C- -q-c-a- -r-A- 1B- 0C- 2D- 3

The correct option is D 3 (a̅+b̅).b̅×c̅/[a̅b̅c̅]+(b̅+c̅).c̅×a̅/[a̅b̅c̅]+(c̅+a̅).a̅×b̅/[a̅b̅c̅] =a̅.(b̅×c̅)+0+b̅.(c̅×a̅)+0+c̅.(a̅×b̅)+0/[a̅b̅c̅] =3 ...

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Byju's Answer

Standard XII

Mathematics

Collinear Vectors

c⃗ ·q⃗+c⃗+a⃗ ...

Question

If $\to a, \to b & \to c$ are 3 non-coplanar vectors and $\to p, \to q & \to r$ are vectors defined by $\to p \frac{\to b \times \to a}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]} a n d \to r = \frac{\to a \times \to b}{[\to a \to b \to c]}$ , and, then the value of $(\to a + \to b) . \to p + (\to b + \to c) . \to q + (\to c + \to a) . \to r =$

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Solution

The correct option is D
3

$(¯ a + ¯ b) . \frac{¯ b \times ¯ c}{[¯ a ¯ b ¯ c]} + (¯ b + ¯ c) . \frac{¯ c \times ¯ a}{[¯ a ¯ b ¯ c]} + (¯ c + ¯ a) . \frac{¯ a \times ¯ b}{[¯ a ¯ b ¯ c]} = \frac{¯ a . (¯ b \times ¯ c) + 0 + ¯ b . (¯ c \times ¯ a) + 0 + ¯ c . (¯ a \times ¯ b) + 0}{[¯ a ¯ b ¯ c]} = 3$

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Similar questions

Q. Let

\to a, \to b

and

\to c

be three non-coplanar vectors and let

\to p, \to q

and

\to r

be the vectors defined by

\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

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

$s i n^{6} A + c o s^{6} A + 3 s i n^{2} A c o s^{2} A =$

Q. If

[\to a \times \to b, \to b \times \to c, \to c \times \to a]

\ then

λ

Q. Evaluate

∣ ∣ ∣ \begin{matrix} cos 15^{\circ} & sin 15^{\circ} sin 75^{\circ} & cos 75^{\circ} \end{matrix} ∣ ∣ ∣

(1 + tan A + sec A)(1 + cot A − cosec A) = ?
(a) 1
(b) 0
(c) 2
(d) 4

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If →a,→b&→c are 3 non-coplanar vectors and →p,→q&→r are vectors defined by →p→b×→a[→a→b→c],→q=→c×→a[→a→b→c] and →r=→a×→b[→a→b→c], and, then the value of (→a+→b).→p+(→b+→c).→q+(→c+→a).→r=

The correct option is D 3 (¯a+¯b).¯b×¯c[¯a¯b¯c]+(¯b+¯c).¯c×¯a[¯a¯b¯c]+(¯c+¯a).¯a×¯b[¯a¯b¯c]=¯a.(¯b×¯c)+0+¯b.(¯c×¯a)+0+¯c.(¯a×¯b)+0[¯a¯b¯c]=3