If the vectors `bar(a),bar(b),bar(c)` are coplanar, then `|[bar(a),bar(b),bar(c)],[bar(a)*bar(a),bar(a)*bar(b),bar(a)*bar(c)],[bar(b)*bar(a),bar(b)*bar(b),bar(b)*bar(c)]|=`

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

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

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

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

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

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

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.

Given four non zero vectors bar a,bar b,bar c and bar d . The vectors bar a,bar b and bar c are coplanar but not collinear pair by pairand vector bar d is not coplanar with vectors bar a,bar b and bar c and hat (bar a bar b) = hat (bar b bar c) = pi/3,(bar d bar b)=beta ,If (bar d bar c)=cos^-1(mcos beta+ncos alpha) then m-n is :

bar(A.barB+barA.B) is equivalent to