Maths Solve the Following Question.(3 Marks)

$\mathrm{LHS}=(\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})$

$=\bar{m} \cdot(\bar{c} \times \bar{d})$

$=(\bar{m} \times \bar{c}) \cdot \bar{d}$

$=[(\bar{a} \times \bar{b}) \times \bar{c}] \cdot \bar{d}$

$=[(\bar{c} \cdot \bar{a}) \bar{b}-(\bar{c} \cdot \bar{b}) \bar{a}] \cdot \bar{d}$

$=(\bar{c} \cdot \bar{a})(\bar{b} \cdot \bar{d})-(\bar{c} \cdot \bar{b})(\bar{a} \cdot \bar{d})$

$\begin{aligned} & =\left|\begin{array}{ll}\bar{c} \cdot \bar{a} & \bar{c} \cdot \bar{b} \\ a \cdot d & \frac{b}{b} \cdot \frac{d}{}\end{array}\right| \\ & =\left|\begin{array}{ll}\bar{a} \cdot \bar{c} & \bar{b} \cdot \bar{c} \\ \bar{a} \cdot d & \bar{b} \cdot d\end{array}\right|\end{aligned}$

[Dot product is commutative]

$=$ RHS.

$\therefore \hat{a}$ is perpendicular to $\hat{b}$ and $\hat{c}$ both

$\therefore \hat{a}$ is parallel to $\hat{b} \times \hat{c}$

$\therefore \hat{a}=\mathrm{m}(\hat{b} \times \hat{c})_t \mathrm{~m}$ is a scalar.

$\begin{aligned} & \therefore|\hat{a}|=|m||\hat{b} \times \hat{c}| \\ & \therefore|\vec{a}|=|m||\hat{b}||\hat{c}| \sin \frac{\pi}{6}\end{aligned}$

$\therefore 1=|m| \times 1 \times 1 \times \frac{1}{2}=\frac{|m|}{2} \quad \ldots[\because|\hat{a}|=|b|=|\hat{c}|=1]$

$\begin{aligned} & \therefore 2=|m| \\ & \therefore m= \pm 2\end{aligned}$

$\therefore \hat{a}= \pm 2(\hat{b} \times \hat{c})$.

Consider $|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|$

$\begin{aligned}=(\bar{b}-\bar{a}) \times(\bar{d}-\bar{c})+(\bar{c}-\bar{b}) \times(\bar{d}-\bar{a})+(\bar{a}-\bar{c}) \times(\bar{d}-\bar{b}) \\ =\bar{b} \times(\bar{d}-\bar{c})-\bar{a} \times(\bar{d}-\bar{c})+\bar{c} \times(\bar{d}-\bar{a})-\bar{b} \times(\bar{d}-\bar{a})+ \\ \bar{a} \times(\bar{d}-\bar{b})-\bar{c} \times(\bar{d}-\bar{b})\end{aligned}$

$\begin{array}{r}=\bar{b} \times \bar{d}-\bar{b} \times \bar{c}-\bar{a} \times \bar{d}+\bar{a} \times \bar{c}+\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{b} \times \bar{d}+ \\ \bar{b} \times \bar{a}+\bar{a} \times \bar{d}-\bar{a} \times \bar{b}-\bar{c} \times \bar{d}+\bar{c} \times \bar{b} \\ =\bar{b} \times \bar{d}-\bar{b} \times \bar{c}-\bar{a} \times \bar{d}-\bar{c} \times \bar{a}+\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{b} \times \bar{d}- \\ \bar{a} \times \bar{b}+\bar{a} \times \bar{d}-\bar{a} \times \bar{b}-\bar{c} \times \bar{d}-\bar{b} \times \bar{c} \\ \ldots[\because \bar{p} \times \bar{q}=-\bar{q} \times \bar{p}]\end{array}$

$-2(\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a})$

$\begin{aligned} & \therefore|\overline{\mathrm{AB}} \times \overline{\mathrm{CD}}+\overline{\mathrm{BC}} \times \overline{\mathrm{AD}}+\overline{\mathrm{CA}} \times \overline{\mathrm{BD}}| \\ & =|-2(\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a})| \\ & =4\left[\frac{1}{2}|\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}|\right]\end{aligned}$

$=4($ area of $\triangle A B C)$

Consider the triangle ABC. Complete the parallelogram ABDC. Vector area of ∆ABC

$=\frac{1}{2}($ vector area of parallelogram ABDC $)$

$=\frac{1}{2}(\overline{A B} \times \overline{A C})$

$\begin{aligned} & =\frac{1}{2}[(\bar{b}-\bar{a}) \times(\bar{c}-\bar{a})] \ldots[\because \overline{\mathrm{AB}}=\bar{b}-\bar{a} \text { and } \overline{\mathrm{AC}}=\bar{c}-\bar{a}] \\ & =\frac{1}{2}[\bar{b} \times \bar{c}-\bar{b} \times \bar{a}-\bar{a} \times \bar{c}+\bar{a} \times \bar{a}] \\ & =\frac{1}{2}[\bar{b} \times \bar{c}+\bar{a} \times \bar{b}+\bar{c} \times \bar{a}+\overline{0}] \\ & =\frac{1}{2}[\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}]\end{aligned}$

$\therefore \alpha=\frac{\pi}{6}, \beta=\frac{\pi}{4}$

Let the line make angle $y$ with $\mathrm{Z}$-axis.

$\begin{aligned} & \because \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \therefore \cos ^2\left(\frac{\pi}{6}\right)+\cos ^2\left(\frac{\pi}{4}\right)+\cos ^2 \gamma=1 \\ & \therefore\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos ^2 \gamma=1\end{aligned}$

$\therefore \cos ^2 \gamma=1-\frac{3}{4}-\frac{1}{2}=-\frac{1}{4}$

This is not possible, because cos γ is real

$\therefore \cos ^2 Y$ cannot be negative.

