If the two diagonals of a parallelogram are $\bar{d_1} = \bar{i} + 2\bar{j} + 3\bar{k}$ and $\bar{d_2} = -2\bar{i} + \bar{j} - 2\bar{k}$,then the area of the parallelogram in square units is

  • A
    $\frac{3}{2} \sqrt{5}$
  • B
    $5 \sqrt{3}$
  • C
    $\frac{5}{2} \sqrt{6}$
  • D
    $3 \sqrt{\frac{5}{2}}$

Explore More

Similar Questions

Let $\overrightarrow{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k}$ and $\overrightarrow{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k}$,where $\alpha \in R$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $\sqrt{15(\alpha^{2} + 4)}$,then the value of $2|\vec{a}|^{2} + (\vec{a} \cdot \vec{b})|\vec{b}|^{2}$ is equal to

If $a$ and $b$ are two non-zero perpendicular vectors,then a vector $y$ satisfying the equations $a \cdot y = c$ (where $c$ is a scalar) and $a \times y = b$ is

If $a \times (b \times c) = 0,$ then

If $a = 2i - 3j - k$ and $b = i + 4j - 2k$,then $a \times b$ is

Difficult
View Solution

Let $\vec{a}$ be a unit vector and $\vec{b}$ be a nonzero vector not parallel to $\vec{a}$. The angles of the triangle,two of whose sides are represented by $\sqrt{3}(\vec{a} \times \vec{b})$ and $\vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$,are

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo