Question

Ifa⃗=i⃗+j⃗+3k⃗;b⃗=2i⃗+3j⃗−5k⃗thena⃗⋅b⃗=

  1. 10
  2. -10
  3. 20
  4. 5

Answer: -10

Solution :-

We have a⃗=i⃗+j⃗+3k⃗andb⃗=2i⃗+3j⃗−5k⃗ then,

a⃗⋅b⃗=(i⃗+j⃗+3k⃗)⋅(2i⃗+3j⃗−5k⃗)=i⃗⋅(2i⃗+3j⃗−5k⃗)+j⃗⋅(2i⃗+3j⃗−5k⃗)+3k⃗⋅(2i⃗+3j⃗−5k⃗)=2⁢(i⃗⋅i⃗)+3⁢(i⃗⋅j⃗)−5⁢(i⃗⋅k⃗)+2⁢(j⃗⋅i⃗)+3⁢(j⃗⋅j⃗)−5⁢(j⃗⋅k⃗)+6⁢(k⃗⋅i⃗)+9⁢(k⃗⋅j⃗)−15⁢(k⃗⋅k⃗)=2⋅1+3⋅0−5⋅0+2⋅0+3⋅1−5⋅0+6⋅0+9⋅0−15⋅1=2+3−15=−10⋅ ∵i⃗⋅i⃗=j⃗⋅j⃗=k⃗⋅k⃗=1and i⃗⋅j⃗=j⃗⋅k⃗=k⃗⋅i⃗=0

also, the scalar product is commutative.

So, correct answer is option-2, →−10

#Vector Algebra #BSEB PYQ 2017 #Mathematics #BSEB PYQs