$\therefore \cos ^2 \gamma$ cannot be negative.

Hence, there is no line in space which makes angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$ with $X$-axis and $Y$-axis,

$|\bar{a}|=\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3$

$\therefore$ unit vector along $\bar{a}$

$=\hat{a}=\frac{\bar{a}}{|\bar{a}|}=\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{3}=\frac{2}{3} \hat{i}+\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}$

$\therefore$ its direction cosines are $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$.

If $\alpha, \beta, y$ are the direction angles, then $\cos \alpha=\frac{2}{3}, \cos \beta=\frac{1}{3}$

$\cos \gamma=\frac{2}{3}$

$\therefore \alpha=\cos ^{-1}\left(\frac{2}{3}\right), \beta=\cos ^{-1}\left(\frac{1}{3}\right), \gamma=\cos ^{-1}\left(\frac{2}{3}\right)$

Hence, direction cosines are $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$ and direction angles

are $\cos ^{-1}\left(\frac{2}{3}\right), \cos ^{-1}\left(\frac{1}{3}\right), \cos ^{-1}\left(\frac{2}{3}\right)$

Then we know that the position vector of the centroid O of the triangle is $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$

Therefore sum of the three vectors $\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$, is

$\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\vec{a}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)+\vec{b}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)+\vec{c}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)$

$=(\vec{a}+\vec{b}+\vec{c})-3\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)$

$=\overrightarrow{0}$

Hence, Sum os the three vectors determined by the medians of a triangle directed from the vertices is zero.

Let $\bar{a}_r, \bar{b}, \bar{c}, \bar{d}$ be the position vectors of the vertices

A, B, C, D of the parallelogram,

where $\bar{a}=\hat{i}+\hat{j}+\hat{k}, \bar{b}=\hat{i}+3 \hat{j}+5 \hat{k}, \bar{c}=7 \hat{i}+9 \hat{j}+11 \hat{k}$.

Since $A B C D$ is a parallelogram, $\overline{\mathrm{AB}}=\overline{\mathrm{DC}}$

$\begin{aligned} & \therefore \bar{b}-\bar{a}=\bar{c}-\bar{d} \\ & \therefore \bar{d}=\bar{a}+\bar{c}-\bar{b}\end{aligned}$

$\begin{aligned} & =(\hat{i}+\hat{j}+\hat{k})+(7 \hat{i}+9 \hat{j}+11 \hat{k})-(\hat{i}+3 \hat{j}+5 \hat{k}) \\ & =7 \hat{i}+7 \hat{j}+7 \hat{k}=7(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

Hence, the position vector of the fourth vertex is $7(\hat{i}+\hat{j}+\hat{k})$.

Since D, E, F are the midpoints of BC, CA, AB respec-tively, by the midpoint formula

$\bar{d}=\frac{\bar{b}+\bar{c}}{2}, \bar{e}=\frac{\bar{c}+\bar{a}}{2}, \bar{f}=\frac{\bar{a}+\bar{b}}{2}$

$\therefore \overline{\mathrm{AD}}+\overline{\mathrm{BE}}+\overline{\mathrm{CF}}=(\bar{d}-\bar{a})+(\bar{e}-\bar{b})+(\bar{f}-\bar{c})$

$=\left(\frac{\bar{b}+c}{2}-\bar{a}\right)+\left(\frac{\bar{c}+\bar{a}}{2}-\bar{b}\right)+\left(\frac{\bar{a}+\bar{b}}{2}-\bar{c}\right)$

$=\frac{1}{2} \bar{b}+\frac{1}{2} \bar{c}-\bar{a}+\frac{1}{2}-\bar{c}+\frac{1}{2} \bar{a}-\bar{b}+\frac{1}{2} \bar{a}+\frac{1}{2} \bar{b}-\bar{c}$

$=(\vec{a}+\bar{b}+\bar{c})-(\bar{a}+\bar{b}+\bar{c})=\overline{0}$

$\overline{\mathrm{a}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}, \overline{\mathrm{b}}=7 \hat{\mathrm{i}}+\hat{\mathrm{k}}, \overline{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\overline{\mathrm{LM}}=\overline{\mathrm{b}}-\overline{\mathrm{a}}=(7 \hat{\mathrm{i}}+\hat{\mathrm{k}})-(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$

$\begin{aligned} & \overline{\mathrm{MN}}=\overline{\mathrm{c}}-\overline{\mathrm{b}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})-(7 \hat{\mathrm{i}}+\hat{\mathrm{k}})=-6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}} \\ & \overline{\mathrm{NL}}=\overline{\mathrm{a}}-\overline{\mathrm{c}}=(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})-(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \\ & \therefore \mathrm{I}(\mathrm{LM})=|\overline{\mathrm{AB}}|=\sqrt{4^2+2^2+4^2}-\sqrt{16+4+16}=\sqrt{36}=6 \\ & \mathrm{I}(\mathrm{MN})=|\overline{\mathrm{MN}}|=\sqrt{(-6)^2+2^2}-\sqrt{36+4}=\sqrt{40}=\sqrt{10 \times 4}=2 \sqrt{10} \\ & I(\mathrm{NL})=|\overline{\mathrm{NL}}|=\sqrt{(2)^2+(-4)^2+(-4)^2}=\sqrt{4+16+16}=\sqrt{36}=6\end{aligned}$

