(a) Show that the area of a parallelogram whose diagonals are represented by a and b is given by 1/2| a × b |. Also find the area of a parallelogram…

Let $\displaystyle A B C D$ be the parallelogram with diagonals $\displaystyle \overrightarrow{A B}=\vec{\mathrm{a}}$ and $\displaystyle \overrightarrow{B D}=\vec{\mathrm{b}}$. $\displaystyle \therefore \overrightarrow{A B}=\frac{1}{2}(\vec{\mathrm{a}}-\vec{\mathrm{b}})$ and $\displaystyle \overrightarrow{A D}=\frac{1}{2}(\vec{\mathrm{a}}+\vec{\mathrm{b}})$ Area of $\displaystyle A B C D$ $\displaystyle =|\overrightarrow{A B} \times \overrightarrow{A D}|$ $\displaystyle =\left|\frac{1}{2}(\vec{\mathrm{a}}-\vec{\mathrm{b}}) \times \frac{1}{2}(\vec{\mathrm{a}}+\vec{\mathrm{b}})\right|$ $\displaystyle =\frac{1}{4}|\vec{\mathrm{a}} \times \vec{\mathrm{a}}+\vec{\mathrm{a}} \times \vec{\mathrm{b}}-\vec{\mathrm{b}} \times \vec{\mathrm{a}}-\vec{\mathrm{b}} \times \vec{\mathrm{b}}|$ Given lines are $\displaystyle \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\displaystyle \vec{r}=(15 \hat{\mathrm{i}}+29 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$ The first line in vector form is $\displaystyle \vec{r}=(8 \hat{\mathrm{i}}-19 \hat{\mathrm{j}}+10 \hat{\mathrm{k}})+\lambda(3 \hat{\mathrm{i}}-16 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$ \[\vec{\mathrm{a}}_{1}=8 \hat{\mathrm{i}}-19 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}, \vec{\mathrm{a}}_{2}=15 \hat{\mathrm{i}}+29 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \] $\displaystyle \vec{\mathrm{b}}_{1}=3 \hat{\mathrm{i}}-16 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \vec{\mathrm{b}}_{2}=3 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$ $\displaystyle \vec{\mathrm{b}}_{1} \times \vec{\mathrm{b}}_{2}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right|=24 \hat{\mathrm{i}}+36 \hat{\mathrm{j}}+72 \hat{\mathrm{k}}$ ∴ Equation of line passing through ( $\displaystyle 1,2,-4$ ) and parallel to $\displaystyle \vec{\mathrm{b}}$ is \[\vec{r}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})+t(24 \hat{\mathrm{i}}+36 \hat{\mathrm{j}}+72 \hat{\mathrm{k}}) \text { or } \vec{r}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})+t^{\prime}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \] Cartesian formof lineis $\displaystyle \frac{x-1}{24}=\frac{y-2}{36}=\frac{z+4}{72}$ or $\displaystyle \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$