Let a - b - c be three non-coplanar vector and p - q - r be vectors defined by the relations p - b - c - a b c - - q - c - a - a b c - - r - a - b - a b c - then the value of the expression a - b - p - b - c - q - c - a - r is equal to

The correct option is C 3 Solve for #160;( a + b #160;) . p = a . b × c #160;[ a b c #160;] #160; + b .( b × c #160;) #160 ...

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Section Formula

Let a , b ,...

Question

Let $\to a, \to b, \to c$ be three non-coplanar vector and $\to p, \to q, \to r$ be vectors defined by the relations $\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}$ then the value of the expression $(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r$ is equal to

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Solution

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Similar questions

a, b

c

p, q

r

p = \frac{b \times c}{a b c}, q = \frac{c \times a}{a b c}

r = \frac{a \times b}{a b c}

(a + b) . p + (b + c) . q + (c + a) . r

\to a, \to b, \to c

\to p = \frac{\to b \times \to a}{[\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 p + (\to b + \to c) . \to q + (\to c + \to a) . \to r =

a, b

c

p, q

r

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

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

(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r

\to a, \to b, \to c

\to p, \to p, \to r

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

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