$I(\mathrm{LM})=6, I(\mathrm{MN})=2 \sqrt{10}, I(\mathrm{NL})=6$

∆LMN is sosceles

$\begin{aligned} & \bar{a}=2 \hat{i}-\hat{j}, \bar{b}=4 \hat{i}+\hat{j}+\hat{k}, \bar{c}=4 \hat{i}-5 \hat{j}+4 \hat{k} \\ & \overline{\mathrm{AB}}=\bar{b}-\bar{a}=(4 \hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j})=2 \hat{i}+2 \hat{j}+\hat{k} \\ & \overline{\mathrm{BC}}=\bar{c}-\bar{b}=(4 \hat{i}-5 \hat{j}+4 \hat{k})-(4 \hat{i}+\hat{j}+\hat{k})=-6 \hat{j}+3 \hat{k} \\ & \overline{\mathrm{CA}}=\bar{a}-\bar{c}=(2 \hat{i}-\hat{j})-(4 \hat{i}-5 \hat{j}+4 \hat{k})=-2 \hat{i}+4 \hat{j}-4 \hat{k}\end{aligned}$

$\begin{aligned} \therefore l(A B) & =|\overline{A B}|=\sqrt{2^2+2^2+1^2} \\ & =\sqrt{4+4+1}=\sqrt{9}=3\end{aligned}$

$\begin{aligned} l(\mathrm{BC}) & =|\overline{\mathrm{BC}}|=\sqrt{(-6)^2+3^2} \\ & =\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\end{aligned}$

$\begin{aligned} l(\mathrm{CA}) & =|\overline{\mathrm{CA}}|=\sqrt{(-2)^2+4^2+(-4)^2} \\ & =\sqrt{4+16+16}=\sqrt{36}=6\end{aligned}$

$\begin{aligned} \therefore[l(\mathrm{AB})]^2+[l(\mathrm{CA})]^2 & =3^2+6^2 \\ & =9+36=45=(3 \sqrt{5})^2 \\ & =[l(\mathrm{BC})]^2\end{aligned}$

∴ ∆ ABC is right angled at A.

Let $\mathrm{ABC}$ be a triangle with $\overline{A B}=\hat{i}+2 \hat{j}, \overline{B C}=\hat{i}+\hat{k}$.

By triangle law of vectors

$\overline{A C}=\overline{A B}+\overline{B C}$

$\begin{aligned} & =(\hat{i}+2 \hat{j})+(\hat{i}+\hat{k})=2 \hat{i}+2 \hat{j}+\hat{k} \\ \therefore & l(\mathrm{AC})=|\overline{\mathrm{AC}}|=\sqrt{2^2+2^2+1^2}=\sqrt{9}=3 \text { units }\end{aligned}$

Hence, the length of third side is 3 units.

$\begin{aligned} & \therefore \overline{\mathrm{AB}}=\overline{\mathrm{DC}}, \overline{\mathrm{AD}}=\overline{\mathrm{BC}} \\ & \overline{\mathrm{AC}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}\end{aligned}$

$\begin{gathered}=\overline{\mathrm{AB}}+\overline{\mathrm{AD}}$\ldots$ (1) \\ \overrightarrow{\mathrm{BD}}=\overline{\mathrm{BA}}+\overline{\mathrm{AD}}=-\overline{\mathrm{AB}}+\overline{\mathrm{AD}}\end{gathered}$

Adding (1) and (2), we get

$2 \overline{\mathrm{AD}}=\overline{\mathrm{AC}}+\overline{\mathrm{BD}}=(2 \hat{i}+3 \hat{j}+4 \hat{k})+(-6 \hat{i}+7 \hat{j}-2 \hat{k})$

$=-4 \hat{i}+10 \hat{j}+2 \hat{k}$

$\therefore \overline{\mathrm{AD}}=\frac{1}{2}(-4 \hat{i}+10 \hat{j}+2 \hat{k})$

$=-2 \hat{i}+5 \hat{j}+\hat{k}$

$\operatorname{From}(1), \overline{\mathrm{AB}}=\overline{\mathrm{AC}}-\overline{\mathrm{AD}}$

$\begin{aligned} & =(2 \hat{i}+3 \hat{j}+4 \hat{k})-(-2 \hat{i}+5 \hat{j}+\hat{k}) \\ & =4 \hat{i}-2 \hat{j}+3 \hat{k}\end{aligned}$

$
\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 2 \\
0 & -5 & -5
\end{array}\right|=15(\hat{i}+\hat{j}-\hat{k})
$
from (1) and (2), we conclude that
$
(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times(\vec{b} \times \vec{c})
$

$=(6-0) \hat{i}-(9-0) \hat{j}+(3-4) \hat{k}$

$=6 \hat{i}-9 \vec{j}-\vec{k}$

$\therefore \bar{a} \times(\bar{b} \times \bar{c})=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 6 & -9 & -1\end{array}\right|$

$=(-2+27) \hat{i}-(-1-18) \hat{j}+(-9-12) \hat{k}$

$=25 \hat{i}+19 \hat{j}-21 \hat{k}$

$\ldots$ (1)

$\begin{aligned} \bar{a} \cdot \bar{c} & =(\hat{i}+2 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+\hat{j}+3 \hat{k}) \\ & =(1)(2)+(2)(1)+(3)(3)\end{aligned}$

$\begin{aligned} & =2+2+9=13 \\ \therefore & (\vec{a} \cdot \bar{c}) \bar{b}=13(3 \hat{i}+2 \hat{j})=39 \hat{i}+26 \hat{j}\end{aligned}$

Also, $\begin{aligned} \bar{a} \cdot \bar{b} & =(\vec{i}+2 \vec{j}+3 \hat{k}) \cdot(3 \vec{i}+2 \hat{j}) \\ & =(1)(3)+(2)(2)+(3)(0)\end{aligned}$

