -2? Watch to know more about the reflection of a point over the coordinate axes.

Question

An aeroplane is flying along the line

\vec{r} = λ (\hat{i} - \hat{j} + \hat{k})

; where '

λ

' is a scalar and another aeroplane is flying along the line

\vec{r} = \hat{i} - \hat{j} + μ (- 2 \hat{j} + \hat{k})

; where '

μ

' is a scalar. At what points on the lines should they reach, so that the distance between them is the shortest? Find the shortest possible distance between them.

Accepted Answer

The equation of two given straight lines in the Cartesian form are
$\frac{x}{1} = \frac{y}{- 1} = \frac{z}{1}$ ...(i)
and $λ$ ...(ii)

The lines are not parallel as direction ratios are not proportional. Let $P$ be a point on straight line (i) and $Q$ be a point on a straight line (ii) such that line $P Q$ is perpendicular to both of the lines.

Let the coordinates of $P$ be $(λ, - λ, λ)$ and that of $Q$ be $(1, - 2 μ - 1, μ)$ ; where ' $λ$ ' and ' $μ$ ' are scalars.

Then the direction ratios of the line $P Q$ are $(λ - 1, - λ + 2 μ + 1, λ - μ)$

Since $P Q$ is perpendicular to straight line (i), we have,

$(λ - 1) . 1 + (- λ + 2 μ + 1) . (- 1) + (λ - μ) . 1 = 0$

$\Rightarrow 3 λ - 3 μ = 2$ ...(iii)

Since, $P Q$ is perpendicular to straight line (ii), we have

$0 . (λ - 1) + (- λ + 2 μ + 1) . (- 2) + (λ - μ) . 1 = 0$

$\Rightarrow 3 λ - 5 μ = 2$ ...(iv)

Solving (iii) and (iv), we get $μ = 0, λ = \frac{2}{3}$

Therefore, the coordinates of $P$ are $(\frac{2}{3}, - \frac{2}{3}, \frac{2}{3})$ and those of $Q$ are $(1, - 1, 0)$ .

So, the required shortest distance is $\sqrt{{(1 - \frac{2}{3})}^{2} + {(- 1 + \frac{2}{3})}^{2} + {(0 - \frac{2}{3})}^{2}} = \sqrt{\frac{2}{3}}$ units.

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