Exercise 6.1 - Chapter 6 Applications of Vector Algebra 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Applications of Vector Algebra
Ex $6.1$
Question 1.
Prove by vector method that if a line is drawn from the centre of a circle of a circle to the midpoint of a chord, then the line is perpendicular to the chord.
Solution:
Let ' $\mathrm{C}$ ' be the mid point of the chord $A B$ Take ' $\mathrm{O}$ ' on the centre of the circle.
Since, $\mathrm{OA}=\mathrm{OB}$ (Radii)
Let $\overrightarrow{\mathrm{OA}}=\vec{a} ; \overrightarrow{\mathrm{OB}}=\vec{b} ; \overrightarrow{\mathrm{OC}}=\frac{\vec{a}+\vec{b}}{2}$
To prove: $\overrightarrow{\mathrm{OC}} \perp^r \overrightarrow{\mathrm{AB}}$
$
\begin{aligned}
\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{AB}} & =\left(\frac{a+\vec{b}}{2}\right) \cdot(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}})=\left(\frac{\vec{b}+\vec{a}}{2}\right) \cdot(\vec{b}-\vec{a}) \\
& =\frac{1}{2}\left[(\vec{b})^2-(\vec{a})^2\right]=\frac{1}{2}\left(\overrightarrow{\mathrm{OB}}^2-\overrightarrow{\mathrm{OA}}^2\right)=\frac{1}{2}\left(\mathrm{OB}^2-\mathrm{OA}^2\right) \\
& =0
\end{aligned}
$
$\therefore \overrightarrow{\mathrm{OC}} \perp^r$ to $\overrightarrow{\mathrm{AB}}$. Hence the result.
Question 2.
Prove by vector method that the median to the base of an isosceles triangle is perpendicular to the base.
Solution:
Let $\mathrm{OAB}$ be an isosceles triangle with $\mathrm{OA}=\mathrm{OB}$
Let $\mathrm{OC}$ be the median to the base $\mathrm{AB}$
$\mathrm{C}$ is the midpoint of $\mathrm{AB}$

Take $\mathrm{O}$ as origin.
- Let $\overrightarrow{\mathrm{OA}}=\vec{a} ; \overrightarrow{\mathrm{OB}}=\vec{b} ; \overrightarrow{\mathrm{OC}}=\frac{\vec{a}+\vec{b}}{2}$
To prove $\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{AB}}=0$
$
\begin{aligned}
\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{AB}} & =\left(\frac{\vec{a}+\vec{b}}{2}\right) \cdot(\overrightarrow{\mathrm{OB}} \overrightarrow{\mathrm{OA}})=\left(\frac{\vec{a}+\vec{b}}{2}\right) \cdot(\vec{b}-\vec{a}) \\
& =\left(\frac{\vec{b}+\vec{a}}{2}\right) \cdot(\vec{b}-\vec{a})=\frac{1}{2}\left[(\vec{b})^2-(\vec{a})^2\right] \\
& =\frac{1}{2}\left[(\overrightarrow{\mathrm{OB}})^2-(\overrightarrow{\mathrm{OA}})^2\right]=\frac{1}{2}\left(\mathrm{OB}^2-\mathrm{OA}^2\right) \\
\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{AB}} & =\frac{1}{2}(0)
\end{aligned}
$
$\therefore \overrightarrow{\mathrm{OC}} \perp^r$ to $\overrightarrow{\mathrm{AB}}$. Hence the result.

Question 3.
Prove by vector method that an angle in a semi-circle is a right angle.
Solution:
Let $\mathrm{AB}$ be the diameter of the circle with centre ' $\mathrm{O}$ '
Let $\mathrm{P}$ be any point on the semi-circle.

- To prove $\left\lfloor\mathrm{APB}=90^{\circ}\right.$
We have $\mathrm{OA}=\mathrm{OB}=\mathrm{OP}$ (radii)
Now
$
\begin{aligned}
\overrightarrow{\mathrm{PA}} & =\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OA}} \\
\overrightarrow{\mathrm{PB}} & =\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OB}} \\
& =\overrightarrow{\mathrm{PO}}-\overrightarrow{\mathrm{OA}} \text { (since } \overrightarrow{\mathrm{OB}}=-\overrightarrow{\mathrm{OA}} \text { ) }
\end{aligned}
$
$
\begin{array}{ll}
\therefore & \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{PB}}=(\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OA}}) \cdot(\overrightarrow{\mathrm{PO}}-\overrightarrow{\mathrm{OA}})=(\overrightarrow{\mathrm{PO}})^2-(\overrightarrow{\mathrm{OA}})^2=(\mathrm{PO})^2-(\mathrm{OA})^2 \\
\therefore \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{PB}}=0
\end{array}
$
This gives $\angle \mathrm{APB}=90^{\circ}$. Hence the result.

