Question

can anybody answer 39 one

Answer 1

Dot product of two perpendicular vectors is zero.
So, p.q=0

=> a^2 + 2a - 3 = 0
=> a^2 - 3a + a -3 = 0 
=> a(a-3) + 1(a-3) = 0
=> (a+1) (a-3) = 0
=> a=-1 ; a=3

Hence the positive value of a=3.
Please mark brainliest. Thanks :)

Answer 2

If two vectors are perpendicular then their dot product=0
So P.Q=0
a×a - 2×a +3×(-1) = 0
=>a^2 - 2a -3=0
=>a^2 -3a + a -3=0
=>a(a-3)+ 1(a-3)=0
=>(a-3)(a+1)=0
=>a=3 or a= –1

So the positive value of a is 3.

HOPE IT HELPS......