$\begin{aligned} & =(1)(3)+(2)(2)+(3)(0) \\ & =3+4+0=7\end{aligned}$

$\therefore(\vec{a} \cdot \bar{b}) \bar{c}=7(2 \hat{i}+\hat{j}+3 \hat{k})=14 \hat{i}+7 \hat{j}+21 \hat{k}$

$\therefore(\bar{a} \cdot \bar{c}) \bar{b}-(\bar{a} \cdot \bar{b}) \bar{c}$

$\begin{aligned} & =(39 \hat{i}+26 \hat{j})-(14 \hat{i}+7 \hat{j}+21 \hat{k}) \\ & =25 \hat{i}+19 \hat{j}-21 \hat{k}\end{aligned}$

$\ldots$ (1)

$\begin{aligned} \bar{a} \cdot \bar{c} & =(\hat{i}+2 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+\hat{j}+3 \hat{k}) \\ & =(1)(2)+(2)(1)+(3)(3)\end{aligned}$

$=2+2+9=13$

$\therefore(\bar{a} \cdot \bar{c}) \bar{b}=13(3 \hat{i}+2 \hat{j})=39 \hat{i}+26 \hat{j}$ Also, $\begin{aligned} \bar{a} \cdot \bar{b} & =(\hat{i}+2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}+2 \hat{j}) \\ & =(1)(3)+(2)(2)+(3)(0) \\ & =3+4+0=7\end{aligned}$ $\therefore(\bar{a} \cdot \bar{b}) \bar{c}=7(2 \hat{i}+\hat{j}+3 \hat{k})=14 \hat{i}+7 \hat{j}+21 \hat{k}$

$\therefore(\bar{a} \cdot \bar{c}) \bar{b}-(\bar{a} \cdot \bar{b}) \bar{c}$

$\begin{aligned} & =(39 \hat{i}+26 \hat{j})-(14 \hat{i}+7 \hat{j}+21 \hat{k}) \\ & =25 \hat{i}+19 \hat{j}-21 \hat{k}\end{aligned}$

$\ldots(2)$

From (1) and (2), we get

$\bar{a} \times(\bar{b} \times \bar{c})=(\bar{a}-\bar{c}) \bar{b}-(\bar{a} \cdot \bar{b}) \bar{c}$

$-\hat{i}+2 \hat{j}+3 \hat{k}_t \bar{b}=3 \hat{i}-2 \hat{j}+\hat{k}_z \bar{c}=2 \hat{i}+\hat{j}+3 \hat{k}$ and

$\bar{d}=-\hat{i}-2 \hat{j}+4 \hat{k}$

$\therefore \overline{\mathrm{AB}}=\bar{b}-\bar{a}=(3 \hat{i}-2 \hat{j}+\hat{k})-(-\hat{i}+2 \hat{j}+3 \hat{k})$

$=4 \hat{i}-4 \hat{j}-2 \hat{k}$

$\overline{\mathrm{AC}}=\bar{c}-\bar{a}=(2 \hat{i}+\hat{j}+3 \hat{k})-(-\hat{i}+2 \hat{j}+3 \hat{k})$

$=3 \hat{i}-\hat{j}$

and $\overline{\mathrm{AD}}=\bar{d}-\bar{a}=(-\hat{i}-2 \hat{j}+4 \hat{k})-(-\hat{i}+2 \vec{j}+3 \hat{k})$

$=-4 \hat{j}+\hat{k}$

$\therefore[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]=\left|\begin{array}{rrr}4 & -4 & -2 \\ 3 & -1 & 0 \\ 0 & -4 & 1\end{array}\right|$

$=4(-1+0)+4(3-0)-2(-12+0)$

$=-4+12+24=32$

$\therefore$ volume of the tetrahedron $=\frac{1}{6}|[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]|$

$=\frac{1}{6}(32)=\frac{16}{3}$ cu units.

$=\vec{i}-\vec{j}$

$\bar{u} \times \bar{r}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 3 & 0 & 1\end{array}\right|$

$\begin{aligned} & =(-2-0) \hat{i}-(1-3) \hat{j}+(0+6) \hat{k} \\ & =-2 \hat{i}+2 \hat{j}+6 \hat{k}\end{aligned}$

$\bar{r} \times \bar{w}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 1 \\ 0 & 1 & -1\end{array}\right|$

$=(0-1) \hat{i}-(-3-0) \hat{j}+(3-0) \hat{k}$

$=-\hat{i}+3 \hat{j}+3 \hat{k}$

$\operatorname{Now}_r(\bar{u}+\bar{w}) \cdot[(\bar{u} \times \bar{r}) \times(\bar{r} \times \bar{w})]=\left|\begin{array}{rrr}1 & -1 & 0 \\ -2 & 2 & 6 \\ -1 & 3 & 3\end{array}\right|$

= 1(6 – 18) + 1 (-6 + 6) + 0
= -12 + 0 + 0 = -12.

the scalar triple product is zero.

$\begin{aligned} {\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right] } & =\bar{a} \cdot(\bar{b} \times \bar{c}) \\ & =\bar{a} \cdot[\bar{b} \times(\bar{a}-2 \bar{b})] \\ & =\bar{a} \cdot(3 \bar{b} \times \bar{a}-2 \bar{b} \times \bar{b}) \\ & =\bar{a} \cdot(3 \bar{b} \times \bar{a}-\overline{0}) \\ & =3 \bar{a} \cdot(\bar{b} \times \bar{a})=3 \times 0=0\end{aligned}$

Alternative Method:

$\begin{aligned} & \bar{c}=3 \bar{a}-2 \bar{b} \\ & \therefore \bar{c} \text { is a linear combination of } \bar{a} \text { and } \bar{b} \\ & \therefore \bar{a}, \bar{b}, \bar{c} \text { are coplanar } \\ & \therefore[\bar{a} \bar{b} \bar{c}]=0 .\end{aligned}$

$\begin{aligned} & \therefore \frac{a}{\left|\begin{array}{cc}1 & -1 \\ -4 & 1\end{array}\right|}=\frac{b}{\left|\begin{array}{cc}-1 & -2 \\ 1 & -2\end{array}\right|}=\frac{c}{\left|\begin{array}{cc}-2 & 1 \\ -3 & -4\end{array}\right|} \\ & \therefore \frac{a}{1-4}=\frac{b}{3+2}=\frac{c}{8+3}\end{aligned}$

$\therefore \frac{a}{-3}=\frac{b}{5}=\frac{c}{11}$

∴ the required direction ratios are -3, 5, 11 Alternative Method:

Let $\bar{a}$ and $\bar{b}$ be the vectors along the lines whose direction ratios are $-2,1,-1$ and $-3,-4,1$

respectively.