Question 4.
Prove by vector method that the diagonals of a rhombus bisect each other at right angles.
Solution:
Let $\mathrm{ABCD}$ be a rhombus

To prove $\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BD}}=0$
We have $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$
Now
$
\begin{aligned}
\overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}} \\
\overrightarrow{\mathrm{BD}} & =\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}} \\
& =\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AB}}(\text { since } \overrightarrow{\mathrm{CD}}=-\overrightarrow{\mathrm{AB}}) \\
\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BD}} & =(\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{AB}}) \cdot(\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AB}}) \\
& =(\overline{\mathrm{BC}})^2-(\overrightarrow{\mathrm{AB}})^2=(\mathrm{BC})^2-(\mathrm{AB})^2 \\
\overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BD}} & =0
\end{aligned}
$
$\therefore \overrightarrow{\mathrm{AC}} \perp^r$ to $\overrightarrow{\mathrm{BD}}$. Hence the result.

Question 5.
Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle

Solution:
Let $\mathrm{ABCD}$ be a parallelogram To prove $\mathrm{ABCD}$ be a rectangle provided the diagonals are equal.

But
$
\begin{aligned}
& \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}} \\
& \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AB}}
\end{aligned}
$
$
\begin{aligned}
&(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}})^2=(\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AB}})^2 \\
&(\overrightarrow{\mathrm{AB}})^2+(\overrightarrow{\mathrm{BC}})^2+2 \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}=(\overrightarrow{\mathrm{BC}})^2+(\overrightarrow{\mathrm{AB}})^2-2 \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}} \\
& 4 \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}=0 \\
& \overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}=0 \\
& \overrightarrow{\mathrm{AB}} \perp^{\mathrm{r}} \text { to } \overrightarrow{\mathrm{BC}} \\
& \Rightarrow \mathrm{ABCD} \text { is a rectangle. }
\end{aligned}
$

Question 6.
Prove by vector method that the area of the quadrilateral $\mathrm{ABCD}$ having diagonals $\mathrm{AC}$ and $\mathrm{BD}$ is $\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|$
Solution:
Vector area of quadrilateral $\mathrm{ABCD}=\{$ Vector area of $\triangle \mathrm{ABC}\}+\{$ Vector area of $\triangle \mathrm{ACD}\}$

$
\begin{aligned}
& =\frac{1}{2}(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}})+\frac{1}{2}(\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AD}}) \\
& =-\frac{1}{2}(\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AB}})+\frac{1}{2}(\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{AD}}) \\
& =\frac{1}{2} \overrightarrow{\mathrm{AC}} \times[-\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}] \\
& =\frac{1}{2} \overrightarrow{\mathrm{AC}} \times[\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AD}}]=\frac{1}{2} \overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}
\end{aligned}
$
$\therefore$ The area of the quadrilateral $\mathrm{ABCD}=\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{BD}}|$

Question 7.
Prove by vector method that the parallelogram on the same base and between the same parallels are equal in area.
Solution:
Let $A B C D$ and $A B C$ ' $D$ ' be two parallelogram between the parallels with same base To prove: Area of $\mathrm{ABCD}=$ Area of $\mathrm{ABC}^{\prime} \mathrm{D}^{\prime}$

$
\text { Area of } \begin{aligned}
\mathrm{ABC}^{\prime} \mathrm{D}^{\prime} & =|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}| \\
& =\left[\overrightarrow{\mathrm{AB}} \times\left(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DD}^{\prime}}\right)\right] \\
& =\left|(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}})+\left(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{DD}^{\prime}}\right)\right| \\
& =|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}|+0 \\
& =\text { Area of } \mathrm{ABCD}
\end{aligned}
$
$
=|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AD}}|+0 \quad \text { since }\left(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{DD}^{\prime}} \text { are parallel }\right) \text {. }
$

Question 8.

