If the vectors a- b- c are coplanar- then the value of a b c a-a a-b a-c b-a b-b b-c -

The correct option is B 0 Given a⃗,b⃗,c⃗ are coplanar if there exists three scalars x,y,z such that #160; xa⃗+yb⃗+zc⃗=0, #160; #16 ...

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Condition for Two Lines to be on the Same Plane

If the vector...

Question

If the vectors $a, b, c$ are coplanar, then the value of
$∣ ∣ ∣ ∣ \begin{matrix} a & b & c a . a & a . b & a . c b . a & b . b & b . c \end{matrix} ∣ ∣ ∣ ∣ =$

1

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- 1

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None of these

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Solution

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\to a, \to b, \to c

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a & \to b & \to c \to a . \to a & \to a . \to b & \to a . \to c \to b . \to a & \to b . \to b & \to b . \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ =

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

A = (1, a, a^{2}), B = (1, b, b^{2}), C = (1, c, c^{2})

a b c

If $¯ a, ¯ b, ¯ c$ are non collinear, non coplanar vectors and $x ¯ a + ¯ b, x ¯ b + ¯ c, x ¯ c + ¯ a$ are coplanar vectors then the value of x is

\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})}

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

\vec{a} + \vec{b} + \vec{c} = \vec{0},

\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} is

- \frac{3}{2}

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