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Question: If \(\mathbf{d} = \lambda(\mathbf{a} \times \mathbf{b}) + \mu(\mathbf{b} \times \mathbf{c}) + \nu(\m...

If d=λ(a×b)+μ(b×c)+ν(c×a)\mathbf{d} = \lambda(\mathbf{a} \times \mathbf{b}) + \mu(\mathbf{b} \times \mathbf{c}) + \nu(\mathbf{c} \times \mathbf{a}) and [abc]=18,\lbrack\mathbf{abc}\rbrack = \frac{1}{8},

then λ+μ+ν\lambda + \mu + \nu is equal to

A

8d.(a+b+c)8\mathbf{d}.(\mathbf{a} + \mathbf{b} + \mathbf{c})

B

8d×(a+b+c)8\mathbf{d} \times (\mathbf{a} + \mathbf{b} + \mathbf{c})

C

d8.(a+b+c)\frac{\mathbf{d}}{8}.(\mathbf{a} + \mathbf{b} + \mathbf{c})

D

d8×(a+b+c)\frac{\mathbf{d}}{8} \times (\mathbf{a} + \mathbf{b} + \mathbf{c})

Answer

8d.(a+b+c)8\mathbf{d}.(\mathbf{a} + \mathbf{b} + \mathbf{c})

Explanation

Solution

d.c=λ(a×b).c+μ(b×c).c+ν(c×a).c\mathbf{d}.\mathbf{c} = \lambda(\mathbf{a} \times \mathbf{b}).\mathbf{c} + \mu(\mathbf{b} \times \mathbf{c}).\mathbf{c} + \nu(\mathbf{c} \times \mathbf{a}).\mathbf{c}

=λ[abc]+0+0=λ[abc]=λ8= \lambda\lbrack\mathbf{abc}\rbrack + 0 + 0 = \lambda\lbrack\mathbf{abc}\rbrack = \frac{\lambda}{8}

Hence λ=8(d.c),\lambda = 8(\mathbf{d}.\mathbf{c}), μ=8(d.a)\mu = 8(\mathbf{d}.\mathbf{a}) and ν=8(d.b)\nu = 8(\mathbf{d}.\mathbf{b})

Therefore, λ+μ+ν=8d.c+8d.a+8d.b\lambda + \mu + \nu = 8\mathbf{d}.\mathbf{c} + 8\mathbf{d}.\mathbf{a} + 8\mathbf{d}.\mathbf{b}

=8d.(a+b+c).= 8\mathbf{d}.(\mathbf{a} + \mathbf{b} + \mathbf{c}).