Find the sine of the angle between the vectors $\overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$.
Here, $a_1=3, a_2=1, a_3=2$ and $b_1=2, b_2=-2, b_3=4$
We know that,
$$\begin{aligned} \cos \theta & =\frac{a_1 b_1+a_2 b_2+a_3 b_3}{\sqrt{a_1^2+a_2^2+a_3^2} \sqrt{b_1^2+b_2^2+b_3^2}} \\ & =\frac{3 \times 2+1 \times(-2)+2 \times 4}{\sqrt{3^2+1^2+2^2} \sqrt{2^2+(-2)^2+4^2}} \\ & =\frac{6-2+8}{\sqrt{14} \sqrt{24}}=\frac{12}{2 \sqrt{14} \sqrt{6}}=\frac{6}{\sqrt{84}}=\frac{6}{2 \sqrt{21}}=\frac{3}{\sqrt{21}} \\ &\therefore \quad \sin \theta=\sqrt{1-\cos ^2 \theta} \\ & =\sqrt{1-\frac{9}{21}}=\sqrt{\frac{12}{21}}=\frac{2 \sqrt{3}}{\sqrt{3} \sqrt{7}}=\frac{2}{\sqrt{7}} \end{aligned} $$
If $A, B, C$ and $D$ are the points with position vectors $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$, $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \quad 2 \hat{\mathbf{i}}-3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ respectively, then find the projection of $\overrightarrow{\mathbf{A B}}$ along $\overrightarrow{\mathbf{C D}}$.
$$\begin{aligned} & \text { Here, } \overrightarrow{O A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \overrightarrow{O B}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \overrightarrow{O C}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} \text { and } \overrightarrow{O D}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \therefore \quad \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(2-1) \hat{\mathbf{i}}+(-1-1) \hat{\mathbf{j}}+(3+1) \hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \\ & \text { and } \\ & \overrightarrow{C D}=\overrightarrow{O D}-\overrightarrow{O C}=(3-2) \hat{\mathbf{i}}+(-2-0) \hat{\mathbf{j}}+(1+3) \hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \end{aligned}$$
$$\begin{aligned} &\text { So, the projection of } \overrightarrow{A B} \text { along }\\ &\begin{aligned} \overrightarrow{C D} & =\overrightarrow{A B} \cdot \frac{\overrightarrow{C D}}{|\overrightarrow{C D}|} \\ & =\frac{(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})}{\sqrt{1^2+2^2+4^2}} \\ & =\frac{1+4+16}{\sqrt{21}}=\frac{21}{\sqrt{21}} \\ & =\sqrt{21} \text { units } \end{aligned} \end{aligned}$$
Using vectors, find the area of the $\triangle A B C$ with vertices $A(1,2,3)$, $B(2,-1,4)$ and $C(4,5,-1)$.
Here,
$$\begin{aligned} \overrightarrow{\mathrm{AB}} & =(2-1) \hat{\mathbf{i}}+(-1-2) \hat{\mathbf{j}}+(4-3) \hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned}$$
and
$$\begin{aligned} \overrightarrow{\mathbf{A C}} & =(4-1) \hat{\mathbf{i}}+(5-2) \hat{\mathbf{j}}+(-1-3) \hat{\mathbf{k}} \\ & =3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \end{aligned}$$
$$\begin{aligned} & \therefore \quad \begin{aligned} \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{array}\right| \\ & =\hat{\mathbf{i}}(12-3)-\hat{\mathbf{j}}(-4-3)+\hat{\mathbf{k}}(3+9) \\ & =9 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} \end{aligned} \\ & \begin{aligned} \text { and } \quad |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathbf{A C}}| & =\sqrt{9^2+7^2+12^2} \\ & =\sqrt{81+49+144} \\ & =\sqrt{274} \\ \therefore \quad \text { Area of } \triangle A B C & =\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}| \\ & =\frac{1}{2} \sqrt{274} \text { sq units } \end{aligned} \end{aligned}$$
Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
Let $A B C D$ and $A B F E$ are parallelograms on the same base $A B$ and between the same parallel lines $A B$ and $D F$.
Here, $A B \| C D$ and $A E \| B F$
Let $$\overrightarrow{A B}=\vec{a} \text { and } \overrightarrow{A D}=\vec{b}$$
$\therefore \quad$ Area of parallelogram $A B C D=\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$
Now, area of parallelogram $A B F F=\overrightarrow{A B} \times \overrightarrow{A E}$
$$ \begin{aligned} &\begin{aligned} & =\overrightarrow{A B} \times(\overrightarrow{A D}+\overrightarrow{D E}) \\ & =\overrightarrow{A B} \times(\overrightarrow{\mathbf{b}}+k \overrightarrow{\mathbf{a}}) \quad[\text { let } \overrightarrow{D E}=k \overrightarrow{\mathbf{a}}, \text { where } k \text { is a scalar }] \\ & =\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}+k \overrightarrow{\mathbf{a}}) \\ & =(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{a}} \times k \overrightarrow{\mathbf{a}}) \\ & =(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+k(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}) \\ & =(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \quad[\because \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}=0] \\ & =\text { Area of parallelogram } A B C D \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$
Prove that in any $\triangle A B C, \cos A=\frac{b^2+c^2-a^2}{2 b c}$, where $a, b$ and $c$ are the magnitudes of the sides opposite to the vertices $A, B$ and $C$, respectively.
Here, components of C are c cos A and c sin A is drawn.
Since, $$\overrightarrow{C D}=b-c \cos A$$
In $\triangle B D C$,
$$a^2=(b-c \cos A)^2+(c \sin A)^2$$
$$\begin{array}{lrl} \Rightarrow & a^2 & =b^2+c^2 \cos ^2 A-2 b c \cos A+c^2 \sin ^2 A \\ \Rightarrow & 2 b c \cos A & =b^2-a^2+c^2\left(\cos ^2 A+\sin ^2 A\right) \\ \therefore & \cos A & =\frac{b^2+c^2-a^2}{2 b c} \end{array}$$