If →a=−^i+^j+^k and →b=2^i+0^j+^k then the vectors →c satisfying the conditions, (i) that it is coplanar with →a and →b (ii) that it is perpendicular to →b, and (iii) that →a.→c=7, is
A
−32^i+52^j+3^k
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B
−3^i+5^j+6^k
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C
−6^i+0^j+5^k
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D
−^i+2^j+2^k
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Solution
The correct option is A−32^i+52^j+3^k Given →a=−^i+^j+^k →b=2^i+0^j+^k Let c = x^i+y^j+z^k If →a,→b,→c are coplanar [→a,→b,→c]=0 ⎡⎢⎣−111201xyz⎤⎥⎦ x+3y-2z=0 - (1) →c is prependicular to →b →c.→b=0 2x+z=0 →a.→c=7 - (2) -x+y+z=7 - (3) Solving (1), (2) and (3) we get →c=12(−3^i+5^j+6^k)