Then $\bar{a}=-2 \hat{i}+\hat{j}-\hat{k}$ and $\bar{b}=-3 \hat{i}-4 \hat{j}+\hat{k}$

The vector perpendicular to both $\bar{a}$ and $\bar{b}$ is given by

$\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & -1 \\ -3 & -4 & 1\end{array}\right|$

$=(1-4) \hat{i}-(-2-3) \hat{j}+(8+3) \hat{k}$

$=-3 \hat{i}+5 \hat{j}+11 \hat{k}$

Hence, the required direction ratios are -3, 5, 11.

Then $\begin{aligned} \bar{a} \times \hat{i} & =(x \hat{i}+y \hat{j}+z \hat{k}) \times \hat{i} \\ & =x(\hat{i} \times \hat{i})+y(\hat{j} \times \hat{i})+z(\hat{k} \times \hat{i}) \\ & =z \hat{j}-y \hat{k} \ldots[\because \hat{i} \times \hat{i}=\overline{0}, \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{i}=\hat{j}]\end{aligned}$

It is given that

$\bar{a} \times \hat{i}+2 \bar{a}-5 \hat{j}=\overline{0}$

$\begin{aligned} & \therefore z \hat{j}-y \hat{k}+2(x \hat{i}+y \hat{j}+z \hat{k})-5 \hat{j}=0 \\ & \therefore z \hat{j}-y \hat{k}+2 x \hat{i}+2 y \hat{j}+2 z \hat{k}-5 \hat{j}=\overline{0} \\ & \therefore 2 x \hat{i}+(2 y+z-5) \hat{j}+(2 z-y) \hat{k}=\overline{0}\end{aligned}$

By equality of vectors

2x = 0 i.e. x = 0 2y + z – 5 = 0 … (1) 2z – y = 0 … (2) From (2), y = 2z Substituting y = 2z in (1), we get 4z + z = 5 ∴ z = 1 ∴ y = 2z = 2(1) = 2 ∴ x = 0, y = 2, z = 1

$\therefore \bar{a}=2 \hat{j}+\hat{k}$

$\therefore \bar{a} \neq \bar{b} \neq \bar{c} \neq \bar{d}$

$\therefore \bar{a}-\bar{d} \neq \overline{0}$ and $\bar{b}-\bar{c} \neq \overline{0}$

$\ldots(1)$

Now, $\bar{a} \times \bar{b}=\bar{c} \times \bar{d}$

$\ldots(2)$

and $\bar{a} \times \bar{c}=\bar{b} \times \bar{d}$

$\ldots(3)$

Subtracting (3) from (2), we get

$\bar{a} \times \bar{b}-\bar{a} \times \bar{c}=\bar{c} \times \bar{d}-\bar{b} \times \bar{d}$

$\begin{aligned} & \therefore \bar{a} \times(\bar{b}-\bar{c})=(\bar{c}-\bar{b}) \times \bar{d}=-(\bar{b}-\bar{c}) \times \bar{d} \\ & \quad=\bar{d} \times(\bar{b}-\bar{c}) \\ & \therefore \bar{a} \times(\bar{b}-\bar{c})-\bar{d} \times(\bar{b}-\bar{c})=\overline{0} \\ & \therefore(\bar{a}-\bar{d}) \times(\bar{b}-\bar{c})=\overline{0}\end{aligned}$

$\therefore \bar{a}-\bar{d}$ and $\bar{b}-\bar{c}$ are parallel to each other.

$\ldots[B y(1)]$

Then $\overline{\mathrm{AC}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}$

and $\begin{aligned} \overline{\mathrm{BD}} & =\overline{\mathrm{BA}}+\overline{\mathrm{AD}}=-\overline{\mathrm{AB}}+\overline{\mathrm{BC}} \quad \ldots[\because \overline{\mathrm{BC}}=\overline{\mathrm{AD}}] \\ & =\overline{\mathrm{BC}}-\overline{\mathrm{AB}}\end{aligned}$

$\therefore \overline{\mathrm{AC}} \times \overline{\mathrm{BD}}=(\overline{\mathrm{AB}}+\overline{\mathrm{BC}}) \times(\overline{\mathrm{BC}}-\overline{\mathrm{AB}})$

$\begin{aligned} & =\overline{\mathrm{AB}} \times(\overline{\mathrm{BC}}-\overline{\mathrm{AB}})+\overline{\mathrm{BC}} \times(\overline{\mathrm{BC}}-\overline{\mathrm{AB}}) \\ & =\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}-\overline{\mathrm{AB}} \times \overline{\mathrm{AB}}+\overline{\mathrm{BC}} \times \overline{\mathrm{BC}}-\overline{\mathrm{BC}} \times \overline{\mathrm{AB}} \\ & =\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}+\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}\end{aligned}$

$\ldots[\overline{\mathrm{AB}} \times \overline{\mathrm{AB}}=\mathrm{BC} \times \overline{\mathrm{BC}}=0$ and $-\overline{\mathrm{BC}} \times \overline{\mathrm{AB}}=\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}]$

$\therefore \overline{\mathrm{AC}} \times \overline{\mathrm{BD}}=2(\overline{\mathrm{AB}} \times \overline{\mathrm{BC}})$

$=2$ (vector area of parallelogram $\mathrm{ABCD})$

$\therefore$ vector area of parallelogram $\mathrm{ABCD}=\frac{1}{2}(\overline{\mathrm{AC}} \times \overline{\mathrm{BD}})$.

