$a = 3i - 5j$ and $b = 6i + 3j$ are two vectors and $c$ is a vector such that $c = a \times b$,then $|a|:|b|:|c|$ is

If $\vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{c}=-\hat{i}+\hat{j}-4 \hat{k}$ and $\vec{d}=\hat{i}+\hat{j}+\hat{k}$,then $|(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})|=$

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the planes defined by the vectors $\vec{i}, \vec{i} + \vec{j}$ and $\vec{i} - \vec{j}, \vec{i} + \vec{k}$. The angle between $\vec{a}$ and the vector $\vec{i} - 2\vec{j} + 2\vec{k}$ is .....

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{c}=\hat{i}-\hat{j}$ and $\vec{d}=\hat{i}+\hat{j}+x \hat{k}$. If $(\vec{a} \times \vec{b}) \times \vec{c}$ is perpendicular to $\vec{d}$, then $x=$

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{(\bar{b}+\bar{c})}{\sqrt{2}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

Let $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Assertion $(A)$ : The identity $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2=2|\overrightarrow{a}|^2$ holds for $\overrightarrow{a}$.
Reason $(R)$ : $\overrightarrow{a} \times \hat{i}=a_3 \hat{j}-a_2 \hat{k}$,$\overrightarrow{a} \times \hat{j}=a_1 \hat{k}-a_3 \hat{i}$,and $\overrightarrow{a} \times \hat{k}=a_2 \hat{i}-a_1 \hat{j}$.
Which of the following is correct?