ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 12 - 11. three dimension geometry

ધારોકે રેખાઓ \(l_1: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}\) અને \(l_2: 3 x+2 y+z-2=0=x-3 y+2 z-13\) સમતલીય છે.જો \(l_1\) પરનું બિંદુ \(P (a, b, c)\) એ બિંદુ \(Q (-4,-3,2)\) થી સૌથી નજીક હોય, તો \(|a|+|b|+|c|=.........\)

  1. A \(12\)
  2. B \(14\)
  3. C \(10\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(10\)

Step-by-step Solution

Detailed explanation

\((3 x+2 y+z-2)+\mu(x-3 y+2 z-13)=0\) \(3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0\) \(9-4 \mu=0\) \(\mu=\frac{9}{4}\) \(4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0\) \(-100+4 \alpha-54+18 \alpha=0\) \(\Rightarrow \alpha=7\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app