The scalar product of the vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of the vectors $\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = \lambda \hat{i} + 2 \hat{j} + 3 \hat{k}$ is equal to $1$. Then,$\lambda =$

Let $ABC$ be a triangle. Let $u = \overrightarrow{AB}$ and $v = \overrightarrow{AC}$. If $D$ is the midpoint of $BC$,then the length of the median of $\triangle ABD$ through the vertex $B$ is:

Let $x \in R$ and $\log_2 x > 0$. Then,the vectors $A = (2, \log_2 x, s)$ and $B = (\log_2 x, s, \log_2 x)$ include an acute angle if

If $\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ are orthogonal,then the value of $\lambda$ is:

The value of $c$ such that for all real $x$, the vectors $\vec{a} = cxi - 6j + 3k$ and $\vec{b} = xi + 2j + 2cxk$ make an obtuse angle is:

Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$,then the cosine of the angle $\alpha$ is given by