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If `vecu=veca-vecb,vecv=veca+vecb and |veca|=|vecb|=2,` then `|vecuxxvecv|` is equal toA. `2sqrt(16-(a.b)^(2))`B. `sqrt(16-(a.b)^(2))`C. `2sqrt(4-(a.b)^(2))`D. `2sqrt(4+(a.b)^(2))` |
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Answer» Correct Answer - A Now, `|uxxv|=(a-b)xx(a+b)=2|=2||axxb|` `[because axxa =bxxb=0]` and `|axxb|^(2)+(a*b)^(2)=(ab sin theta)^(2)+(ab cos theta)^(2)=a^(2)b^(2)` `|axxb|=sqrt(a^(2)b^(2)-(a*b)^(2))` So`|uxxv|=2|axxb|=2sqrt(a^(2)b^(2)-(a*b)^(2))` `=2sqrt(2^(2)2^(2)-(a*b)^(2))` `=2sqrt(16-(a*b)^(2))[because |a|=|b|=2]` |
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