marks] Let 
and
be vectors in the three dimensional space.
of 
is (b) 3.
[Reason:
.]
is (a) 3.
[Reason: 
.]
is:
(d) 
.
[Reason:
be
the equation of the motion of a particle.
.
[Reason: The velocity is
and hence the speed is
which is at t=1.]
, the acceleration is
(e) 
.
[Reason: The acceleration is
,
which is 
at .]
( 
) in cylindrical coordinates is: (d) r=2z.
[Reason: For cylindrical coordinates 
, we have 
and 
and hence 
. Therefore
becomes 
. Taking square roots on both sides and
noticing that , we have r=2z.]
. [Reason: Subsituting the change of coordinates
formulae 
, 
and
into , we have
, which gives 
. Since
, we have 
. So, taking square roots on both sides
of , we have .]
, 
and 
by finding a nonzero
normal vector to this plane first.
,
and
is
a normal vector. Thus, if P=(x,y,z) is a point on the plane, we have
, or
3(x-1)+(y-2)-4(z-3)=0. Simplifying the last identity, we have
which is the required equation for the plane.
, 
and
for the function 
. (Here
,
and
.)
and
at the point 
.
and
. So the required tangent plane is
z=1+5(x-2)+(-6)(y-1). Simplifying, we have
