Find a vector of magnitude √2 units and coplanar with vectors 3i−j−k and i+j−2k and perpendicular to vector 2i+2j+k.
A
r=±15(−3i+5j−4k)
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B
r=±15(3i−5j+4k)
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C
r=±14(2i−2j+4k)
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D
r=±14(−2i+2j−4k)
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Solution
The correct option is Br=±15(3i−5j+4k) r=t{a×(b×c)} represents a vector coplanar with b and c and perpendicular to a. r=t[(a⋅c)b−(a⋅b)c] Substitute the values of a.c and a.b ∴r=t(2b−3c)=t[2(3i−j−k)−3(i+j−2k)] ⇒r=t(3i−5j+4k). Take mod of both sides. |r|=|t|√9+25+16 ⇒√2=|t|.5√2|t|=15∴t=±15 ∴ Required vector is ±15(3i−5j+4k)