Add: `t-8tz,3tz-z,z-t`

Simplify combining like terms: -z^2+13z^2-5z+7z^3-15z

If the straight line, x =1 + s, y = 3- lambda s, z = 1+ lambda s and x=t/2, y=1+t, z=2-t , with parameters s and t and respectively, are coplanar, then lambda equals:

Add: 2x(z-x-y) and 2y(z-y-x)

Factorize the expressions and divide them as directed: 4yz(z^2+6z-16):-2y(z+8)

Let z_(1) and z_(2) be two distinct complex numbers and let z=(1-t)z_(1)+tz_(2) for some real number t with 0 lt t lt 1 . If Arg (w) denotes the principal argument of a non zero complex number w , then

For what values of k the set of equations 2x- 3y + 6z - 5t = 3, y -4z + t=1 , 4x-5y+8z-9t = k has infinite solution and no solution.

Let C:vecr(t)=x(t)hati+y(t)hatj+z(t)hatk be a differentiable curve, i.e. exists lim_(hto0) (vecr(t+h)-vecr(t))/(h) AA t therefore vecr'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk vecr'(t) is tangent to the curve C at the point P[x(t),y(t),z(t)] and points in the direction of increasing t . The tangent vector to vecr(t)=(2t^(2))hati+(1-t)hatj+(3t^(2)+2)hatk at (2,0,5) is:

Let C:vecr(t)=x(t)hati+y(t)hatj+z(t)hatk be a differentiable curve i.e., exists lim_(hto0) (vecr(t+h)-vecr(t))/(h) forall t therefore vecr'(t)=x'(t)hati+y'(t)hatj+z'(t)hatk vecr'(t) is tangent to the curve C at the point P[x(t),y(t),z(t)], vecr'(t) points in the direction of increasing t . The point P on the curve vecr(t)=(1-2t)hati+t^(2)hatj+2e^(2(t-1))hatk at which the tangent vector vecr'(t) is parallel to the radius vector vecr(t) is:

If |z_(1)|=|z_(2)|=|z_(3)|=1 and z_(1)+z_(2)+z_(3)=0 , then z_(1),z_(2),z_(3) are vertices of