यदि दो सदिश `bar(a)` और `bar(b)` इस प्रकार हैं कि `|bar(a)+bar(b)|=|bar(b)|`, तब सिद्ध कीजिए कि `bar(a)+2bar(b)`, सदिश `bar(a)` पर लम्ब है |

If |bar(a)+bar(b)|<|bar(a)-bar(b)| then (bar(a),bar(b)) is

If (bar(a), bar(b))=60^(@)" then "(-bar(a), -bar(b))=

bar(a) , bar(b) and bar(c) are three vectors such that bar(a) + bar(b) + bar(c) = bar(0) and |bar(a)| =2, |bar(b)| =3, |bar(c)| =5 ,then bar(a) . bar(b) + bar(b) . bar(c) + bar(c) . bar(a) equals

If |bar(a)|=2, |bar(b)|=3, (bar(a), bar(b))= pi//6 , then find (bar(a) xx bar(b))^(2)

Compute bar(a) xx (bar(b) + bar(c )) + bar(b) xx (bar(c ) +bar(a)) + bar(c ) xx (bar(a)+bar(b))

The non-zero vectors bar(a) , bar(b) , bar(c ) holds | (bar(a) .bar(b) ).bar( c) |=| bar(a) ||bar(b)|| bar( c) if

The points 2bar(a)+3bar(b)+bar(c), bar(a)+bar(b), 6bar(a)+11bar(b)+5bar(c) are

If |bar(a)| = 2, |bar(b)| = 5, and bar(a).bar(b) = 8 then |bar(a) - bar(b)| =__________