Skip to content
Exercise 6.4 · Q5

Q.Find the acute angle between the following lines.

(i) r⃗=(4i^−j^)+t(i^+2j^−2k^), r⃗=(i^−2j^+4k^)+s(−i^−2j^+2k^)\vec r=(4\hat i-\hat j)+t(\hat i+2\hat j-2\hat k),\ \vec r=(\hat i-2\hat j+4\hat k)+s(-\hat i-2\hat j+2\hat k)
(ii) x+43=y−74=z+55, r⃗=4k^+t(2i^+j^+k^)\dfrac{x+4}{3}=\dfrac{y-7}{4}=\dfrac{z+5}{5},\ \vec r=4\hat k+t(2\hat i+\hat j+\hat k)
(iii) 2x=3y=−z2x=3y=-z and 6x=−y=−4z6x=-y=-4z.
Puducherry TnboardTextbookSubjectiveImportance★★★★★
23% · 37/162 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

In each part, extract direction ratios from the given form, then apply cos⁡θ=∣b⃗⋅d⃗∣b⃗∣∣d⃗∣∣\cos\theta=\left|\dfrac{\vec b\cdot\vec d}{|\vec b||\vec d|}\right|.

Part (i). Directions b⃗=(1,2,−2), d⃗=(−1,−2,2)\vec b=(1,2,-2),\ \vec d=(-1,-2,2). Since d⃗=−b⃗\vec d=-\vec b, the lines are parallel: θ=0∘\theta=0^\circ.

Part (ii). Line 1: x+43=y−74=z+55\dfrac{x+4}3=\dfrac{y-7}4=\dfrac{z+5}5, direction (3,4,5)(3,4,5), ∣b⃗∣=9+16+25=50=52|\vec b|=\sqrt{9+16+25}=\sqrt{50}=5\sqrt2. Line 2: direction (2,1,1)(2,1,1), ∣d⃗∣=4+1+1=6|\vec d|=\sqrt{4+1+1}=\sqrt6.

cos⁡θ=∣3(2)+4(1)+5(1)∣52⋅6=∣6+4+5∣512=155⋅23=15103=323=32.\cos\theta=\frac{|3(2)+4(1)+5(1)|}{5\sqrt2\cdot\sqrt6}=\frac{|6+4+5|}{5\sqrt{12}}=\frac{15}{5\cdot2\sqrt3}=\frac{15}{10\sqrt3}=\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}.

θ=cos⁡−1(32)=30∘=π6.\theta=\cos^{-1}\left(\frac{\sqrt3}2\right)=30^\circ=\frac\pi6. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.