If a-b-c- and p-q-r- are reciprocal system of vectors- then a-p-b-q-c-r- equals

The correct option is C #160;0 We have reciprocal system #160;p= b× c [ abc ] #160; ,q= c× a [ abc ] #160; ,r= a× b [ abc ] #160; #160;∴ ...

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Standard XII

Mathematics

Invertible Element Binary Operation

If a⃗,b⃗,c⃗...

Question

If $\to a, \to b, \to c$ and $\to p, \to q, \to r$ are reciprocal system of vectors, then $(\to a \times \to p) + (\to b \times \to q) + (\to c \times \to r)$ equals

[\to a \to b \to c]

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(\to p + \to q + \to r)

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\to 0

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(\to a + \to b + \to c)

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Solution

The correct option is C $\to 0$
We have reciprocal system
$p = \frac{b \times c}{[a b c]}, q = \frac{c \times a}{[a b c]}, r = \frac{a \times b}{[a b c]}$

$∴ a \times p = \frac{1}{[a b c]} a \times (b \times c),$

$b \times q = \frac{1}{[a b c]} b \times (c \times a),$

$c \times r = \frac{1}{[a b c]} c \times (a \times b) .$

$∴ a \times p + b \times q + c \times r$

$= \frac{1}{[a b c]} {a \times (b \times c) + b \times (c \times a) + c \times (a \times b)} = 0.$

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Similar questions

\to a, \to b, \to c

are non coplanar vectors and

\to p, \to q, \to r

are defined as

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to b & \to c & \to a \end{matrix}]}

\to q = \frac{\to c \times \to a}{[\begin{matrix} \to c & \to a & \to b \end{matrix}]}

\to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

, then

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

is equal to

Q. If

\to a, \to b, \to c

are three non coplanar vectors,

\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

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

Q. If

¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

are three non-coplanar vectors,

¯ ¯ ¯ p, ¯ ¯ ¯ q, ¯ ¯ ¯ r

are non-zero vectors such that

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to q = \frac{\to c \times \to a}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

, then the value of the expression

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

Q. Let

\to a, \to b

and

\to c

be three non-zero and non-coplanar vectors and

\to p, \to q

and

\to r

be three vectors given by

\to p = \to a + \to b - 2 \to c, \to q = 3 \to a - 2 \to b + \to c

and

\to r = \to a - 4 \to b + 2 \to c

. If the volume of the parallelopiped determined by

\to a, \to b

and

\to c

V_{1}

and the volume of the parallelopiped determined by

\to p, \to q

and

\to r

V_{2}

then

V_{2} : V_{1} =

Q. If

| \to a | = ∣ ∣ \to b ∣ ∣ = | \to c | = 3

and

\to a . \to b . = \to b . \to c = \to c . \to a = 4.

\to p, \to q, \to r

is the reciprocal system of

\to a, \to b, \to c,

then

85

[\to p \to q \to r]

is equal to

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If →a,→b,→c and →p,→q,→r are reciprocal system of vectors, then (→a×→p)+(→b×→q)+(→c×→r) equals

The correct option is C →0We have reciprocal systemp=b×c[abc],q=c×a[abc],r=a×b[abc]∴a×p=1[abc]a×(b×c),b×q=1[abc]b×(c×a),c×r=1[abc]c×(a×b).∴a×p+b×q+c×r=1[abc]{a×(b×c)+b×(c×a)+c×(a×b)}=0.

If $\to a, \to b, \to c$ and $\to p, \to q, \to r$ are reciprocal system of vectors, then $(\to a \times \to p) + (\to b \times \to q) + (\to c \times \to r)$ equals