If a and b are not perpendicular to each other and r - a - b - a - r - c -0 then r is equal to

The correct option is B #160; b ( b . c #160; a . c #160; #160;) . a Since r × a = b × a ⇒( r b #160;) × a =0 ⇒ r b is ...

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Applications of Cross Product

If a and ...

Question

If $\to a$ and $\to b$ are not perpendicular to each other and $\to r \times \to a = \to b \times \to a, \to r . \to c = 0$ then $\to r$ is equal to

\to a - \to c

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\to b + x \to a

for all scalar

c

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\to b - ⎛ ⎝ \frac{\to b . \to c}{\to a . \to c} ⎞ ⎠ . \to a

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None of these

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\to a

\to b

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

\to r

\to a, \to b, \to c

\to r

\to a \times {(\to r - \to b) \times \to a} + \to b \times {(\to r - \to c) \times \to b} + \to c \times {(\to r - \to a) \times \to c} = \to 0

\to r

a, b

a

b

a

r

r \cdot a = α,

α, a \times r = b

y = \frac{1}{1 + x^{a - b} +^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}

\frac{d y}{d x}

{(a + b + c)}^{x^{a + b + c - 1}}

If $¯ a, ¯ b$ are not perpendicular to each other and $¯ r \times ¯ b = ¯ c \times ¯ b, ¯ r . ¯ a = 0, t h e n ¯ r =$

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