In a trapezium,the vector $\overrightarrow{BC} = \lambda \overrightarrow{AD}$. We will then find that $p = \overrightarrow{AC} + \overrightarrow{BD}$ is collinear with $\overrightarrow{AD}$. If $p = \mu \overrightarrow{AD}$,then

$a = 4 \hat{i} + 3 \hat{j}$ 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:

$a$ and $b$ are unit vectors such that $a+2b$ is also a unit vector. If $\theta$ is the angle between $a$ and $b$,then $\sin \theta + \cos^3 \theta + \tan^5 \theta$ is equal to

If $\vec{a}=\hat{i}+(\tan \theta) \hat{j}+\left(\frac{3}{\sqrt{\sin \frac{\theta}{2}}}\right) \hat{k}$ and $\vec{b}=\tan \theta(\hat{j}-\hat{i})-\left(2 \sqrt{\sin \frac{\theta}{2}}\right) \hat{k}$ are orthogonal vectors and $\vec{c}=(\sin 2 \theta) \hat{i}-2 \hat{j}+2 \hat{k}$ makes an obtuse angle with $X$-axis, then $\theta=$

Let $\bar{a}, \bar{b}, \bar{c}$ be three unit vectors satisfying $|\bar{a}-\bar{b}|^2+|\bar{a}-\bar{c}|^2=10$. Then
Statement $(I)$ : $|\bar{a}+2 \bar{b}|^2+|2 \bar{a}+\bar{c}|^2=2$.
Statement $(II)$ : $|2 \bar{a}+3 \bar{b}|^2+|3 \bar{a}+2 \bar{c}|^2=10$.
Which of the above statements is (are) true?

If $\bar{a}=\hat{i}+2 \hat{j}-3 \hat{k}$,$\bar{b}=3 \hat{i}-\hat{j}+2 \hat{k}$,$\bar{c}=\hat{i}+3 \hat{j}+\hat{k}$ and $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then $\lambda=$