$\therefore$ the direction cosines of $\overline{A B}$ are

$l=\frac{-2}{\sqrt{(-2)^2+2^2+1^2}}=\frac{-2}{3}$,

$m=\frac{2}{\sqrt{(-2)^2+2^2+1^2}}=\frac{2}{3}$,

$n=\frac{1}{\sqrt{(-2)^2+2^2+1^2}}=\frac{1}{3}$.

i.e. $l=\frac{-2}{3}, m=\frac{2}{3}, n=\frac{1}{3}$

The coordinates of the points which are at a distance of d units from the point (x1, y1, z1) are given by (x1 ± ld, y1 ± md, z1 ± nd)

Here $x_1=4, y_1=1, z_1=5, d=6,1=-\frac{2}{3}, m=\frac{2}{3}, n=\frac{1}{3}$

∴ the coordinates of the requited points are

$\begin{aligned} & \left(4 \pm\left(-\frac{2}{3}\right) 6,1 \pm \frac{2}{3}(6), 5 \pm \frac{1}{3}(6)\right) \\ & \text { i.e. }(4-4,1+4,5+2) \text { and }(4+4,1-4,5-2) \\ & \text { i.e. }(0,5,7) \text { and }(8,-3,3) \text {. }\end{aligned}$

Then $l=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{4}{\sqrt{4^2+(-12)^2+(18)^2}}$

$=\frac{4}{\sqrt{16+144+324}}=\frac{4}{22}=\frac{2}{11}$

$m=\frac{b}{\sqrt{a^2+b^2+c^2}}=\frac{-12}{\sqrt{4^2+(-12)^2+(18)^2}}$

$=\frac{-12}{\sqrt{16+144+324}}=\frac{-12}{22}=\frac{-6}{11}$

and $n=\frac{c}{\sqrt{a^2+b^2+c^2}}=\frac{18}{\sqrt{4^2+(-12)^2+(18)^2}}$

$=\frac{18}{\sqrt{16+144+324}}=\frac{18}{22}=\frac{9}{11}$

Hence, the direction cosines of the line are $\frac{2}{11}, \frac{-6}{11}, \frac{9}{11}$.

$\begin{aligned} & \therefore \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \therefore \cos ^2 45^{\circ}+\cos ^2 60^{\circ}+\cos ^2 \gamma=1\end{aligned}$

$\begin{aligned} & \therefore\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)+\cos ^2 \gamma=1 \\ & \therefore \cos ^2 \gamma=1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\end{aligned}$

$\therefore \cos \gamma= \pm \frac{1}{2}$

$\therefore \cos \gamma=\frac{1}{2}$ or $\cos \gamma=-\frac{1}{2}$

$\therefore \cos \gamma=\cos \frac{\pi}{3}$ or $\cos \gamma=-\cos \frac{\pi}{3}$

$=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}$

$\therefore \gamma=\frac{\pi}{3}$ or $\gamma=\frac{2 \pi}{3}$

Hence, the third direction angle is $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$.

Let $\overline{\mathrm{AC}}=\overline{\mathrm{CB}}=\bar{a}$ and $\overline{\mathrm{CP}}=\bar{r}$.

Then $|\bar{a}|=|\bar{r}| \ldots(1)$

$\overline{\mathrm{AP}}=\overline{\mathrm{AC}}+\overline{\mathrm{CP}}=\bar{a}+\bar{r}=\bar{r}+\bar{a}$

$\overline{\mathrm{BP}}=\overline{\mathrm{BC}}+\overline{\mathrm{CP}}=-\overline{\mathrm{CB}}+\overline{\mathrm{CP}}=-\bar{a}+\bar{r}$

$\begin{aligned} \therefore \overline{\mathrm{AP}} \cdot \overline{\mathrm{BP}}= & (\bar{r}+\bar{a}) \cdot(\bar{r}-\bar{a}) \\ & \bar{r} \cdot \bar{r}-\bar{r} \cdot \bar{a}+\bar{a} \cdot \bar{r}-\bar{a} \cdot \bar{a}\end{aligned}$

$=|\vec{r}|^2-|\bar{a}|^2=0 \quad \ldots(\because \bar{r} \cdot \bar{a}=\bar{a} \cdot \bar{r})$

$\therefore \overline{\mathrm{AP}} \perp \overline{\mathrm{BP}} \quad \therefore \angle \mathrm{APB}$ is a right angle.

Hence, the angle subtended on a semicircle is the right angle.

$\because \overline{\mathrm{AC}}, \overline{\mathrm{BD}}$ are non-zero vectors

$\therefore \overline{\mathrm{AC}}$ is perpendicular to $\overline{\mathrm{BD}}$

Hence, the diagonals are perpendicular.

Consider $\bar{a} \cdot \bar{b}=(x c \hat{i}-6 \hat{j}+3 \hat{k}) \cdot(x \hat{i}+2 \hat{j}+2 c x \hat{k})$

$=(x c)(x)+(-6)(2)+(3)(2 c x)$

$=c x^2-12+6 c x$

$=c x^2+6 c x-12$

If the angle between $\bar{a}$ and $\bar{b}$ is obtuse, $\bar{a} \cdot \bar{b}<0$.

