Let $\vec{AB} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{AD} = \hat{i} + 2 \hat{j} + \lambda \hat{k}$, $\lambda \in R$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram $ABCD$ be of length $1$ unit. If $\alpha, \beta$, where $\alpha > \beta$, are the roots of the equation $\lambda^2 x^2 - 6 \lambda x + 5 = 0$, then $2 \alpha - \beta$ is equal to

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?

The vector $\vec{a} = (\alpha, 2, \beta)$ lies in the plane of the vectors $\vec{b} = (1, 1, 0)$ and $\vec{c} = (0, 1, 1)$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives the possible values of $\alpha$ and $\beta$?

$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

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 $|x-y|=$

Two adjacent sides of a parallelogram are represented by $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $\vec{AD}$ is rotated by an acute angle $\theta$ in the plane of the parallelogram such that $\vec{AD}$ becomes $\vec{AD'}$. If $\vec{AD'}$ is perpendicular to $\vec{AB}$,find $\cos \theta$.