The vectors a- and b- are not perpendicular and c- and d- are two vectors satisfying- b-c-b-d- and a-d-0- then the vector d- is equal to

The correct option is D c⃗ (a⃗.c⃗a⃗.b⃗)b⃗ Given that #160; a⃗.b⃗≠ 0 a⃗.d⃗=0 b⃗×c⃗=b⃗×d⃗ So #160; b⃗×(c⃗ d⃗)=0 b⃗ is parallel t ...

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Byju's Answer

Standard XII

Chemistry

Mathematics of Dipole Moment

The vectors ...

Question

The vectors $\to a$ and $\to b$ are not perpendicular and $\to c$ and $\to d$ are two vectors satisfying: $\to b \times \to c = \to b \times \to d$ and $\to a . \to d = 0$ , then the vector $\to d$ is equal to

\to b - (\frac{\to b . \to c}{\to a . \to b}) \to c

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\to c + (\frac{\to a . \to c}{\to a . \to b}) \to b

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\to b + (\frac{\to b . \to c}{\to a . \to b}) \to c

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\to c - (\frac{\to a . \to c}{\to a . \to b}) \to b

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Solution

The correct option is D $\to c - (\frac{\to a . \to c}{\to a . \to b}) \to b$
Given that
$\to a . \to b \neq 0$
$\to a . \to d = 0$
$\to b \times \to c = \to b \times \to d$
So
$\to b \times (\to c - \to d) = 0$
$\to b$ is parallel to $\to c - \to d$
$∴ \to c - \to d = λ \to b$
Taking dot product with $\to a$
$\to a . \to c - 0 = λ \to a . \to b$
$λ = \frac{\to a . \to c}{\to a . \to b}$
So, $\to d = \to c - (\frac{\to a . \to c}{\to a . \to b}) \to b$

Hence, option $^{'} D^{'}$ is correct.

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The vectors →a and →b are not perpendicular and →c and →d are two vectors satisfying: →b×→c=→b×→d and →a.→d=0, then the vector →d is equal to

The vectors $\to a$ and $\to b$ are not perpendicular and $\to c$ and $\to d$ are two vectors satisfying: $\to b \times \to c = \to b \times \to d$ and $\to a . \to d = 0$ , then the vector $\to d$ is equal to