Let →a=6→i−3→j−6→k and →d=→i+→j+→k.Suppose that →a=→b+→c where →b is parallel to →d and →c is perpendicular to →d. Then →c is
A
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B
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C
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D
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Solution
The correct option is B →a=6→i−3→j−6→k→d=→i+→j+→k→a=→b+→c...............(1)→b=λ→d&→c.→d=0→b=λ^i+λ^j+λ^kFrom(1)6^i−3^j−6^k=λ^i+λ^j+λ^k+→c→c=(6−λ)^i−(3+λ)^j−(6+λ)^k→c.→d=0⇒λ=−1→c=7^i−2^j−5^k