Question

if A is constant vector and R=xi+yj+zk then grad( A.R) is ?????​

Answer 1

Step-by-step explanation:

1. A . xi+yj+zk
2. A.xi+A.yj+A.zk
3. Ax+Ay+Az

Answer 2

Answer:

Final Answer.

Step-by-step explanation:

Given,

A is a constant vector

R is a position vector (R=xi+yj+zk)

Let vector A be $a_{1}$i + $a_{2}$j + $a_{3}$k

⇒ A.r = $a_{1}$x + $a_{2}$y + $a_{3}$z

We know that,

∇(A.r)=(∂a⋅r/∂x, ∂a⋅r/∂y, ∂a⋅r/∂z)

We also see that,

∂a⋅r/∂x = ∂($a_{1}$x + $a_{2}$y + $a_{3}$z)/∂x = $a_{1}$

Similarly for $a_{2}$ and $a_{3}$,

∂a⋅r/∂y = ∂($a_{1}$x + $a_{2}$y + $a_{3}$z)/∂y = $a_{2}$

∂a⋅r/∂z = ∂($a_{1}$x + $a_{2}$y + $a_{3}$z)/∂z = $a_{3}$

To Find, The Gradient

Thus,

∇(A.r) = ($a_{1}$, $a_{2}$, $a_{3}$)

As taken above,

$a_{1}$i + $a_{2}$j + $a_{3}$k = A

hence, the final answer will be

Gradient of (A.r) is A.

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