Skip to content
Question of 68

Q.If the lines 1−x3=7y−142λ=z−32\dfrac{1-x}{3}=\dfrac{7y-14}{2\lambda}=\dfrac{z-3}{2} and 7−7x3λ=y−51=6−z5\dfrac{7-7x}{3\lambda}=\dfrac{y-5}{1}=\dfrac{6-z}{5} are mutually perpendicular, find the value of λ\lambda.

Tripura TbseHigher Secondary (+2 Stage) Examination 2025Subjective· 3mImportance★★★★★
0% · 0/68 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 →

Read off each line's direction ratios from its symmetric form, then use the perpendicularity condition a1a2+b1b2+c1c2=0a_1a_2+b_1b_2+c_1c_2=0.

Line 1: 1−x3=7y−142λ=z−32\dfrac{1-x}{3}=\dfrac{7y-14}{2\lambda}=\dfrac{z-3}{2}. Rewrite each part in the standard x−x0a\dfrac{x-x_0}{a} form:

  • 1−x3=x−1−3⇒a1=−3\dfrac{1-x}{3}=\dfrac{x-1}{-3}\Rightarrow a_1=-3
  • 7y−142λ=7(y−2)2λ=y−22λ/7⇒b1=2λ7\dfrac{7y-14}{2\lambda}=\dfrac{7(y-2)}{2\lambda}=\dfrac{y-2}{2\lambda/7}\Rightarrow b_1=\dfrac{2\lambda}{7}
  • z−32⇒c1=2\dfrac{z-3}{2}\Rightarrow c_1=2

Line 2: 7−7x3λ=y−51=6−z5\dfrac{7-7x}{3\lambda}=\dfrac{y-5}{1}=\dfrac{6-z}{5}.

  • 7−7x3λ=−7(x−1)3λ=x−1−3λ/7⇒a2=−3λ7\dfrac{7-7x}{3\lambda}=\dfrac{-7(x-1)}{3\lambda}=\dfrac{x-1}{-3\lambda/7}\Rightarrow a_2=-\dfrac{3\lambda}{7}
  • y−51⇒b2=1\dfrac{y-5}{1}\Rightarrow b_2=1 …

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.