If $\mathrm{G}$ is the centroid of a $A A B C$, prove that.
(area of $\triangle \mathrm{GAB})=($ area of $\triangle \mathrm{GBC})=($ area of $\triangle \mathrm{GAC})=\frac{1}{3}$ [area of $\triangle \mathrm{ABC}$ ]
Solution:

$
\begin{aligned}
& \text { Area of } \triangle \mathrm{GAB}=\frac{1}{2}(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AG}}) \\
& =\frac{1}{2}[(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}) \times(\overrightarrow{\mathrm{OG}}-\overrightarrow{\mathrm{OA}})] \\
& =\frac{1}{2}\left\{(\vec{b}-\vec{a}) \times\left[\frac{\vec{a}+\vec{b}+\vec{c}}{3}-a\right]\right\} \\
& =\frac{1}{2}\left\{(\vec{b}-\vec{a}) \times\left[\frac{\vec{b}+\vec{c}-2 \vec{a}}{3}\right]\right\} \\
& =\frac{1}{6}\{\vec{b} \times \vec{c}-2 \vec{b} \times \vec{a}-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}\}=\frac{1}{6}\{\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}\} \\
& \left.=\frac{1}{3} \times \frac{1}{2}\{\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}\}=\frac{1}{3} \text { [area of } \triangle \mathrm{ABC}\right] \\
&
\end{aligned}
$
Similarly we can prove
Area of $\Delta \mathrm{GBC}=$ Area of $\triangle \mathrm{GAC}=\frac{1}{3}$ [Area of $\triangle \mathrm{ABC}$ ]

Question 9.
Using vector method, prove that $\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$.
Solution:
- Take two points $\mathrm{A}$ and $\mathrm{B}$ on the unit circle with centre as origin ' $o$ '. so

$|\overrightarrow{\mathrm{OA}}|=|\overrightarrow{\mathrm{OB}}|=1$

$\text { - }\lfloor\mathrm{AO} x=\alpha ;\lfloor\mathrm{BO} x=\beta \Rightarrow\lfloor\mathrm{AOB}=\alpha-\beta$
- Let $\vec{i}$ and $\vec{j}$ be the unit vectors along the $x, y$ respectively.
- The co-ordinates of A and B be $(\cos \alpha, \sin \alpha)$ and $(\cos \beta, \sin \beta)$ respectively.
$
\begin{aligned}
& \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{OL}}+\overrightarrow{\mathrm{LA}} \\
& \overrightarrow{\mathrm{OA}}=\cos \alpha \vec{i}+\sin \alpha \vec{j} \\
& \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OM}}+\overrightarrow{\mathrm{MB}} \quad \Rightarrow \overrightarrow{\mathrm{OB}}=\cos \beta \vec{i}-\sin \beta \vec{j}
\end{aligned}
$
So, $\quad \overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OA}}=\cos \alpha \cos \beta+\sin \alpha \sin \beta$
But $\quad \overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OA}}=|\overrightarrow{\mathrm{OB}}||\overrightarrow{\mathrm{OA}}| \cos (\alpha+\beta)$
$
\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OA}}=\cos (\alpha+\beta)
$
From (1) and (2), we get
$
\cos (\alpha+\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta
$

Question 10.
Prove by vector method that $\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$.
Solution:
Take two points $A$ and $B$ on the unit circle with centre as origin ' $O$ ', so $|\overrightarrow{\mathrm{OA}}|=|\overrightarrow{\mathrm{OB}}|=1$

