If a-b-c-d are non-zero vectors satisfying a-b-c-b-d-0 and c-d-0 - then d- a-d-d-2 is always equal to

Given a=b+c ⋯(i) b×d=0,c·d=0 Taking cross product of d in (i), we get ⇒a×d=b×d+c×d=c×d (∵b×d=0) Now ⇒d× (a×d)=d× (c×d) =(d·d)c (c·d)d =|d|^2c (∵c·d=0) ∴ nbs ...

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Standard XII

Mathematics

Test for Coplanarity

If a,b,c,d ar...

Question

If $\to a, \to b, \to c, \to d$ are non-zero vectors satisfying $\to a = \to b + \to c$ , $\to b \times \to d = \to 0$ and $\to c \cdot \to d = 0$ , then $\frac{\to d \times (\to a \times \to d)}{| \to d |^{2}}$ is always equal to

\to a

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\to d

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\to b

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\to c

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Solution

The correct option is D $\to c$
Given $\to a = \to b + \to c \dots (i)$
$\to b \times \to d = \to 0, \to c \cdot \to d = 0$
Taking cross product of $\to d$ in $(i),$ we get
$\Rightarrow \to a \times \to d = \to b \times \to d + \to c \times \to d = \to c \times \to d$
$(∵ \to b \times \to d = \to 0)$
Now $\Rightarrow \to d \times (\to a \times \to d) = \to d \times (\to c \times \to d)$
$= (\to d \cdot \to d) \to c - (\to c \cdot \to d) \to d = | \to d |^{2} \to c (∵ \to c \cdot \to d = 0) ∴ \to c = \frac{\to d \times (\to a \times \to d)}{| \to d |^{2}}$

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Similar questions

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Test for Coplanarity

Standard XII Mathematics

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If →a,→b,→c,→d are non-zero vectors satisfying →a=→b+→c,→b×→d=→0 and →c⋅→d=0 , then →d×(→a×→d)|→d|2 is always equal to

If $\to a, \to b, \to c, \to d$ are non-zero vectors satisfying $\to a = \to b + \to c$ , $\to b \times \to d = \to 0$ and $\to c \cdot \to d = 0$ , then $\frac{\to d \times (\to a \times \to d)}{| \to d |^{2}}$ is always equal to