Question

Find the foot of the perpendicular from the point (-1,2,3) to the line
x−1/2= y+4/3 = z-3/1

Answer 1

Step-by-step explanation:

Given line $\tt{\green{\dfrac{x-1}{2}=\dfrac{y+4}{3}=\dfrac{z-3}{1}}}$

Vector equation of the given line is

$\sf{ \blue{ \vec{r} = ( \hat{i} - 4 \hat{j} + 3 \hat{k}) + \lambda(2\hat{i} + 3\hat{j} + \hat{k})}}$

Let foot of perpendicular be $\sf{\equiv\,(1+2\lambda\,,\,-4+3\lambda\,,\,3+\lambda)}$

DR of line joining the given point and its foot of perpendicular is

$\sf{\equiv\,(-2-2\lambda\,,\,6-3\lambda\,,\,-\lambda)}$

Now, the line joining the given point and its foot of perpendicular is perpendicular to the given line, so,

$\sf{2(-2-2\lambda) + 3(6-3\lambda) + 1(-\lambda) = 0}$

$\sf{ \implies - 4-4\lambda + 18-9\lambda -\lambda= 0}$

$\sf{ \implies + 14-4\lambda -9\lambda -\lambda= 0}$

$\sf{ \implies 14-14\lambda= 0}$

$\sf{ \implies \lambda= 1}$

Hence, coordinates of foot of perpendicular is

$\sf{ \equiv(1+2\,,\,-4+3\,,\,3+1)}$

$\sf{ \equiv(3\,,\,-1\,,\,4)}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given equation of line l is

$\rm :\longmapsto\:\dfrac{x - 1}{2} = \dfrac{y + 4}{3} = \dfrac{z - 3}{1}$

Let assume that

$\rm :\longmapsto\:\dfrac{x - 1}{2} = \dfrac{y + 4}{3} = \dfrac{z - 3}{1} = k$

Any point on the line l, say P is

$\rm :\longmapsto\:x = 2k + 1$

$\rm :\longmapsto\:y = 3k - 4$

$\rm :\longmapsto\:z = k + 3$

So,

$\rm :\longmapsto\:Coordinates \: of \: P = (2k + 1, 3k - 4, k + 3)$

Also, Direction ratios of line l is (2, 3, 1)

Let assume that the coordinate (- 1, 2, 3) is represented as N.

So, Direction ratios of line segment PN is given by

$\rm \:  =  \: (2k + 1 + 1, 3k - 4 - 2, k + 3 - 3)$

$\rm \:  =  \: (2k +2, 3k - 6, k)$

Now, as it is given that

Line l and PN are perpendicular.

So, (Direction ratios of line l) . (Direction ratios of PN) = 0

$\rm :\longmapsto\:\: 2(2k +2) + 3(3k - 6) + k = 0$

$\rm :\longmapsto\:\: 4k +4 + 9k - 18 + k = 0$

$\rm :\longmapsto\:\: 14k - 14 = 0$

$\rm :\longmapsto\:\: 14k = 14$

$\bf\implies \:k = 1$

Hence, Coordinates of point P is

$\rm :\longmapsto\:Coordinates \: of \: P = (2 + 1, 3 - 4, 1 + 3)$

$\rm :\longmapsto\:Coordinates \: of \: P = (3, - 1, 4)$

So, Coordinates of foot of perpendicular P is

$\rm :\longmapsto\:\boxed{ \tt{ \: Coordinates \: of \: P = (3, - 1, 4) \: }}$

More to know :-

Let us consider two equation of lines

$\rm :\longmapsto\:\vec{r} = \vec{a_1} + \: k \: \vec{b_1}$

and

$\rm :\longmapsto\:\vec{r} = \vec{a_2} + \: m \: \vec{b_2}$

Then

1. Two lines are parallel iff

$\boxed{ \tt{ \: \vec{b_1} \times \vec{b_2} = 0 \: }}$

or

$\boxed{ \tt{ \: \vec{b_1} = \alpha \vec{b_2}\: }}$

2. Two lines are perpendicular iff

$\boxed{ \tt{ \: \vec{b_1}.\vec{b_2} = 0 \: }}$