$\begin{aligned} & \therefore c x^2+6 c x-12<0 \\ & \therefore c x^2+6 c x<12 \\ & \therefore c\left(x^2+6 x\right)<12 \\ & \therefore c<\frac{12}{x^2+6 x}\end{aligned}$

$\therefore c<\frac{12}{x^2+6 x}$

$\therefore c<\frac{12}{\left(x^2+6 x+9\right)-9}=\frac{12}{(x+3)^2-9}$

$\therefore c<\min \left\{\frac{12}{(x+3)^2-9}\right\}$

Now, $\frac{12}{(x+3)^2-9}$ is minimum if $(x+3)^2-9$ is maximum

$\begin{aligned} & \text { i.e. }(x+3)^2-9=\infty-9=\infty \\ & \therefore c<\min \left\{\frac{12}{\infty}\right\}=0 \\ & \therefore c<0 .\end{aligned}$

Hence, the angle between a and b is obtuse if c < 0.

$\therefore \bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{a}=0$

...(1)

$\begin{aligned} \mathrm{LHS} & =(\bar{a}+\bar{b})^2 \\ & =(\bar{a}+\bar{b}) \cdot(\bar{a}+\bar{b}) \\ & =\bar{a} \cdot(\bar{a}+\bar{b})+\bar{b} \cdot(\bar{a}+\bar{b}) \\ & =\bar{a} \cdot \bar{a}+\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{a}+\bar{b} \cdot \bar{b} \\ & =\bar{a} \cdot \bar{a}+0+0+\bar{b} \cdot \bar{b} \\ & =|\bar{a}|^2+|\bar{b}|^2\end{aligned}$

$\ldots$... [By (1)]

$\operatorname{RHS}=(\bar{a}-\bar{b})^2$

$\begin{aligned} & =(\bar{a}-\bar{b}) \cdot(\bar{a}-\bar{b}) \\ & =\bar{a} \cdot(\bar{a}-\bar{b})+\bar{b} \cdot(\bar{a}-\bar{b}) \\ & =\bar{a} \cdot \bar{a}-\bar{a} \cdot \bar{b}-\bar{b} \cdot \bar{a}+\bar{b} \cdot \bar{b} \\ & =\bar{a} \cdot \bar{a}+\bar{b} \cdot \bar{b} \\ & =|\bar{a}|^2+|\bar{b}|^2\end{aligned}$

$\ldots[$ By (1)]

Hence, $(\bar{a}+\bar{b})^2=(\bar{a}-\bar{b})^2$.

∴ LHS = RHS

Hence, $(\bar{a}+\bar{b})^2=(\bar{a}-\bar{b})^2$.

$\bar{v}=\hat{i}+2 \hat{j}-2 \hat{k}$

Then $\bar{u} \times \bar{v}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 2 & -2\end{array}\right|$

$=(-2+4) \hat{i}+(-4+2) \hat{j}+(4-1) \hat{k}$

$=2 \hat{i}-2 \hat{j}+3 \hat{k}$

$\therefore|\bar{u} \times \bar{v}|=\sqrt{(2)^2+(-2)^2+(3)^2}$

$\begin{aligned} & =\sqrt{4+4+9} \\ & =\sqrt{17}\end{aligned}$

$\begin{aligned} & = \pm \frac{\bar{u} \times \overline{\mathrm{v}}}{|\overline{\mathrm{u}} \times \overline{\mathrm{v}}|}= \pm \frac{2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{k}}{\sqrt{17}} \\ & = \pm\left(\frac{2}{\sqrt{17}} \hat{\mathrm{i}}-\frac{2}{\sqrt{17}} \hat{\mathrm{j}}+\frac{3}{\sqrt{17}} \hat{\mathrm{k}}\right)\end{aligned}$

Then $\bar{a}=3 \hat{i}+\hat{j}+4 \hat{k}, \bar{b}=-4 \hat{i}+5 \hat{j}-3 \hat{k}$ and $\bar{g}=-\hat{i}+2 \hat{j}+\hat{k}$

Since G is the centroid of the ∆ABC, by the centroid formula,

$\bar{g}=\frac{\bar{a}+\bar{b}+\bar{c}}{3} \quad \therefore 3 \bar{g}=\bar{a}+\bar{b}+\bar{c}$

$\begin{aligned} & \therefore 3(-\hat{i}+2 \hat{j}+\hat{k})=(3 \hat{i}+\hat{j}+4 \hat{k})+(-4 \hat{i}+5 \hat{j}-3 \hat{k})+\bar{c} \\ & \therefore-3 \hat{i}+6 \hat{j}+3 \hat{k}=(-\hat{i}+6 \hat{j}+\vec{k})+\bar{c}\end{aligned}$

$\therefore \bar{c}=(-3 \hat{i}+6 \hat{j}+3 \hat{k})-(-\hat{i}+6 \hat{j}+\hat{k})$

$\therefore \bar{c}=-2 \hat{i}+0 \cdot \hat{j}+2 \hat{k}$

∴ the coordinates of third vertex C are (-2, 0, 2).

Since C divides AB in the ratio 3 : 2,

$\begin{aligned} \bar{c} & =\frac{3(\bar{a}-3 \bar{b})+2(6 \bar{a}+2 \bar{b})}{3+2} \\ & =\frac{3 \bar{a}-9 \bar{b}+12 \bar{a}+4 \bar{b}}{5}\end{aligned}$

$=\frac{1}{5}(15 \bar{a}-5 \bar{b})=3 \bar{a}-\bar{b}$

Hence, the position vector of $C$ is $3 \bar{a}-\bar{b}$.

