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.
If $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{a}}=0, \overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{b}}=0$ and $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{c}}=0$ for some non-zero vector $\overrightarrow{\mathbf{r}}$, then the value of $\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$ is ......... .
Since, $\overrightarrow{\mathbf{r}}$ is a non-zero vector. So, we can say that $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are in a same plane.
$$\therefore \quad \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=0$$
[since, angle between $\overrightarrow{\mathbf{a}}, \overrightarrow{\boldsymbol{b}}$ and $\overrightarrow{\mathbf{c}}$ are zero i.e., $\theta=0$ ]
The vectors $\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}-2 \hat{\mathbf{k}}$ are the adjacent sides of a parallelogram. The angle between its diagonals is ............. .
$$\begin{array}{ll} \text { We have, } & \overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}-2 \hat{\mathbf{k}} \\ \therefore & \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}} \text { and } \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{array}$$
Now, let $\theta$ is the acute angle between the diagonals $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}$.
$$ \begin{array}{ll} \therefore \quad \cos \theta & =\frac{(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}})}{|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|} \\ & =\frac{(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \cdot(4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})}{\sqrt{8} \sqrt{16+4+16}}=\frac{8+4}{2 \sqrt{2} \cdot 6}=\frac{1}{\sqrt{2}} \\ &\therefore \quad \theta =\frac{\pi}{4} \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \end{array}$$
The values of $k$, for which $|k \overrightarrow{\mathbf{a}}|<\overrightarrow{\mathbf{a}} \mid$ and $k \overrightarrow{\mathbf{a}}+\frac{1}{2} \overrightarrow{\mathbf{a}}$ is parallel to $\overrightarrow{\mathbf{a}}$ holds true are ............. .
We have, $|k \overrightarrow{\mathbf{a}}|< |\overrightarrow{\mathbf{a}}|$ and $k \overrightarrow{\mathbf{a}}+\frac{1}{2} \overrightarrow{\mathbf{a}}$ is parallel to $\overrightarrow{\mathbf{a}}$.
$$\begin{array}{ll} \therefore & |k \overrightarrow{\mathbf{a}}|< |\overrightarrow{\mathbf{a}}| \Rightarrow|k||\overrightarrow{\mathbf{a}}|< |\overrightarrow{\mathbf{a}}| \\ \Rightarrow & |k|< 1 \Rightarrow-1< k<1 \end{array}$$
Also, since $k \overrightarrow{\mathbf{a}}+\frac{1}{2} \overrightarrow{\mathbf{a}}$ is parallel to $\overrightarrow{\mathbf{a}}$, then we see that at $k=\frac{-1}{2}, k \overrightarrow{\mathbf{a}}+\frac{1}{2} \overrightarrow{\mathbf{a}}$ becomes a null vector and then it will not be parallel to $\overrightarrow{\mathbf{a}}$.
So, $k \overrightarrow{\mathbf{a}}+\frac{1}{2} \overrightarrow{\mathbf{a}}$ is parallel to $\overrightarrow{\mathbf{a}}$ holds true when $\left.k \in\right]-1,1\left[k \neq \frac{-1}{2}\right.$.