Question

solve this numerical please.​

Answer 1

Let $\vec a,\vec b,\vec c,\vec d$ be vectors pointing to $A,B,C,D$ respectively.

Then

$AB = \vec b - \vec a = 4\,\vec\imath - 3\,\vec k$

$BC = \vec c - \vec b = -2\,\vec\jmath$

$CD = \vec d - \vec c = 2\,\vec\jmath - 4\,\vec k$

and the resultant vector is

$\vec a + \vec b + \vec c + \vec d = -\vec\imath + 2\,\vec\jmath + 3\,\vec k$

We want to find $\vec a$, the displacement from the origin to $A$.

By elimination, we have

$(\vec b - \vec a) + (\vec c - \vec b) = \vec c - \vec a \\\\ \implies \vec c - \vec a = (4\,\vec\imath - 3\,\vec k) + (-2\,\vec\jmath) = 4\,\vec\imath - 2\,\vec\jmath - 3\,\vec k$

$(\vec c - \vec a) + (\vec d - \vec c) = \vec d - \vec a \\\\ \implies \vec d - \vec a = (4\,\vec\imath - 2\,\vec\jmath - 3\,\vec k) + (2\,\vec\jmath - 4\,\vec k) = 4\,\vec\imath - 7\vec k$

Then in the resultant equation, we have

$\vec a + \vec b + \vec c + \vec d = -\vec\imath + 2\,\vec\jmath + 3\,\vec k \\\\ \vec a + (\vec b - \vec a) + (\vec c - \vec a) + (\vec d - \vec a) = -3\vec a -\vec\imath + 2\,\vec\jmath + 3\,\vec k$

Solve for $\vec a$.

$4\vec a + (4\,\vec\imath - 3\,\vec k) + (4\,\vec\imath - 2\,\vec\jmath - 3\,\vec k) + (4\,\vec\imath - 7\vec k) = -\vec\imath + 2\,\vec\jmath + 3\,\vec k \\\\ 4 \vec a + 12\,\vec\imath - 2\,\vec\jmath - 13\,\vec k = -\vec\imath + 2\,\vec\jmath + 3\,\vec k \\\\ 4\vec a = -13\,\vec\imath + 4\,\vec\jmath + 16\,\vec k \\\\ \boxed{\vec a = -\dfrac{13}4\,\vec\imath + \vec\jmath + 4\,\vec k}$