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 $\left.k \in\right]-1,1\left[k \neq \frac{-1}{2}\right.$.
The value of the expression $|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2$ is .................. .
$$\begin{aligned} |\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2 & =|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2 \sin ^2 \theta+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2 \\ & =|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2\left(1-\cos ^2 \theta\right)+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2 \\ & =|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2-|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2 \cos ^2 \theta+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2 \\ & =|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2 \\ |\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2+(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}})^2 & =|\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2 \end{aligned}$$
If $|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2+|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|^2=144$ and $|\overrightarrow{\mathbf{a}}|=4$, then $|\overrightarrow{\mathbf{b}}|$ is equal to ......... .
$$\begin{array}{ll} \because & |\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2+|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|^2=144=|\overrightarrow{\mathbf{a}}|^2 \cdot|\overrightarrow{\mathbf{b}}|^2 \\ \Rightarrow & |\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2=144 \\ \Rightarrow & |\overrightarrow{\mathbf{b}}|^2=\frac{144}{|\overrightarrow{\mathbf{a}}|^2}=\frac{144}{16}=9 \\ \therefore & |\overrightarrow{\mathbf{b}}|=3 \end{array}$$
If $\overrightarrow{\mathbf{a}}$ is any non-zero vector, then $(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}) \cdot \hat{\mathbf{i}}+(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{j}}) \cdot \hat{\mathbf{j}}+(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{k}}) \hat{\mathbf{k}}$ is equal to ........... .
Let $$\overrightarrow{\mathbf{a}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}$$
$$\therefore \quad \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}=\mathrm{a}_1, \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{j}}=a_2 \text { and } \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{k}}=a_3$$
$\therefore \quad(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{i}}) \hat{\mathbf{i}}+(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{j}}) \hat{\mathbf{j}}+(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{k}}) \hat{\mathbf{k}}=a_1 \hat{\mathbf{i}}+a_2 \hat{\mathbf{j}}+a_3 \hat{\mathbf{k}}=\overrightarrow{\mathbf{a}}$
If $|\overrightarrow{\mathbf{a}}|=|\overrightarrow{\mathbf{b}}|$, then necessarily it implies $\overrightarrow{\mathbf{a}}= \pm \overrightarrow{\mathbf{b}}$.