Question

If $i+j-k$ and $2i-3j+k$ are adjacent sides of a parallelogram, then the lengths of its diagonals are

• A. $\sqrt{3},\sqrt{14}$
• B. $\sqrt{13},\sqrt{14}$
• C. $\sqrt{21},\sqrt{3}$
• D. $\sqrt{21},\sqrt{13}$

Answer 1

Answer: (D)

Let $AB=i+j-k$
and $BC=2i-3j+k$

The diagonal $AC=AB+BC$
$=(i+j-k)+(2-3j+k)$
$=3i-2j$
The length of diagonal
$AC=\sqrt{(3{)}^2+(-2{)}^2}=\sqrt{9+4}=\sqrt{13}$
The diagonal $DB=AB-AD$
$=(i+j-k)-(2i-3j+k)$
$=-i+4j-2k$
The length of diagonal
$DB=\sqrt{1+16+4}=\sqrt{21}$