A vector (→d) is equally inclined to three vectors →a=^i−^j+^k,→b=2^i+^j and →c=3^j−2^k. Let →x,→y,→z be three vectors in the plane of →a,→b;→b,→c;→c,→a, respectively. Then which of the following is INCORRECT?
A
→x.→d=0
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B
→y.→d=0
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C
→z.→d=0
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D
none of these
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Solution
The correct option is D none of these →a=^i−^j+^k, →b=2^i+^j →c=3^j−2^k As, [→a→b→c]=∣∣
∣∣1−1121003−2∣∣
∣∣=0 Therefore, →a,→b and →c are coplanar vectors.
As →d is equally inclined to →a,→b and →c, so →d should be normal to the plane containing vector →a,→b,→c