In the cube of side 'a' shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be:

Question

In the cube of side 'a' shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be: (A) (1/2) a ( hati - hatk)  (B) (1/2) a ( hatj - hati)  (C) (1/2) a ( hatk - hati)  (D) (1/2) a ( hatj - hatk)

Tardigrade Admin · Accepted Answer

vecrg = (a/2) hati + (a/2) hatk vecrH =(a/2) hatj + (a/2) hatk vec rH - vecrg = (a/2) ( hatj - hati)

Q. In the cube of side $^{'} a^{'}$ shown in the figure, the vector from the central point of the face $A BO D$ to the central point of the face $BEFO$ will be:

$\frac{1}{2} a (\hat{i} - \hat{k})$

$\frac{1}{2} a (\hat{j} - \hat{i})$

$\frac{1}{2} a (\hat{k} - \hat{i})$

$\frac{1}{2} a (\hat{j} - \hat{k})$

$r_{g} = \frac{a}{2} \hat{i} + \frac{a}{2} \hat{k}$
$r_{H} = \frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}$
$r_{H} - r_{g} = \frac{a}{2} (\hat{j} - \hat{i})$

Q. In the cube of side ′a′ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be:

21​a(i^−k^)

21​a(j^​−i^)

21​a(k^−i^)

21​a(j^​−k^)