Question

If the two lines
$l1:\frac{(x−2)}{3}=\frac{(y+1)}{−2},z=2$
and
$l2:\frac{(x−1)}{1}=\frac{(2y+3)}{α}=\frac{(z+5)}{2}$
are perpendicular, then an angle between the lines l2 and
$l3:\frac{(1−x)}{3}=\frac{(2y−1)}{−4}=\frac{z}{4}$
is

• A. $cos^{-1}(\frac{29}{4})$
• B. $sec^{-1}(\frac{29}{4})$
• C. $cos^{-1}(\frac{2}{29})$
• D. $cos^{-1}(\frac{2}{\sqrt29})$

Answer 1

Answer: (B)

$sec^{-1}(\frac{29}{4})$

Answer 2

The correct answer is (B) : $sec^{-1}(\frac{29}{4})$
∵ l1 and l2 are perpendicular, so
$3×1+(−2)(\frac{α}{2})+0×2=0$
⇒ α = 3
Now angle between l2 and l3,
$cos⁡θ=\frac{1(−3)+\frac{α}{2}(−2)+2(4)}{(\sqrt{1+\frac{α^2}{4}}+4\sqrt{9+4+16}}$
$⇒cos⁡θ=\frac{\frac{2}{29}}{2}$
$⇒θ=cos^{−1}⁡(\frac{4}{29})=sec−1⁡(\frac{29}{4})$