For non-zero vectors $a$ and $b$, if $|a+b| < |a-b|$, then $a$ and $b$ are

If $a$ and $b$ are unit vectors and $\alpha$ is the angle between them, then $a+b$ is a unit vector when $\cos \alpha=$

If $a, b, c$ are non-coplanar vectors,then the point of intersection of the line passing through the points $2a+3b-c$ and $3a+4b-2c$ with the line joining the points $a-2b+3c$ and $a-6b+6c$ is

Let $\vec{\alpha}=4 \hat{i}+3 \hat{j}+5 \hat{k}$ and $\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}$. Let $\vec{\beta}_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$,then the value of $5 \vec{\beta}_2 \cdot(\hat{i}+\hat{j}+\hat{k})$ is

The points $O, A, B, C, D$ are such that $\overrightarrow{OA} = a, \overrightarrow{OB} = b, \overrightarrow{OC} = 2a + 3b$ and $\overrightarrow{OD} = a - 2b$. If $|a| = 3|b|$,then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is

If $a = x^2 \hat{i} + x \hat{j} + 3 \hat{k}$ and $b = x \hat{i} - 4 \hat{j} + 2 \hat{k}$ and $a \cdot b > 6$,then: