Q49 / 100 · NTA Abhyas (JEE Main) 2020 · Mathematics · Vector Algebra

Let p→=2i^+j^-2k^,q→=i^+j^. If r→ is a vector such that p→.r→=r→, r→-p→=22 and the angle between p→×q→ and r→ is π6, then the value of p→×q→×r→ is equal to

  • Correct answer
    A
    32
  • B
    12
  • C
    332
  • D
    3

Solution

p→×q→×r→=p→×q→r→sin⁡30o
=|p→×q→||r→|12 … 1
p→×q→=i^j^k^21-2110
=2i^-2j^+k^
⇒p→×q→=3
Given, r→-p→2=8
⇒r→2+p→2-2r→.p→=8
⇒r→2+9-2r→=8
⇒r→=1
from 1,

p→×q→×r→=32

Hence, the correct answer is option: 'A'.

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