For unit vectors b c and any vector a- the value of a - b - a - c- b - c- b - c is equal to

The correct option is D none of these {{ a× c+b× a+b× c }×( b× c ) #160;} .( b+c ) ={( a× c ) ×( b× c ) ×( b× a ) ×( b× c ) #160;} .( b+c ) ...

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Byju's Answer

Standard XII

Physics

Differentiation

For unit vect...

Question

For unit vectors $b$ & $c$ and any vector $a$ , the value of ${(a + b) \times (a + c) \times (b \times c)} \cdot (b + c)$ is equal to

{| a |}^{2}

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2 {| a |}^{2}

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3 {| a |}^{2}

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none of these

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Solution

The correct option is D none of these
${{a \times c + b \times a + b \times c} \times (b \times c)} . (b + c)$
$= {(a \times c) \times (b \times c) \times (b \times a) \times (b \times c)} . (b + c)$
$= [(a . (b \times c)) c - (c . (b \times c)) a + (b \times (b \times c)) a - (a (b \times c) b)] . (b + c)$
$= [(a . (b \times c)) (c - b) . (b + c)] = (a . (b \times c)) [{| c |}^{2} = {| b |}^{2}] = 0$ $[∵ | b | = | c | = 1]$

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For unit vectors b & c and any vector a, the value of {(a+b)×(a+c)×(b×c)}⋅(b+c) is equal to

The correct option is D none of these{{a×c+b×a+b×c}×(b×c)}.(b+c)={(a×c)×(b×c)×(b×a)×(b×c)}.(b+c)=[(a.(b×c))c−(c.(b×c))a+(b×(b×c))a−(a(b×c)b)].(b+c)=[(a.(b×c))(c−b).(b+c)]=(a.(b×c))[|c|2=|b|2]=0 [∵|b|=|c|=1]

For unit vectors $b$ & $c$ and any vector $a$ , the value of ${(a + b) \times (a + c) \times (b \times c)} \cdot (b + c)$ is equal to