Question

If the vectors $3i−4j−k$ and $2i+3j−6k$ represent the diagonals of a rhombus, then the length of the side of the rhombus is

• A. 15
• B. $15\sqrt{3}$
• C. $\frac{5\sqrt{3}}{2}$
• D. $\frac{15\sqrt{3}}{2}$
• E. $\frac{17\sqrt{3}}{2}$

Answer 1

Answer: (C)

Given, $AC=3i−4j−k$
and $BD=2i+3j−6k$

Let sides of a rhombus be $a$ and $b$. In $ΔABC$
$AC=AB+BC$
$⇒3i−4j−k=a+b⋯(i)$
$BD=BA+AD$
$⇒2i+3j−6k=−a+b...(ii)$
On adding Eqs. (i) and (ii), we get
$5i−j−7k=2b$
$⇒âˆ\pounds 2bâˆ\pounds =\sqrt{5^2+1^2+7^2}$
$⇒âˆ\pounds bâˆ\pounds =\frac{\sqrt{25+1+49}}{2}$
$⇒âˆ\pounds bâˆ\pounds =\frac{\sqrt{75}}{2}=\frac{5\sqrt{3}}{2}$