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Question : 72 of 120
Marks:
+1,
-0
Solution:
Given,
3p+2q=i+j+k......(i)
and
3p−2q=i−j−k.....(ii)
On adding Eqs. (i) and (ii), we get
6p=2i ∴p= On subtracting Eq. (ii) from Eq. (i), we get
4q=2(j+k)⇒q=(j+k) Let
θ be the angle between
p and
q, then
p⋅q=|p|q|cosθ ⇒cosθ= ∴θ=90∘=
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