Let $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{b} = 6\hat{i} + 3\hat{j} + 3\hat{k}$. Then the square of the area of the triangle with adjacent sides determined by the vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is :

The moment of a force represented by $\overrightarrow{F} = i + 2j + 3k$ about the point $P(2, -1, 1)$ is given by the cross product of the position vector $\overrightarrow{r}$ and the force vector $\overrightarrow{F}$. If the origin is $O(0, 0, 0)$,then $\overrightarrow{r} = 2i - j + k$. Calculate the moment $\overrightarrow{M} = \overrightarrow{r} \times \overrightarrow{F}$.

Let $\overrightarrow{OA}=2 \overrightarrow{a}$,$\overrightarrow{OB}=6 \overrightarrow{a}+5 \overrightarrow{b}$ and $\overrightarrow{OC}=3 \overrightarrow{b}$,where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units,then the area (in sq. units) of the quadrilateral $OABC$ is equal to :

The adjacent sides of a parallelogram are $\bar{a} = \hat{j} + 2\hat{k}$ and $\bar{b} = \hat{i} + 2\hat{j}$. Find its area.

If a non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ and the plane determined by the vectors $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$,then the angle between the vectors $\vec{a}$ and $\hat{i}+\hat{j}+\hat{k}$ is

Let $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Assertion $(A)$ : The identity $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2=2|\overrightarrow{a}|^2$ holds for $\overrightarrow{a}$.
Reason $(R)$ : $\overrightarrow{a} \times \hat{i}=a_3 \hat{j}-a_2 \hat{k}$,$\overrightarrow{a} \times \hat{j}=a_1 \hat{k}-a_3 \hat{i}$,and $\overrightarrow{a} \times \hat{k}=a_2 \hat{i}-a_1 \hat{j}$.
Which of the following is correct?