The value of $x$ for which the angle between the vectors $a = -3i + xj + k$ and $b = xi + 2xj + k$ is acute and the angle between $b$ and the $x$-axis lies between $\pi/2$ and $\pi$ satisfies:

If $a \cdot i = 4$,then $(a \times j) \cdot (2j - 3k) = $

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three vectors such that $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi}{2}$,then:

The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the line $\vec{r} = 3\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ is:

Let $\overrightarrow{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$,where $\alpha, \beta \in R$. Let a vector $\overrightarrow{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ and $|\vec{b}|^2=6$. If $\vec{a} \cdot \vec{b}=3 \sqrt{2}$,then the value of $(\alpha^2+\beta^2)|\vec{a} \times \vec{b}|^2$ is equal to