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 $\left.k \in\right]-1,1\left[k \neq \frac{-1}{2}\right.$.
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 ......... .
$$\begin{array}{ll} \because & |\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|^2+|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}|^2=144=|\overrightarrow{\mathbf{a}}|^2 \cdot|\overrightarrow{\mathbf{b}}|^2 \\ \Rightarrow & |\overrightarrow{\mathbf{a}}|^2|\overrightarrow{\mathbf{b}}|^2=144 \\ \Rightarrow & |\overrightarrow{\mathbf{b}}|^2=\frac{144}{|\overrightarrow{\mathbf{a}}|^2}=\frac{144}{16}=9 \\ \therefore & |\overrightarrow{\mathbf{b}}|=3 \end{array}$$