∵x'(π/4)=-√2/2,y'(π/4)=√2/2,z'(π/4)=2
∴所求切线方程是(x-√2/2)/(-√2/2)=(y-√2/2)/(√2/2)=(z-π/2)/(2)
所求法平面方程是(-√2/2)(x-√2/2)+(√2/2)(y-√2/2)+2(z-π/2)=0
∵x'(π/4)=-√2/2,y'(π/4)=√2/2,z'(π/4)=2
∴所求切线方程是(x-√2/2)/(-√2/2)=(y-√2/2)/(√2/2)=(z-π/2)/(2)
所求法平面方程是(-√2/2)(x-√2/2)+(√2/2)(y-√2/2)+2(z-π/2)=0