The projection vector of $\overrightarrow{\mathbf{a}}$ on $\overrightarrow{\mathbf{b}}$ is
If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are three vectors such that $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{0}$ and $|\overrightarrow{\mathbf{a}}|=2$, $|\overrightarrow{\mathbf{b}}|=3$ and $|\overrightarrow{\mathbf{c}}|=5$, then the value of $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}$ is
If $|\overrightarrow{\mathbf{a}}|=4$ and $-3 \leq \lambda \leq 2$, then the range of $|\lambda \overrightarrow{\mathbf{a}}|$ is
The number of vectors of unit length perpendicular to the vectors $\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
The vector $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$ bisects the angle between the non-collinear vectors $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, if...... .
$$\begin{aligned} &\text { If vector } \vec{a}+\vec{b} \text { bisects the angle between the non-collinear vectors, then }\\ &\begin{aligned} \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) & =|\overrightarrow{\mathbf{a}}| \cdot \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \mid \cos \theta \\ \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) & =a \sqrt{a^2+b^2} \cos \theta \\ \Rightarrow \quad \cos \theta & =\frac{\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})}{a \sqrt{a^2+b^2}}\quad\text{.... (i)} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text{and}\quad & \overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=|\overrightarrow{\mathbf{b}}| \cdot|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}| \cos \theta \\ & \overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=b \sqrt{a^2+b^2} \cos \theta \end{aligned}$$
[since, $\theta$ should be same]
$$\Rightarrow \quad \cos \theta=\frac{\overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})}{b \sqrt{a^2+b^2}}\quad\text{.... (ii)}$$
From Eqs. (i) and (ii),
$$\frac{\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})}{a \sqrt{a^2+b^2}}=\frac{\overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})}{b \sqrt{a^2+b^2}} \Rightarrow \frac{\overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|}=\frac{\overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|}$$
$\therefore \hat{a}=\hat{b} \Rightarrow \vec{a}$ and $\overrightarrow{\mathrm{b}}$ are equal vectors.