Question

➯ Solve⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀​

Answer 1

Answer:

Hope this is OK for you completely

Answer 2

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

☆ Let assume that the direction ratios of the line which passes through the point (1, 2, - 4) is (a, b, c) respectively.

☆ Now it is given that the line passes through the point (1, 2, - 4) is perpendicular to the two lines,

$\rm :\longmapsto\:\dfrac{x - 8}{3} = \dfrac{y + 9}{ - 16} = \dfrac{z - 10}{7} - - - (1)$

and

$\rm :\longmapsto\:\dfrac{x - 15}{3} = \dfrac{y - 29}{8} = \dfrac{z - 5}{ - 5} - - - (2)$

So,

• Direction ratios of line (1) is (3, - 16, 7)

and

• Direction ratios of line (2) is (3, 8, - 5).

☆Hence,

Using condition for perpendicular lines, we have

$\rm :\longmapsto\:3a - 16b + 7c = 0 - - - (3)$

and

$\rm :\longmapsto\:3a + 8b - 5c = 0 - - - (4)$

☆ On solving equation (3) and (4), using cross multiplication method, we get

$\rm :\longmapsto\:\dfrac{a}{80 - 56} = \dfrac{b}{21 + 15} = \dfrac{c}{24 + 48}$

$\rm :\longmapsto\:\dfrac{a}{24} = \dfrac{b}{36} = \dfrac{c}{72}$

$\rm :\longmapsto\:\dfrac{a}{2} = \dfrac{b}{3} = \dfrac{c}{6} = k$

$\rm :\longmapsto\:a = 2k$

$\rm :\longmapsto\:b = 3k$

$\rm :\longmapsto\:c = 6k$

Hence,

• Direction ratios of line is (2k, 3k, 6k) = (2, 3, 6).

☆ So the required equation of line passes through the point (1, 2, - 4) and having direction ratios (2, 3, 6) is

$\bf :\longmapsto\:\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z + 4}{6}$

and

☆ In Vector form is

$\bf:\longmapsto\: \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} \: + \: \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$

Additional Information :-

Let us consider two lines,

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

and

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

then,

1. Two lines are perpendicular iff

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

2. Two lines are parallel iff

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

3. Angle between 2 lines is

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