$a = 4 \hat{i} + 3 \hat{j}$ and $b$ are two vectors in the $XOY$ plane, and $a$ is perpendicular to $b$. $A$ vector $c$ lying in the same plane and having projections $1$ and $2$ on $a$ and $b$ respectively is:

If the vectors $a\hat{i} + 2\hat{j} + 3\hat{k}$ and $-\hat{i} + 5\hat{j} + a\hat{k}$ are perpendicular to each other,then $a = $

Let $a=\hat{i}+2 \hat{j}-2 \hat{k}$ and $b=2 \hat{i}-\hat{j}-2 \hat{k}$. If the orthogonal projection vector of $a$ on $b$ is $x$ and the orthogonal projection vector of $b$ on $a$ is $y$, then $|x-y|=$

Three vectors of magnitudes $a, 2a, 3a$ are along the directions of the diagonals of $3$ adjacent faces of a cube that meet at a point. The magnitude of the sum of these vectors is: (in $a$)

If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=2 \hat{i}+3 \hat{j}+2 \hat{k}$ and $c$ is a vector perpendicular to $b$,then $\left\{\frac{a \cdot(b \times c)}{|b \times c|^2}\right\}(b \times c)+\left\{\frac{a \cdot b}{|b|^2}\right\} b+\left\{\frac{a \cdot c}{|c|^2}\right\} c$ is equal to:

Let $\overrightarrow{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{b} = 7\hat{i} + \hat{j} - 6\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{r} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,then $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k})$ is equal to: