If the angle between the vectors $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$ is obtuse,then $\lambda \in$

  • A
    $\left(0, \frac{1}{2}\right]$
  • B
    $\left(0, \frac{1}{2}\right)$
  • C
    $\left[0, \frac{1}{2}\right]$
  • D
    $\left[0, \frac{1}{2}\right)$

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Consider the following Assertion $(A)$ and Reason $(R)$:
Assertion $(A)$: The two lines $\bar{r}=\bar{a}+t(\bar{b})$ and $\bar{r}=\bar{b}+s(\bar{a})$ intersect each other.
Reason $(R)$: The shortest distance between the lines $\bar{r}=\bar{p}+t(\bar{q})$ and $\bar{r}=\bar{c}+s(\bar{d})$ is equal to the length of the projection of the vector $(\bar{p}-\bar{c})$ on $(\bar{q} \times \bar{d})$.
The correct answer is:

If the lengths of three vectors $\bar{a}, \bar{b}$ and $\bar{c}$ are $5, 12, 13$ units respectively,and each one is perpendicular to the sum of the other two,then $|\bar{a}+\bar{b}+\bar{c}| = \dots$

If the angle $\theta$ between the vectors $\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$ is such that $90^{\circ} < \theta < 180^{\circ}$,then $x$ lies in the interval

If $G(\vec{g}), H(\vec{h})$ and $P(\vec{p})$ are the centroid,orthocenter,and circumcenter of a triangle respectively,and $x \vec{p} + y \vec{h} + z \vec{g} = 0$,then $(x, y, z) = $

Let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be two sides of a triangle. The median $\overrightarrow{AM}$ is perpendicular to the angle bisector $\overrightarrow{OL}$ and $|\overrightarrow{AM}|:|\overrightarrow{OL}|=1:2$. The angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is

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