Find r by solving vector equations r- b - a- b- r- c - 0 provided that c is not perpendicular to b-

The correct option is C r = a (a· cb· c)b Given r × b=a × b ⇒ (r a)× b=0 Which means the angle between vectors r a and b is 0 , ...

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Equation of a Plane Passing through Three Points

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Question

Find $r$ by solving vector equations $r \times b = a \times b, r \cdot c = 0$ provided that $c$ is not perpendicular to $b$ .

r = a + (\frac{a \cdot c}{b \cdot c}) b

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r = b + (\frac{a \cdot c}{b \cdot c}) a

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r = a - (\frac{a \cdot c}{b \cdot c}) b

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r = b - (\frac{a \cdot c}{b \cdot c}) a

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Similar questions

\to r \times \to a = \to b \times \to a

\to r \cdot \to c = 0

\to a \cdot \to c \neq 0

(\to a \cdot \to c) (\to r \times \to b) + (\to a \times \to r)

\to r \times \to b = \to a \times \to b, \to r . \to c = 0

\to c

\to b

\to A, \to B

\to C

\to A \cdot \to B = 0

\to A \cdot \to C = 0

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a \cdot b = a \cdot c

a \times b = a \times c,

\to a, \to b and \to c are unit vectors such that \to a . \to b = \to a . \to c = 0 and the angle between \to b and \to c is \frac{π}{6}, then prove that

(i) \to a = \pm 2 (\to b \times \to c) (i i) [\to a + \to b \to b + \to c \to c + \to a] = \pm 1.

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