Exercise 6.2-Additional Problems - Chapter 6 - Applications of Vector Algebra - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
If the edges $\vec{a}=-3 \vec{i}+7 \vec{j}+5 \vec{k}, \vec{b}=-5 \vec{i}+7 \vec{j}-3 \vec{k}, \vec{c}=7 \vec{i}-5 \vec{j}-3 \vec{k}$ meet a vertex, find the volume of the parallelepiped.
Solution:
Volume of the parallelepiped $=$
$
[\vec{a}, \vec{b}, \vec{c}]=\left|\begin{array}{rrr}
-3 & 7 & 5 \\
-5 & 7 & -3 \\
7 & -5 & -3
\end{array}\right|=-264
$
The volume cannot be negative
$\therefore$ Volume of parallelepiped $=264 \mathrm{cu}$. units

Question 2.
If $\vec{x} \cdot \vec{a}=0, \vec{x} \cdot \vec{b}=0, \vec{x} \cdot \vec{c}=0$ and $\vec{x} \neq \overrightarrow{0}$ then show that $\vec{a}, \vec{b}, \vec{c}$ are coplanar.
Solution:
$\vec{x} \cdot \vec{a}=0, \vec{x} \cdot \vec{b}=0$ implies $\vec{a}$ and $\vec{b}$ are $\perp \mathrm{r}$ to $\vec{x}$
$\therefore \vec{a} \times \vec{b}$ is parallel to $\vec{x}$
$
\begin{aligned}
& \vec{x}=\lambda(\vec{a} \times \vec{b}) \\
& \text { Now } \quad \vec{x} \cdot \vec{c}=0 \quad \Rightarrow \lambda(\vec{a} \times \vec{b}) \cdot \vec{c}=0 \\
& \Rightarrow \quad\lceil\vec{a} \vec{b} \vec{c}\rceil=0 \quad \Rightarrow \vec{a}, \vec{b}, \vec{c} \text { are coplanar. } \\
&
\end{aligned}
$

Question 3.
The volume of a parallelepiped whose edges are represented by $-12 \vec{i}+\lambda k, 3 \vec{j}-\vec{k}, 2 \vec{i}+\vec{j}-15 \vec{k}$ is 546 . Find the value of $\lambda$.
Solution:
$
[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{rrr}
-12 & 0 & \lambda \\
0 & 3 & -1 \\
2 & 1 & -15
\end{array}\right|
$
Volume of the parallelepiped $=$
$
\begin{aligned}
& =-12[-45+1]-00+\lambda[0-6]=-12(-44)-6 \lambda \\
& =528-6 \lambda=546 \text { (given) } \\
& \Rightarrow-6 \lambda=546-528=18 \\
& \therefore \lambda=\frac{18}{-6}=-3
\end{aligned}
$
Question 4.
Prove that $|\vec{a} \vec{b} \vec{c}|=$ abc if and only if $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular.
Solution:

Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors.
$
\begin{aligned}
\text { i.e., } \vec{a} \cdot \vec{b} & =\vec{b} \cdot \vec{c}=\vec{a} \cdot \vec{c}=0 \\
\text { Now }[\vec{a} \vec{b} \vec{c}] & =\vec{a} \cdot(\vec{b} \times \vec{c})=|\vec{a}||\vec{b} \times \vec{c}| \cos \phi
\end{aligned}
$
where $\phi$ is the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$.
Squaring on both sides $[\vec{a}, \vec{b}, \vec{c}]^2=|\vec{a}|^2|\vec{b}|^2|\vec{c}|^2 \frac{1}{4}$
(Since $|\vec{c}|=1$ )
$
[\vec{a}, \vec{b}, \vec{c}]^2=\frac{1}{4}|\vec{a}|^2|\vec{b}|^2
$

Question 5.
Show that the points $(1,3,1),(1,1,-1),(-1,1,1),(2,2,-1)$ are lying on the same plane.
Solution:
To prove that the points $A, B, C, D$ are coplanar, we have to prove that $[\overrightarrow{\mathrm{AB}} \overrightarrow{\mathrm{AC}} \overrightarrow{\mathrm{AD}}]=0$.
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\vec{i}+\vec{j}-\vec{k})-(\vec{i}+3 \vec{j}+\vec{k})=-2 \vec{j}-2 \vec{k} \\
\overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(-\vec{i}+\vec{j}+\vec{k})-(\vec{i}+3 \vec{j}+\vec{k})=-2 \vec{i}-2 \vec{j} \\
\overrightarrow{\mathrm{AD}} & =\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OA}}=(2 \vec{i}+2 \vec{j}-\vec{k})-(\vec{i}+3 \vec{j}+\vec{k})=\vec{i}-\vec{j}-2 \vec{k} \\
{[\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AD}}] } & =\left|\begin{array}{rrr}
0 & -2 & -2 \\
-2 & -2 & 0 \\
1 & -1 & -2
\end{array}\right|=0()+2[4-0]-2[2+2]=8-8=0
\end{aligned}
$
$\Rightarrow$ The points A, B, C, D are lying on the same plane.
Question 6.
If $\vec{a}=2 \vec{i}+3 \vec{j}-5 \vec{k}, \vec{b}=-\vec{i}+\vec{j}+2 \vec{k}$ and $\vec{c}=4 \vec{i}-2 \vec{j}+3 \vec{k}$, show that $(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times(\vec{b} \times \vec{c})$.
Solution:

$\begin{aligned}
&\begin{aligned}
& \vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 3 & -5 \\
-1 & 1 & 2
\end{array}\right|=\vec{i}(6+5)-\vec{j}(4-5)+\vec{k}(2+3)=11 \vec{i}+\vec{j}+5 \vec{k} \\
& \text { So, }(\vec{a} \times \vec{b}) \times \vec{c}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
11 & 1 & 5 \\
4 & -2 & 3
\end{array}\right|=\vec{i}(3+10)-\vec{j}(33-20)+\vec{k}(-22-4) \\
& =13 \vec{i}-13 \vec{j}-26 \vec{k} \\
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-1 & 1 & 2 \\
4 & -2 & 3
\end{array}\right|=\vec{i}(3+4)-\vec{j}(-3-8)+\vec{k}(2-4)=7 \vec{i}+11 \vec{j}-2 \vec{k} \\
& \text { So, } \vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 3 & -5 \\
7 & 11 & -2
\end{array}\right|=\vec{i}(-6+55)-\vec{j}(-4+35)+\vec{k}(22-21) \\
& =49 \vec{i}-31 \vec{j}+\vec{k} \\
&
\end{aligned}\\
&\text { (1) } \neq(2) \quad \Rightarrow(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times(\vec{b} \times \vec{c})
\end{aligned}$