$a=3 \hat{i}+\hat{j}-\hat{k}, b=\hat{i}-4 \hat{j}+5 \hat{k}, c=4 \hat{i}+5 \hat{j}-\hat{k}$ are three vectors and a vector $r$ is perpendicular to both the vectors $b$ and $c$. If $r \cdot a=9$,then $r=$

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$, then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to:

The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$,where $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ is

Show that $(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$.

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 $(\lambda_1, \lambda_2, \lambda_3)=$

If $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ are unit vectors such that $\bar{a} \cdot \bar{b} = \frac{1}{2}$,$\bar{c} \cdot \bar{d} = \frac{1}{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c} \times \bar{d}$ is $\frac{\pi}{6}$,then the value of $|[\bar{a} \bar{b} \bar{d}] \bar{c} - [\bar{a} \bar{b} \bar{c}] \bar{d}| = $