$
\lfloor\mathrm{AOX}=\alpha ; \quad \mathrm{BOX}=\beta \Rightarrow \mathrm{AOB}=\alpha+\beta
$
Let $\vec{i}$ and $\vec{j}$ be the unit vectors along the $x, y$ direction respectively.
The co-ordinates of $\mathrm{A}$ and $\mathrm{B}$ be $(\cos \alpha, \sin \alpha)$ and $(\cos \beta,-\sin \beta)$ respectively.
$
\begin{aligned}
\overrightarrow{\mathrm{OA}} & =\overrightarrow{\mathrm{OL}}+\overrightarrow{\mathrm{LA}} \\
\overrightarrow{\mathrm{OA}} & =\cos \alpha \vec{i}+\sin \alpha \vec{j} \\
\overrightarrow{\mathrm{OB}} & =\overrightarrow{\mathrm{OM}}+\overrightarrow{\mathrm{MB}} \\
\overrightarrow{\mathrm{OB}} & =\cos \beta \vec{i}-\sin \beta \vec{j}
\end{aligned}
$
So, $\quad \overrightarrow{\mathrm{OB}} \times \overrightarrow{\mathrm{OA}}=(\sin \alpha \cos \beta+\cos \alpha \sin \beta) \vec{k}$
But, $\quad \overrightarrow{\mathrm{OB}} \times \overrightarrow{\mathrm{OA}}=|\overrightarrow{\mathrm{OB}}||\overrightarrow{\mathrm{OA}}| \sin (\alpha+\beta) \vec{k}$
$
\overrightarrow{\mathrm{OB}} \times \overrightarrow{\mathrm{OA}}=\sin (\alpha+\beta) \vec{k}
$
From (1) \& (2), we get
$
\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta
$

Question 11.
A particle acted on by constant forces $8 \vec{i}+2 \vec{j}-6 \vec{k}$ and $\overrightarrow{6 i}+2 \vec{j}-2 \vec{k}$ is displaced from the point $(1,2,3)$ to the point $(5,4,1)$. Find the total work done by the forces.
Solution:
$
\begin{aligned}
& \overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_1+\overrightarrow{\mathrm{F}}_2 \\
& \overrightarrow{\mathrm{F}}=(8 \vec{i}+2 \vec{j}-6 \vec{k})+(6 \vec{i}+2 \vec{j}-2 \vec{k}) \\
& \overrightarrow{\mathrm{F}}=(14 \vec{i}+4 \vec{j}-8 \vec{k}) \\
& \vec{d}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\
& \vec{d}=(5 \vec{i}+4 \vec{j}+\vec{k})-(\vec{i}+2 \vec{j}+3 \vec{k}) \\
& \vec{d}=(4 \vec{i}+2 \vec{j}-2 \vec{k})
\end{aligned}
$
From (1) \& (2), we get
Work done by the force $=\overrightarrow{\mathrm{F}} \cdot \vec{d}=56+8+16=80$ units.

Question 12 .
Forces of magnitude $5 \sqrt{2}$ and $10 \sqrt{2}$ units acting in the directions $(3 \vec{i}+4 \vec{j}+5 \vec{k})$ and $(10 \vec{i}+6 \vec{j}-8 \vec{k})$, respectively, act on a particle which is displaced from the point with position vector $(\vec{i}-3 \vec{j}-2 \vec{k})$ to the point with position vector $(\overrightarrow{6 i}+\vec{j}-3 \vec{k})$. Find the work done by the forces.
Solution:
$
\begin{aligned}
& \vec{a}=3 \vec{i}+4 \vec{j}+5 \vec{k} \quad\left(\text { since } \hat{a}=\frac{\vec{a}}{|\vec{a}|}\right) \\
& |\vec{a}|=\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2} \\
& \overrightarrow{\mathrm{F}}_1=5 \sqrt{2} \hat{a}=5 \sqrt{2} \frac{(3 \vec{i}+4 \vec{j}+5 \vec{k})}{5 \sqrt{2}}=3 \vec{i}+4 \vec{j}+5 \vec{k} \\
& \vec{b}=10 \vec{i}+6 \vec{j}-8 \vec{k} \\
& |\vec{b}|=\sqrt{100+36+64}=\sqrt{200}=10 \sqrt{2} \\
& \overrightarrow{\mathrm{F}_2}=10 \sqrt{2} \hat{b} \\
& =10 \sqrt{2} \frac{(10 \vec{i}+6 \vec{j}-8 \vec{k})}{10 \sqrt{2}}\left(\text { since } \hat{b}=\frac{\vec{b}}{|\vec{b}|}\right) \\
& =10 \vec{i}+6 \vec{j}-8 \vec{k} \\
& \therefore \text { So, } \quad \overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_1+\overrightarrow{\mathrm{F}}_2 \\
& \overrightarrow{\mathrm{F}}=(3 \vec{i}+4 \vec{j}+5 \vec{k})+(10 \vec{i}+6 \vec{j}-8 \vec{k}) \\
& \overrightarrow{\mathrm{F}}=13 \vec{i}+10 \vec{j}-3 \vec{k} \\
& \vec{d}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\
& \vec{d}=(6 \vec{i}+\vec{j}-3 \vec{k})-(4 \vec{i}-3 \vec{j}-2 \vec{k}) \\
& \vec{d}=2 \vec{i}+4 \vec{j}-\vec{k} \\
&
\end{aligned}
$
From (1) \& (2), we get
Work done by the force $=\overrightarrow{\mathrm{F}} \cdot \vec{d}=(13 \vec{i}+10 \vec{j}-3 \vec{k}) \cdot(2 \vec{i}+4 \vec{j}-\vec{k})=26+40+3=69$ Units

