Let a-b-c be non coplanar vectors- If p-b-a-abc- - q-c-a-abc- r-a-b-abc- then a-b-p-b-c-q-c-a-r-

The correct option is D 3 a.p=a.(b×c)[abc]=[abc][abc]=1 b.q=b.(c×a)[abc]=[abc][abc]=1 c.r=c.(a×b)[abc]=[abc][abc]=1 ∴a.p=b.q=c.r=1 con ...

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Condition for Two Lines to be on the Same Plane

Let a,b,c b...

Question

Let $\to a, \to b, \to c$ be non coplanar vectors. If $\to p = \frac{\to b \times \to a}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}$ then $(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r =$

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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, \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 p + (\to b + \to c) . \to q + (\to c + \to a) . \to 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)

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

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