Question

the line joining (1, 2,3) and(-3, -2, 4) meet the yz plane at the point​

Answer 1

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

We know that

Line joining the points (a, b, c) and (d, e, f) is given by

$\rm :\longmapsto\:\dfrac{x - a}{d - a} = \dfrac{y - b}{e - b} = \dfrac{z - c}{f - c} = k$

Thus,

Equation of line joining the points (1, 2, 3) and (- 3, - 2, 4) is given by

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

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

So, Let assume that P be any point on the line joining the points (1, 2, 3) and (- 3, - 2, 4) when line crosses the yz - plane, whose co-ordinates is given by

$\rm :\longmapsto\:x = - 4k + 1$

$\rm :\longmapsto\:y = - 4k + 2$

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

Hence,

Coordinates of P ( - 4k + 1, - 4k + 2, k + 3).

Since, the line crosses the yz - plane.

$\rm :\longmapsto\:x - coordinate = 0$

$\rm :\longmapsto\: - 4k + 1 = 0$

$\rm :\longmapsto\: - 4k = - 1$

$\rm :\implies\:k = \dfrac{1}{4}$

So,

$\rm :\longmapsto\:x = - 4k + 1 = 0$

$\rm :\longmapsto\:y = - 4k + 2 = - 1 + 2 = 1$

$\rm :\longmapsto\:z = k + 3 = \dfrac{1}{4} + 3 = \dfrac{13}{4}$

Hence,

Coordinates of the point P is given by

$\boxed{ \bf{Coordinates \: P \: is \: \bigg(0, \: 1, \: \dfrac{13}{4} \bigg) \: }}$

Additional Information :-

Let us consider two lines

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

and

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

then

1. Two lines are parallel iff

$\rm :\longmapsto\:\vec{b_1} = k \: \vec{b_2}$

2. Two lines are perpendicular iff

$\rm :\longmapsto\:\vec{b_1} \: . \: \vec{b_2} = 0$

3. Angle between two lines is given by

$\rm :\longmapsto\:cos \theta \: = \dfrac{\vec{b_1}.\vec{b_2}}{ |\vec{b_1}| \: |\vec{b_2}| }$