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:

The value of $b$ such that the scalar product of the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ with the unit vector parallel to the sum of the vectors $\vec{u} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{v} = b\hat{i} + 2\hat{j} + 3\hat{k}$ is $1$, is...

If $|a| = |b| = 1$ and $|a + b| = \sqrt{3}$,then the value of $(3a - 4b) \cdot (2a + 5b)$ is

Let $\overrightarrow{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{k}$,and $\overrightarrow{c} = \hat{i} + 2\hat{j} - 3\hat{k}$ be three given vectors. If $\overrightarrow{r}$ is a vector such that $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{c} \times \overrightarrow{a}$ and $\overrightarrow{r} \cdot \overrightarrow{b} = 0$,then $|\overrightarrow{r}|$ is equal to:

The angle between the vectors $\bar{a} = 6 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and $\bar{b} = 4 \hat{i} - 4 \hat{j} + 2 \hat{k}$ is . . . . . . .

Find the distance of the point $P(-\hat{i} + 2\hat{j} + 6\hat{k})$ from the line passing through the point $A(2\hat{i} + 3\hat{j} - 4\hat{k})$ and parallel to the vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.