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

|a|=1=|b| a·b=0c=xa+yb+z(a×b)a·c=x(1)+0+0 ⇒ x=a·cb·c=0+y(1)+0 ⇒ y=b·c(a×b)·c=0+0+z[a×b]2=z(1) (1)x+y+z=c·(a+b+a×b)

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

|a|=1=|b| a·b=0c=xa+yb+z(a×b)a·c=x(1)+0+0 ⇒ x=a·cb·c=0+y(1)+0 ⇒ y=b·c(a×b)·c=0+0+z[a×b]2=z(1) (1)x+y+z=c·(a+b+a×b)

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

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

$a . c$

answer is C.

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Detailed Solution

$\begin{array}{rcl} | a | & = & 1 = | b | a \cdot b = 0 \\ c & = & x a + y b + z (a \times b) \\ a \cdot c & = & x (1) + 0 + 0 \Rightarrow x = a \cdot c \\ b \cdot c & = & 0 + y (1) + 0 \Rightarrow y = b \cdot c \\ (a \times b) \cdot c & = & 0 + 0 + z {[a \times b]}^{2} = z (1) (1) \\ x + y + z & = & c \cdot (a + b + a \times b) \end{array}$