Three vectors →A=a→i+→j+→k,→B=→i+b→j+→k,→C=→i+→j+c→k are mutually perpendicular (→i,→j,→k are unit vectors along X,Y,Z axis respectively). The respective values of a,b and c are
A
0,0,0
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B
−12,−12,−12
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C
1,−1,1
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D
12,12,12
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Solution
The correct option is B−12,−12,−12 a+b+1=0 ------- (1) 1+b+c=0 --------(2) a+1+c=0 ---------(3) -------------------- 2(a+b+c)+3=0 a+b+c=−32 ---- (4) ⇒−1+c=−32 using (1) c=−12b=−12a=−12