Let $\vec{a} = \hat{i} + 4\hat{j} $, $\vec{b} = 4\hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - 2\hat{k}$. If $\vec{d}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ such that $\vec{c} \cdot \vec{d} = 16$, then $|\vec{d}|$ is equal to

If $\vec{a} = 2\hat{j} - \hat{k}$, $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{c} = -\hat{i} + \hat{k}$ are three vectors, then the area (in sq. units) of the parallelogram whose diagonals are $(\vec{b} + \vec{c})$ and $(\vec{a} + \vec{c})$ is

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, where $\vec{a}$ and $\vec{b}$ are unit vectors and $|\vec{c}|=2$, then the angle between the vectors $\vec{b}$ and $\vec{c}$ is :

If $|\vec{a}| = |\vec{b}| = |\vec{a} + \vec{b}| = 1$, then $|\vec{a} - \vec{b}|$ is equal to:

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}$, then which of the following statements are correct?

(A) $|\vec{a}| = \sqrt{14}$

(B) $|\vec{b}| = 29$

(C) $\vec{a} \cdot \vec{b} = 8$

(D) Angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{-8}{\sqrt{406}}\right)$

Choose the correct answer from the options given below:

The vector $\{(a^2+a-2)\hat{i}+(a^2-4)\hat{j}-(a^2-5a+6)\hat{k}\}$ makes acute angles with coordinate axes, then values of $a$ lies in :

Arrange the vectors in descending order of their magnitudes.

(A) $\hat{i} + \hat{j} + \hat{k}$

(B) $2\hat{i} - 3\hat{j}$

(C) $\frac{1}{2}\hat{i} - \frac{1}{3}\hat{j}$

(D) $2\hat{i} - \hat{k}$

Choose the correct answer from the options given below :