$\bar{l}=7 \hat{i}-6 \hat{j}+12 \hat{k}, \bar{n}=5 \hat{i}+4 \hat{j}-2 \hat{k}$

If $\mathrm{M}(\bar{m})$ is the midpoint of $\mathrm{LN}$, by midpoint formula,

$\bar{m}=\frac{\bar{l}+\bar{n}}{2}=\frac{(7 \hat{i}-6 \hat{j}+12 \hat{k})+(5 \hat{i}+4 \hat{j}-2 \hat{k})}{2}$

$=\frac{1}{2}(12 \hat{i}-2 \hat{j}+10 \hat{k})=6 \hat{i}-\hat{j}+5 \hat{k}$

∴ coordinates of M = (6, -1, 5).

Hence, position vector of $\mathrm{M}$ is $6 \hat{i}-\hat{j}+5 \hat{k}$ and the coordinates of $\mathrm{M}$ are $(6,-1,5)$.

$\begin{aligned} & l(\mathrm{AB})=\sqrt{(1-1)^2+(1-0)^2+(0-1)^2}=\sqrt{0+1+1}=\sqrt{2} \\ & l(\mathrm{BC})=\sqrt{(1-0)^2+(0-1)^2+(1-1)^2}=\sqrt{1+1+0}=\sqrt{2} \\ & l(\mathrm{CA})=\sqrt{(0-1)^2+(1-1)^2+(1-0)^2}=\sqrt{1+0+1}=\sqrt{2}\end{aligned}$

$\therefore l(\mathrm{AB})=l(\mathrm{BC})=l(\mathrm{CA})$

$\therefore$ the triangle is equilateral

$\begin{aligned} \therefore \text { its area } & =\frac{\sqrt{3}}{4}(\text { side })^2 \\ & =\frac{\sqrt{3}}{4}(\sqrt{2})^2 \\ & =\frac{\sqrt{3}}{2} \text { sq units }\end{aligned}$

Adding equations (1) and (2), we get

$\begin{aligned} & \bar{a}+\bar{b}=(\bar{c}-\bar{d})+(\bar{c}+\bar{d})=2 \bar{c} \\ & \therefore \bar{c}=\frac{1}{2}(\bar{a}+\bar{b})=\frac{1}{2} \bar{a}+\frac{1}{2} \bar{b}\end{aligned}$

Substituting for $\bar{c}$ in (2), we get

$\bar{d}=\bar{b}-\bar{c}=\bar{b}-\frac{1}{2}(\bar{a}+\bar{b})$

$\begin{aligned} & =\frac{2 \bar{b}-(\bar{a}+\bar{b})}{2} \\ & =\frac{1}{2}(\bar{b}-\bar{a})=\frac{1}{2} \bar{b}-\frac{1}{2} \bar{a}\end{aligned}$

Hence, $\bar{c}=\frac{1}{2} \bar{a}+\frac{1}{2} \bar{b}$ and $\bar{d}=\frac{1}{2} \bar{b}-\frac{1}{2}-\bar{a}$.

ABCDEF is a regular hexagon.

$\therefore \overline{\mathrm{AB}}=\overline{\mathrm{ED}}$ and $\overline{\mathrm{AF}}=\overline{\mathrm{CD}}$

∴ by the triangle law of addition of vectors,

$\begin{aligned} & \overline{\mathrm{AC}}+\overline{\mathrm{AF}}=\overline{\mathrm{AC}}+\overline{\mathrm{CD}}=\overline{\mathrm{AD}} \\ & \overline{\mathrm{AE}}+\overline{\mathrm{AB}}=\overline{\mathrm{AE}}+\overline{\mathrm{ED}}=\overline{\mathrm{AD}} \\ & \therefore \mathrm{LHS}=\overline{\mathrm{AB}}+\overline{\mathrm{AC}}+\overline{\mathrm{AD}}+\overline{\mathrm{AE}}+\overline{\mathrm{AF}}\end{aligned}$

$\begin{aligned} & =\overline{\mathrm{AD}}+(\overline{\mathrm{AC}}+\overline{\mathrm{AF}})+(\overline{\mathrm{AE}}+\overline{\mathrm{AB}}) \\ & =\overline{\mathrm{AD}}+\overline{\mathrm{AD}}+\overline{\mathrm{AD}} \\ & =3 \overline{\mathrm{AD}}=3(2 \overline{\mathrm{AO}}) \quad \ldots[\mathrm{O} \text { is midpoint of } \mathrm{AD}] \\ & =6 \overline{\mathrm{AO}} .\end{aligned}$

Given : $\overline{\mathrm{PQ}}=2 \bar{a}_r \overline{\mathrm{QR}}=2 \bar{b}$

$\begin{aligned} & \text { (i) } \overline{\mathrm{PR}}=\overline{\mathrm{PQ}}+\overline{\mathrm{QR}} \\ & =2 \bar{a}+2 \bar{a} .\end{aligned}$

Let $\overline{\mathrm{AB}}=\bar{a}_r \overline{\mathrm{BC}}=\bar{b}$

Then $\overline{\mathrm{AC}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=\mathrm{a}+b$

Given: $|\bar{a}|=|\overline{\mathrm{AB}}|=1(\mathrm{AB})=24$ and

$|\bar{b}|=|\overline{\mathrm{BC}}|=\mid(\mathrm{BC})=7$

$\begin{aligned} & \therefore \angle A B C=90^{\circ} \\ & \therefore[I(A C)]^2=[I(A B)]^2+[I(B C)]^2\end{aligned}$

$\begin{aligned} & =(24)^2+(7)^2=625 \\ & \therefore|(\mathrm{AC})=25 \therefore| \overline{\mathrm{AC}} \mid=25 \\ & \therefore|\bar{a}+\bar{b}|=|\overline{\mathrm{AC}}|=25 .\end{aligned}$