Question 13.
Find the magnitude and direction cosines of the torque of a force represented by $3 \vec{i}+4 \vec{j}-5 \vec{k}$ about the point with position vector $2 \vec{i}-3 \vec{j}+4 \vec{k}$ acting through a point whose position vector is $\overrightarrow{4 i}+2 \vec{j}-3 \vec{k}$.
Solution:
$\begin{aligned} & \overrightarrow{\mathrm{F}}=-3 \vec{i}+6 \vec{j}-6 \vec{k} \\ & \vec{r}=(\text { through the point })-(\text { about the point }) \\ & \vec{r}=(4 \vec{i}+2 \vec{j}-3 \vec{k})-(2 \vec{i}-3 \vec{j}+4 \vec{k}) \\ & \vec{r}=2 \vec{i}+5 \vec{j}-7 \vec{k} \\ & \text { Torque }(\overrightarrow{\mathrm{M}})=\vec{r} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 2 & 5 & -7 \\ 3 & 4 & -5\end{array}\right| \\ &=\vec{i}[-25+28]-\vec{j}[-10+21]+\vec{k}[8-15] \\ & \overrightarrow{\mathrm{M}}=3 \vec{i}-11 \vec{j}-7 \vec{k} \\ & \text { Magnitude }=|\overrightarrow{\mathrm{M}}|=\sqrt{9+121+49}=\sqrt{179} \\ & \text { Direction cosines }\left(\frac{3}{\sqrt{179}}, \frac{-11}{\sqrt{179}}, \frac{-7}{\sqrt{179}}\right)\end{aligned}$

Question 14.
Find the torque of the resultant of the three forces represented by $-3 \vec{i}+6 \vec{j}-3 \vec{k}, \overrightarrow{4} \vec{i}-10 \vec{j}+12 \vec{k}$ and $\overrightarrow{4 i}+7 \vec{j}$ acting at the point with position vector $8 \vec{i}-\overrightarrow{6} \vec{j}-4 \vec{k}$, about the point with position vector $18 \vec{i}+3 \vec{j}-9 \vec{k}$
Solution:
$\begin{aligned} \overrightarrow{\mathrm{F}}_1 & =-3 \vec{i}+6 \vec{j}-6 \vec{k} \\ \overrightarrow{\mathrm{F}_2} & =4 \vec{i}-10 \vec{j}+12 \vec{k} \text { and } \overrightarrow{\mathrm{F}}_3=4 \vec{i}+7 \vec{j} \\ \overrightarrow{\mathrm{F}} & =\overrightarrow{\mathrm{F}}_1+\overrightarrow{\mathrm{F}}_2+\overrightarrow{\mathrm{F}}_3 \\ \overrightarrow{\mathrm{F}} & =5 \vec{i}+3 \vec{j}+9 \vec{k} \\ \vec{r} & =(\text { act at the point })-(\text { about the point })=(8 \vec{i}-6 \vec{j}-4 \vec{k})-(18 \vec{i}+3 \vec{j}-9 \vec{k}) \\ \vec{r} & =-10 \vec{i}-9 \vec{j}+5 \vec{k} \\ \text { Torque }(\overrightarrow{\mathrm{M}}) & =\vec{r} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -10 & -9 & 5 \\ 5 & 3 & 9\end{array}\right| \\ & =[-81-15]-\vec{j}[-90-25]+\vec{k}[-30+45] \\ & =-96 \vec{i}+115 \vec{j}+15 \vec{k}\end{aligned}$