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Question

# If the vector →b=3^j+4^k is written as the sum of a vector →b1, parallel to →a=^i+^j and a vector →b2, perpendicular to →a, then →b1×→b2 is equal to.

A
3^i3^j+9^k
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B
6^i+6^j92^k
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C
6^i6^j+92^k
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D
3^i+3^j9^k
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Solution

## The correct option is B 6^i−6^j+92^kGiven, →a=^i+^jSince, →b1 is parallel to →a, we can say →b1=k(^i+^j)Let →b2=p^i+q^j+r^kSince →a⊥→b2,→a.→b2=0⇒(p^i+q^j+r^k)⋅(^i+^j)=0⇒p+q=0Since, →b=→b1+→b2∴3^j+4^k=k(^i+^j)+(p^i−p^j+r^k)Comparing the components, we get0=k+p ....(1)3=k−p ....(2)4=r ....(3)Adding (1) and (2),2k=3⇒k=32From (1), p=−k=−32⇒k=32 and p=−32Hence, →b1=32^i+32^j and →b2=−32^i+32^j+4^k⇒→b1×→b2 =∣∣ ∣ ∣ ∣ ∣∣^i^j^k32320−32324∣∣ ∣ ∣ ∣ ∣∣=^i(6)−^j(6)+^k(94+94)=6^i−6^j+92^k

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