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Vectors a and b are such that |a|=1,|b|=4 and a.b=2." If " c=2axxb-3b, then the angle between b and c is |
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Answer» `(pi)/(6)` ` and |c|^(2)=(2axxb-3b)*(2axxb-3b)` `=4|axxb|^(2)+9|b|^(2)=4*12+9*16=192` `rArr |c|=8sqrt(3)` Again now `b*c = b*(2axxb-3b)2(b*axxb)-3|b|^(2)` `=-3|b|^(2)=-48` `therefore cos theta=(b*c)/(|b||c|)=-(48)/(4*8sqrt(3))=-(SQRT(3))/(2)` `therefore theta = (5pi)/(6)` |
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