Question

4) Given that vector field V =(x^ 2 -y^ 2 +2xz)i +(xz-xy+yz)j + (z^ 2 +x^ 2 )k find curl V. Show that the vectors given by curl V at P1{0}(1, 2, - 3) and P2_{1}(2, 3, 12) are orthogonal.

Answer 1

hence the two vectors are orthogonal.

Answer 2

Answer:

The curl V is equal to  $-(x+y)i+(y+z)k$.

Curl V at  P₁(1,2, - 3) and P₂(2, 3, 12) are orthogonal to each other.

Step-by-step explanation:

Given: $V= (x^{2}-y^{2}+2xz)i + (xz-xy+yz)j + (z^{2}+x^{2})k$

Curl V = determinant of V

∴ Curl V =  $\left[\begin{array}{ccc}i&j&k\\\frac{d}{dx} &\frac{d}{dy} &\frac{d}{dz} \\x^{2}-y^{2}+2xz&xz-xy+yz&z^{2}+x^{2}\end{array}\right]$

(consider $\frac{d}{dx}$ as partial derivative w.r.t. x)

$= i[\frac{d}{dy}(z^{2}+x^{2})-\frac{d}{dz}(xz-xy+yz)] +j[\frac{d}{dz} (x^{2}-y^{2}+2xz)- \frac{d}{dx}(z^{2}+x^{2})]+k[\frac{d}{dx}(xz-xy+yz)-\frac{d}{dy} (x^{2}-y^{2}+2xz)]$

 $=i[-(x+y)]+j(2x-2x)+k(z-y+2y)]\\=-(x+y)i+(y+z)k$

Therefore, curl V at P₁(1,2, - 3)  $=-(1+2)i+(2-3)k=-3i-k$

Curl V at P₂(2, 3, 12) $= -(2+3)i+(3+12)k=-5i+15k$

Curl V at P₁,  P₂ will be orthogonal if their dot product is zero.

$(-3i-k).(-5i+15k)=(-3i)(-5i)+(-k)(15k)=15-15=0$

Hence, Curl V at  P₁ and P₂ are orthogonal.