The direction ratios of the line which is perpendicular to the lines ( x - 7/2) = (y +17/-3)= (z - 6/1) and ( x + 5/1) = (y +3/2)= (z - 4/-2) are

Question

The direction ratios of the line which is perpendicular to the lines ( x - 7/2) = (y +17/-3)= (z - 6/1) and ( x + 5/1) = (y +3/2)= (z - 4/-2) are (A) < 4, 5, 7> (B) < 4, -5, 7 > (C) < 4,-5,-7 > (D) < -4, 5, 7 >

Tardigrade Admin · Accepted Answer

Let d.rs of line are < a, b, c >, then 2a-3b + C = 0 and a + 2b - 2c = 0 ⇒ : : (a/6 -2) = (b/1 + 4)= (c/4 + 2) : : ⇒ : (a/4) = (b/5) = (c/7) ∴ d.rs of line are < 4, 5, 7 >

Q. The direction ratios of the line which is perpendicular to the lines $\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1}$ and $\frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}$ are

< 4, 5, 7>

< 4, -5, 7 >

< 4,-5,-7 >

< -4, 5, 7 >

Let d.rs of line are $< a, b, c >$ , then
$2 a - 3 b + C = 0$
and $a + 2 b - 2 c = 0$
$\Rightarrow \frac{a}{6 - 2} = \frac{b}{1 + 4} = \frac{c}{4 + 2} \Rightarrow \frac{a}{4} = \frac{b}{5} = \frac{c}{7}$
$∴$ d.rs of line are $< 4, 5, 7 >$

Q. The direction ratios of the line which is perpendicular to the lines 2x−7​=−3y+17​=1z−6​ and 1x+5​=2y+3​=−2z−4​ are