The square of the distance of the point (-2,-8,6) from the line (x-1)/(1)=(y-1)/(2)=(z)/(-…

Question

The square of the distance of the point $(-2,-8,6)$ from the line $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}$ along the line $\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}$ is equal to:

Accepted Answer

Let the given point be $P(-2,-8,6)$. We are looking for the distance of point $P$ from the first line, measured parallel to the second line.

Let the first line be ${L}_{1}$:

$\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}$

And the second line be ${L}_{2}$:

$\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}$

The direction ratios of the line ${L}_{2}$ are $(1,-1,2)$. Since we need to measure the distance parallel to ${L}_{2}$, we will form the equation of a new line, say ${L}_{3}$, which passes through the point $P(-2,-8,6)$ and is parallel to ${L}_{2}$.

The direction ratios of ${L}_{3}$ will be the same as those of ${L}_{2}$. Using the point-direction form from 3D geometry, the equation of the line ${L}_{3}$ is:

$\frac{x-(-2)}{1}=\frac{y-(-8)}{-1}=\frac{z-6}{2}$

$\frac{x+2}{1}=\frac{y+8}{-1}=\frac{z-6}{2}$

Let the line ${L}_{3}$ intersect the line ${L}_{1}$ at a point $Q$. To find the coordinates of $Q$, we can take a general point on ${L}_{1}$.

Let:

$\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}=\lambda $

From this, the coordinates of any general point on ${L}_{1}$ can be expressed in terms of $\lambda $:

$x=\lambda +1$

$y=2\lambda +1$

$z=-\lambda $

So, we can assume the intersection point is $Q(\lambda +1,2\lambda +1,-\lambda )$.

Since $Q$ also lies on the line ${L}_{3}$, its coordinates must satisfy the equation of ${L}_{3}$. Substituting the coordinates of $Q$ into the equation of ${L}_{3}$, we get:

$\frac{(\lambda +1)+2}{1}=\frac{(2\lambda +1)+8}{-1}=\frac{-\lambda -6}{2}$

$\frac{\lambda +3}{1}=\frac{2\lambda +9}{-1}=\frac{-\lambda -6}{2}$

Now, let us solve for $\lambda $ by equating the first two expressions:

$\lambda +3=-(2\lambda +9)$

$\lambda +3=-2\lambda -9$

$3\lambda =-12$

$\lambda =-4$

We can verify this value by substituting $\lambda =-4$ into the third expression:

$\frac{-(-4)-6}{2}=\frac{4-6}{2}=\frac{-2}{2}=-1$

This matches the first part $(-4)+3=-1$. Thus, $\lambda =-4$ is correct.

Now, substitute $\lambda =-4$ back into our assumption to find the exact coordinates of $Q$:

$x=(-4)+1=-3$

$y=2(-4)+1=-7$

$z=-(-4)=4$

So, the point of intersection is $Q(-3,-7,4)$.

We are asked to find the square of the distance between point $P$ and the line ${L}_{1}$ measured along ${L}_{3}$. This naturally corresponds to the square of the distance between points $P(-2,-8,6)$ and $Q(-3,-7,4)$.

Using the distance formula ${d}^{2}=({x}_{2}-{x}_{1}{)}^{2}+({y}_{2}-{y}_{1}{)}^{2}+({z}_{2}-{z}_{1}{)}^{2}$:

$P{Q}^{2}=(-3-(-2){)}^{2}+(-7-(-8){)}^{2}+(4-6{)}^{2}$

$P{Q}^{2}=(-3+2{)}^{2}+(-7+8{)}^{2}+(-2{)}^{2}$

$P{Q}^{2}=(-1{)}^{2}+(1{)}^{2}+(-2{)}^{2}$

$P{Q}^{2}=1+1+4$

$P{Q}^{2}=6$

The square of the required distance is $6$.

Option B

Solutions · 1