If lines (x-1/-3)=(y-2/2k)=(z-3/2) and (x-1/3k)=(y-5/1)=(z-6/-5) are mutually perpendicular, then k is equal to

Question

If lines (x-1/-3)=(y-2/2k)=(z-3/2) and (x-1/3k)=(y-5/1)=(z-6/-5) are mutually perpendicular, then k is equal to (A) -(10/7) (B) -(7/10) (C) -10 (D) -7

Tardigrade Admin · Accepted Answer

Lines (x-1/-3)=(y-2/2k)=(z-3/2) and (x-1/3k)=(y-5/1)=(z-6/-5) are perpendicular if a1a2+b1b2+c1c2=0 ⇒ -3(3k)+2k+2(-5)=0 ⇒ k-=(10/7)

Q. If lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}$
and $\frac{x - 1}{3 k} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$
are mutually perpendicular, then $k$ is equal to

$- \frac{10}{7}$

$- \frac{7}{10}$

$- 10$

$- 7$

Lines $\frac{x - 1}{- 3} = \frac{y - 2}{2 k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3 k} = \frac{y - 5}{1} = \frac{z - 6}{- 5}$
are perpendicular if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$
$\Rightarrow - 3 (3 k) + 2 k + 2 (- 5) = 0$
$\Rightarrow k - = \frac{10}{7}$

Q. If lines −3x−1​=2ky−2​=2z−3​ and 3kx−1​=1y−5​=−5z−6​ are mutually perpendicular, then k is equal to

−710​

−107​

−10

−7