Vector Algebra · Physics · MHT CET

The angle subtended by the vector $A=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}$ with the $X$-axis is

What is the angle between resultant of $A+B$ and $\mathbf{A} \times \mathbf{B}$.

The $$x, y$$ components of vector $$\mathbf{P}$$ have magnitudes 1 and 3 and $$x, y$$ components of resultant of $$\mathbf{P}$$ and $$\mathbf{Q}$$ have magnitudes 5 and 6, respectively. What is the magnitude of $$\mathbf{Q}$$ ?

The resultant of two vector $$\mathbf{A}$$ and $$\mathbf{B}$$ is $$\mathbf{C}$$. If the magnitude of $$\mathbf{B}$$ is doubled, the new resultant vector becomes perpendicular to A. Then, the magnitude of $$\mathbf{C}$$ is

Two vectors of same magnitude have a resultant equal to either of the two vectors. The angle between two vectors is

The vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$ are at right. angles to each other. This is possible under the condition

A vector $P$ has $X$ and $Y$ components of magnitude 2 units and 4 units respectively. A vector $Q$ along negative $X$-axis has magnitude 6 units. The vector $(\mathbf{Q}-\mathbf{P})$ will be

$\mathbf{P}$ and $\mathbf{Q}$ are two non-zero vectors inclined to each other at an angle ' $\theta$ '. ' $p$ ' and ' $q$ ' are unit vectors along $\mathbf{P}$ and $\mathbf{Q}$ respectively. The component of $\mathbf{Q}$ in the direction of $\mathbf{Q}$ will be

The resultant $\mathbf{R}$ of $\mathbf{P}$ and $\mathbf{Q}$ is perpendicular to $\mathbf{P}$. Also $|\mathbf{P}|=|\mathbf{R}|$. The angle between $\mathbf{P}$ and $\mathbf{Q}$ is $\left[\tan 45^{\circ}=1\right]$

If $\sqrt{A^2+B^2}$ represents the magnitude of resultant of two vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$, then the angle between two vectors is