Exercise 6.1-Additional Problems - Chapter 6 - Applications of Vector Algebra - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
The work done by the force $\overrightarrow{\mathrm{F}}=a \vec{i}+\vec{j}+\vec{k}$ in moving the point of application from $(1,1,1)$ to (2, 2, 2) along a straight line is given to be 5 units. Find the value of a.
Solution:
$
\overrightarrow{\mathrm{F}}=a \vec{i}+\vec{j}+\vec{k} ; \overrightarrow{\mathrm{OA}}=\vec{i}+\vec{j}+\vec{k} ; \overrightarrow{\mathrm{OB}}=2 \vec{i}+2 \vec{j}+2 \vec{k}
$
Work done $=5$ units
$
\begin{aligned}
\vec{d} & =\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\vec{i}+\vec{j}+\vec{k} \\
\text { Work done } & =\overrightarrow{\mathrm{F}} \cdot \vec{d} \\
5 & =(a \vec{i}+\vec{j}+\vec{k}) \cdot(\vec{i}+\vec{j}+\vec{k}) \\
5 & =a+1+1 \quad \Rightarrow a=3
\end{aligned}
$

Question 2.
If the position vectors of three points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are respectively $\vec{i}+2 \vec{j}+3 \vec{k}, 4 \vec{i}+\vec{j}+5 \vec{k}$ and $7(\vec{i}+\vec{k})$. Find $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}$. Interpret the result geometrically.
Solution:
$
\begin{aligned}
\overrightarrow{\mathrm{OA}} & =\vec{i}+2 \vec{j}+3 \vec{k} ; \overrightarrow{\mathrm{OB}}=4 \vec{i}+\vec{j}+5 \vec{k} ; \overrightarrow{\mathrm{OC}}=7 \vec{i}+7 \vec{k} \\
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(4 \vec{i}+\vec{j}+5 \vec{k})-(\vec{i}+2 \vec{j}+3 \vec{k}) \\
\overrightarrow{\mathrm{AB}} & =3 \vec{i}-\vec{j}+2 \vec{k} \\
\overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=6 \vec{i}-2 \vec{j}+4 \vec{k} \\
\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & -1 & 2 \\
6 & -2 & 4
\end{array}\right|=\overrightarrow{0}
\end{aligned}
$
The vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AC}}$ are parallel. But they have the point $\mathrm{A}$ as a common point. $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AC}}$ are along the same line.
$\therefore \mathrm{A}, \mathrm{B}, \mathrm{C}$ are collinear.

Question 3.
A force given by and $3 \vec{i}+2 \vec{j}-4 \vec{k}$ is applied at the point $(1,-1,2)$. Find the moment of the force about the point $(2,-1,3)$.
Solution:
$
\begin{aligned}
\overrightarrow{\mathrm{F}} & =3 \vec{i}+2 \vec{j}-4 \vec{k} \\
\overrightarrow{\mathrm{OP}} & =\vec{i}-\vec{j}+2 \vec{k} \\
\overrightarrow{\mathrm{OA}} & =2 \vec{i}-\vec{j}+3 \vec{k} \\
\vec{r} & =\overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{OA}}=(\vec{i}-\vec{j}+2 \vec{k})-(2 \vec{i}-\vec{j}+3 \vec{k}) \\
\vec{r} & =-\vec{i}-\vec{k}
\end{aligned}
$
The moment $\overrightarrow{\mathrm{M}}$ of the force $\overrightarrow{\mathrm{F}}$ about the point $\mathrm{A}$ is given by
$
\overrightarrow{\mathrm{M}}=\vec{r} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{rrr}
\vec{i} & \vec{j} & \vec{k} \\
-1 & 0 & -1 \\
3 & 2 & -4
\end{array}\right|=2 \vec{i}-7 \vec{j}-2 \vec{k}
$

Question 4.
Show that the area of a parallelogram having diagonals $3 \vec{i}+\vec{j}-2 \vec{k}$ and $\vec{i}-3 \vec{j}+4 \vec{k}$ is $5 \sqrt{3}$

Solution:
Let $\vec{d}_1=3 \vec{i}+\vec{j}-2 \vec{k}$ and $\vec{d}_2=\vec{i}-3 \vec{j}+4 \vec{k}$
$
\begin{aligned}
& \text { Area of the parallelogram }=\frac{1}{2}\left|\vec{d}_1 \times \vec{d}_2\right| \\
& \qquad \vec{d}_1 \times \vec{d}_2=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & 1 & -2 \\
1 & -3 & 4
\end{array}\right|=-2 \vec{i}-14 \vec{j}-10 \vec{k} \\
& \Rightarrow \quad\left|\vec{d}_1 \times \vec{d}_2\right|=\sqrt{(-2)^2+(-14)^2+(-10)^2}=\sqrt{300}=10 \sqrt{3}
\end{aligned}
$
Area of the parallelogram $=\frac{1}{2}\left|\vec{d}_1 \times \vec{d}_2\right|=\frac{1}{2} 10 \sqrt{3}=5 \sqrt{3}$ Sq. units.