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. If →r=3→x+4→y+5→z, then the value of →d.→r is
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Solution
→a=^i−^j+^k, →b=2^i+^j and →c=3^j−2^k Since [→a→b→c]=∣∣
∣∣1−1121003−2∣∣
∣∣=0 ∴→a,→b and →c are coplanar vectors. Further, since →d is equally inclined to →a,→b and →c, we have ∴→d.→a=→d.→b=→d.→c=0 ∴→d.→x=→d.→y=→d.→z=0 ∴→d.→r=0