If a - b - c are non-coplanar vectors r - a - r - b - r - c -0- show that r is a zero vector-

Given r⃗ .a⃗ =0;r⃗ .b⃗ =0;r⃗ .c⃗ =0 Let r⃗ is the point of intersection of three planes r⃗ .a⃗ r⃗ .b⃗ =0⇒r⃗( a⃗ .b⃗ #160; ) =0...(i) Similarly ...

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Condition for Coplanarity of Four Points

If a , b , ...

Question

If $\to a, \to b, \to c$ are non-coplanar vectors $\to r . \to a = \to r . \to b = \to r . \to c = 0$ , show that $\to r$ is a zero vector.

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¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

- ¯ ¯ ¯ a + 3 ¯ ¯ b - 5 ¯ ¯ c, - ¯ ¯ ¯ a + ¯ ¯ b + ¯ ¯ c

2 ¯ ¯ ¯ a - 3 ¯ ¯ b + ¯ ¯ c

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

\vec{d} \cdot \vec{a} = \vec{d} \cdot \vec{b} = \vec{d} \cdot \vec{c} = 0,

\vec{d}

\to a, \to b, \to c

\to r

\to r . \to a = 1, \to r . \to b = 2

\to r . \to c = 3.

[\to a, \to b, \to c] = 1

\to r

\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)

A, b, c

| (a \times b) . c | = | a | | b | | c | i f a . b = b . c = c . a = 0

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