Let the vectors $a, b, c$ and $d$ be such that $(a \times b) \times (c \times d) = 0$. Let $P_1$ and $P_2$ be planes determined by the pairs of vectors $(a, b)$ and $(c, d)$ respectively. Then the angle between $P_1$ and $P_2$ is:

Let $\vec{a} = \alpha \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \alpha \hat{k}$,where $\alpha > 0$. If the projection of $\vec{a} \times \vec{b}$ on the vector $\vec{c} = -\hat{i} + 2 \hat{j} - 2 \hat{k}$ is $30$,then $\alpha$ is equal to:

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

Let $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$,$\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}$ be a vector such that $|\bar{c}-\bar{a}|=4$,$|(\bar{a} \times \bar{b}) \times \bar{c}|=3$ and the angle between $\bar{c}$ and $\bar{a} \times \bar{b}$ is $\frac{\pi}{6}$,then $\bar{a} \cdot \bar{c}$ is equal to

Let $\vec \alpha = 3\hat i + \hat j$ and $\vec \beta = 2\hat i - \hat j + 3\hat k.$ If $\vec \beta = \vec \beta _1 - \vec \beta _2,$ where $\vec \beta _1$ is parallel to $\vec \alpha$ and $\vec \beta _2$ is perpendicular to $\vec \alpha,$ then $\vec \beta _1 \times \vec \beta _2$ is equal to

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{a} \times \vec{c} = \vec{b}$, $\vec{a} \cdot \vec{c} = 3$, then $\vec{c}$ is