If (2,-1,2) and ( K ,-3,-5) are the triads of direction ratios of two lines and the angle between the lines is 60°, then

Question

If (2,-1,2) and ( K ,-3,-5) are the triads of direction ratios of two lines and the angle between the lines is 60°, then (A) K 2-56 K -208=0 (B) 5 K 2-110 K +112=0 (C) 7 K 2-112 K -110=0 (D) 7 K 2-112 K +110=0

Tardigrade Admin · Accepted Answer

a=(2 hati- hatj+2 hatk) b=(k hati-3 hatj-5 hatk) a ⋅ b=|a||b| cos θ ∵ cos 60°=(1/2) (2 hati- hatj+2 hatk)(k hati-3 hatj-5 hatk)=3 √k2+34((1/2)) 2(2 k-7)=3 √k2+34 ⇒ 4(4 k2+49-28 k)=9(k2+34) 7 k2-112 k-110=0

Q. If $(2, - 1, 2)$ and $(K, - 3, - 5)$ are the triads of direction ratios of two lines and the angle between the lines is $6 0^{\circ}$ , then

$K^{2} - 56 K - 208 = 0$

$5 K^{2} - 110 K + 112 = 0$

$7 K^{2} - 112 K - 110 = 0$

$7 K^{2} - 112 K + 110 = 0$

Q. If (2,−1,2) and (K,−3,−5) are the triads of direction ratios of two lines and the angle between the lines is 60∘, then

K2−56K−208=0

5K2−110K+112=0

7K2−112K−110=0

7K2−112K+110=0

a=(2i^−j^​+2k^) b=(ki^−3j^​−5k^) a⋅b=∣a∣∣b∣cosθ ∵cos60∘=21​ (2i^−j^​+2k^)(ki^−3j^​−5k^)=3k2+34​(21​) 2(2k−7)=3k2+34​ ⇒4(4k2+49−28k)=9(k2+34) 7k2−112k−110=0

Q. If $(2, - 1, 2)$ and $(K, - 3, - 5)$ are the triads of direction ratios of two lines and the angle between the lines is $6 0^{\circ}$ , then

$K^{2} - 56 K - 208 = 0$

$5 K^{2} - 110 K + 112 = 0$

$7 K^{2} - 112 K - 110 = 0$

$7 K^{2} - 112 K + 110 = 0$