If `bar(a)=bar(i)+bar(j)+bar(k),bar(b)=2bar(i)-bar(j)+3bar(k)` ,and `bar(c)=bar(i)-2bar(j)+bar(k)` then a unit vector parallel to `2bar(a)-bar(b)+3bar(c)` is

(bar(a)*bar(i))^(2)+(bar(a)*bar(j))^(2)+(bar(a)*bar(k))^(2)=

Evaluate : 2.bar(5)-0.bar(35)

Evaluate : 2.bar(7)+1.bar(3)

Evaluate 3.bar(2)-0.bar(16)

Show that points with p.v bar(a)-2bar(b)+3bar(c ),-2bar(a)+3bar(b)-bar(c ),4bar(a)-7bar(b)+7bar(c ) are collinear. It is given that vectors bara,barb,bar c are non-coplanar.

In a regular hexagon ABCDEF, bar(AB) + bar(AC)+bar(AD)+ bar(AE) + bar(AF)=k bar(AD) then k is equal to

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that bar(OP)*bar(OQ)+bar(OR)*bar(OS)=bar(OR)*bar(OP)+bar(OQ)*bar(OS)=bar(OQ)*bar(OR)+bar(OP)*bar(OS) Then the triangle PQR has S as its

Evaluate : 1.bar(45)+0.bar(3)

In the Boolean algebra bar(A).bar(B) equals

In triangle ABC if bar(AB) =u/(|bar u|)-v/(|bar v|) and bar(AC)=(2bar u)/(|baru|) where |bar u| != |bar |v| then