If the area of the parallelogram with $a$ and $b$ as two adjacent sides is $15$ sq units,then the area of the parallelogram having $3a+2b$ and $a+3b$ as two adjacent sides in sq units is

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-7\hat{j}+\hat{k}$.

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\pi / 3$,then $\vec{a}$ is equal to

If $|\vec{a}|=10, |\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$,then the value of $|\vec{a} \times \vec{b}|$ is

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=\hat{i}-\hat{j}$ and if $6 \hat{i}+2 \hat{j}+3 \hat{k}=\lambda_1(\vec{a} \times \vec{b})+\lambda_2(\vec{b} \times \vec{c})+\lambda_3(\vec{c} \times \vec{a})$,then $(\lambda_1, \lambda_2, \lambda_3)=$