Let a¯,b¯ are two unit vectors such that a¯. b¯=0. If then c¯=xa¯×yb¯+za¯×b¯ then x+y+z=

|a¯|=|b¯|=1 (a¯,b¯)=90∘c¯⋅a¯=x,c¯b¯=y,c¯⋅(a¯×b¯)=z(a¯×b¯)2=z x+y+z=(a¯+b¯+a¯×b¯)⋅c¯

Let a¯,b¯ are two unit vectors such that a¯. b¯=0. If then c¯=xa¯×yb¯+za¯×b¯ then x+y+z=

|a¯|=|b¯|=1 (a¯,b¯)=90∘c¯⋅a¯=x,c¯b¯=y,c¯⋅(a¯×b¯)=z(a¯×b¯)2=z x+y+z=(a¯+b¯+a¯×b¯)⋅c¯

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Let $\bar{a}, \bar{b}$ are two unit vectors such that $\bar{a} . \bar{b} = 0$ . If then $\bar{c} = x \bar{a} \times y \bar{b} + z (\bar{a} \times \bar{b})$ then $x + y + z =$

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$\{\bar{a} + \bar{b} + (\bar{a} \times \bar{b})\} . \bar{c}$

$\bar{a} . \bar{c}$

answer is C.

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Detailed Solution

$\begin{array}{l} | \bar{a} | = | \bar{b} | = 1 (\bar{a}, \bar{b}) = 90^{\circ} \\ \bar{c} \cdot \bar{a} = x, \bar{c} \bar{b} = y, \\ \bar{c} \cdot (\bar{a} \times \bar{b}) = z (\bar{a} \times \bar{b})^{2} = z \end{array} x + y + z = (\bar{a} + \bar{b} + \bar{a} \times \bar{b}) \cdot \bar{c}$