Let $a = i + 2j + k$,$b = i - j + k$,$c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has projection $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$, respectively is . . . . . . .

If $|\vec{a}|=2, |\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$,then $|\vec{a} \cdot \vec{b}|$ is equal to :

The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors $\bar{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\bar{c}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}$ is $1$. Then the value of $\lambda$ is:

Let $a$ and $b$ be linearly independent vectors such that $|a| = \sqrt{3}$, $|b| = 3$, and $|a - b| = 4$. If $a \times (2i + 2j - k) = (2i + 2j - k) \times b$ and $|(a + b) \cdot (3i + 4j + 2k)| = \sqrt{\lambda}$, then $\lambda = ...$

If $\theta$ is the angle between the unit vectors $\mathbf{a}$ and $\mathbf{b}$,then $\cos \frac{\theta}{2} = $