If a- b- c are non-coplanar vectors and d- is a unit vector- then the value of - a - d- b-c - b - d- c-a - c - d- a-b - is equal to

The correct option is C | [ a b c] #160;| Since a,b,c are non coplanar, a× b,b× c,c× a are also non coplanar, d can be expressed uniq ...

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Cross Product of Two Vectors

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Question

If $\to a, \to b, \to c$ are non-coplanar vectors and $^d$ is a unit vector, then the value of $∣ ∣ ∣ (\to a .^d) (\to b \times \to c) + (\to b .^d) (\to c \times \to a) + (\to c .^d) (\to a \times \to b) ∣ ∣ ∣$ is equal to

1

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- 1

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∣ ∣ ∣ [\to a \to b \to c] ∣ ∣ ∣

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none of these

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Solution

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Similar questions

\vec{a,} \vec{b,} \vec{c}

\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}} + \frac{\vec{b} \cdot (\vec{a} \times \vec{c})}{\vec{c} \cdot (\vec{a} \times \vec{b})}

a, b, c

\frac{(a + 2 b - c) . (a - b) \times (a - b - c)}{[a b c]}

\to a, \to b, \to c

\to p, \to q, \to r

\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]},

(\to a + \to b + \to c), (\to p + \to q + \to r)

If \to a, \to b and \to c

\to a \times \to c

\to a \times (\to b \times \to c),

[\to a \times (\to b \times \to c)] \times \to c

∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & 1 + a^{3} b & b^{2} & 1 + b^{3} c & c^{2} & 1 + c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

(1, a, a^{2}), (1, b, b^{2})

(1, c, c^{2})

a b c

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