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Question
Let →α=3^i+^j and →β=2^i−^j+3^k. If →β=→β1−β2, where →β1 is parallel to →α and →β2 is perpendicular to →α, then →β1×→β2 is equal to?
Let →α=3^i+^j and →β=2^i−^j+3^k. If →β=→β1−β2, where →β1 is parallel to →α and →β2 is perpendicular to →α, then →β1×→β2 is equal to?
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Solution
The correct option is C 12(−3^i+9^j+5^k)
→α=3^i+^j→β=2^i−^j+3^k→β=→β1−→β2→β1=λ(3^i+^j),→β2=λ(3^i+^j)−2^ii+j−3^k→β2⋅→α=0(3λ−2)⋅3+(λ+1)=09λ−6+λ+1=0λ=12⇒→β1=32^i+12^j⇒→β2=−12^i+32^j−3^kNow →β1×→β2=∣∣
∣
∣
∣
∣∣^i^j^k32120−1232−3∣∣
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∣
∣
∣∣=^i(−32−0)−^j(−92−0)+^k(94+14)=−32^i+92^j+52^k=12(−3^i+9^j+5^k)Aliter:→β=→β1−→β2⇒→β⋅^α=→β1⋅^α=|→β1|⇒→β1=(→β⋅^α)^α⇒→β2=(→β⋅^α)^α−→β⇒→β1×→β2=−(→β⋅^α)^α×→β=−510(3^i+^j×(2^i−^j+3^k)=12(−3^i+9^j+5^k).
→α=3^i+^j
→β=2^i−^j+3^k
→β=→β1−→β2
→β1=λ(3^i+^j),→β2=λ(3^i+^j)−2^ii+j−3^k
→β2⋅→α=0
(3λ−2)⋅3+(λ+1)=0
9λ−6+λ+1=0
λ=12
⇒→β1=32^i+12^j
⇒→β2=−12^i+32^j−3^k
Now →β1×→β2=∣∣
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∣
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∣∣^i^j^k32120−1232−3∣∣
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∣
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∣∣
=^i(−32−0)−^j(−92−0)+^k(94+14)
=−32^i+92^j+52^k
=12(−3^i+9^j+5^k)
Aliter:
→β=→β1−→β2
⇒→β⋅^α=→β1⋅^α=|→β1|
⇒→β1=(→β⋅^α)^α
⇒→β2=(→β⋅^α)^α−→β
⇒→β1×→β2=−(→β⋅^α)^α×→β
=−510(3^i+^j×(2^i−^j+3^k)
=12(−3^i+9^j+5^k).
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