ABCDEF is a regular hexagon where centre O is the origin. If the position vectors of A and B are hat i - hat j +2 hat k and 2 hat i + hat j - hat k respectively, then BC is equal to

Question

ABCDEF is a regular hexagon where centre O is the origin. If the position vectors of A and B are hat i - hat j +2 hat k and 2 hat i + hat j - hat k respectively, then BC is equal to (A)  hat i + hat j -2 hat k  (B) - hat i + hat j -2 hat k  (C) 3 hat i +3 hat j -4 hat k  (D) None of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/e4c4a4a6f4df953f24d1ed44d1cb2fdb-.png" /> OA = hat i - hat j +2 hat k , OB =2 hat i + hat j - hat k OC = AB = OB - OA = hat i +2 hat j -3 hat k BC = OC - OB =( hat i +2 hat j -3 hat k )-(2 hat i + hat j - hat k ) =- hat i + hat j -2 hat k

Q. ABCDEF is a regular hexagon where centre $O$ is the origin. If the position vectors of $A$ and $B$ are $\hat{i} - \hat{j} + 2 \hat{k}$ and $2 \hat{i} + \hat{j} - \hat{k}$ respectively, then $BC$ is equal to

$\hat{i} + \hat{j} - 2 \hat{k}$

$- \hat{i} + \hat{j} - 2 \hat{k}$

$3 \hat{i} + 3 \hat{j} - 4 \hat{k}$

None of these

$O A = \hat{i} - \hat{j} + 2 \hat{k}, OB = 2 \hat{i} + \hat{j} - \hat{k}$
$OC = A B = OB - O A = \hat{i} + 2 \hat{j} - 3 \hat{k}$
$BC = OC - OB = (\hat{i} + 2 \hat{j} - 3 \hat{k}) - (2 \hat{i} + \hat{j} - \hat{k})$
$= - \hat{i} + \hat{j} - 2 \hat{k}$

Q. ABCDEF is a regular hexagon where centre O is the origin. If the position vectors of A and B are i^−j^​+2k^ and 2i^+j^​−k^ respectively, then BC is equal to

i^+j^​−2k^

−i^+j^​−2k^

3i^+3j^​−4k^