Exercise 6.3 - Chapter 6 Applications of Vector Algebra 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.3$
Question 1.
If $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{c}=3 \hat{i}+2 \hat{j}+\hat{k}$, find $($ i) $(\vec{a} \times \vec{b}) \times \vec{c}$ (ii) $\vec{a} \times(\vec{b} \times \vec{c})$
Solution:
Given $\quad \vec{a}=\vec{i}-2 \vec{j}+3 \vec{k}, \vec{b}=2 \vec{i}+\vec{j}-2 \vec{k}, \vec{c}=3 \vec{i}+2 \vec{j}+\vec{k}$
(i) $(\vec{a} \times \vec{b}) \times \vec{c}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}=(3-4+3) \vec{b}-(6+2-2) \vec{a}=2 \vec{b}-6 \vec{a}$ $=4 \vec{i}+2 \vec{j}-4 \vec{k}-6 \vec{i}+12 \vec{j}-18 \vec{k}=-2 \vec{i}+14 \vec{j}-22 \vec{k}$
(ii) $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$
$
\begin{aligned}
& =(3-4+3) \vec{b}-(2-2-6) \vec{c}=2 \vec{b}+6 \vec{c} \\
& =4 \vec{i}+2 \vec{j}-4 \vec{k}+18 \vec{i}+12 \vec{j}+6 \vec{k}=22 \vec{i}+14 \vec{j}+2 \vec{k}
\end{aligned}
$
Question 2.
For any vector $\vec{a}$, prove that $\hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})=2 \vec{a}$.
Solution:
Let $\quad \vec{a}=a_1 \vec{i}+a_2 \vec{j}+a_3 \vec{k}$
$
\vec{i} \times(\vec{a} \times \vec{i})=(\vec{i} \cdot \vec{i}) \vec{a}-(\vec{i} \cdot \vec{a}) \vec{i}=\vec{a}-a_1 \vec{i}
$
Similarly, $\vec{j} \times(\vec{a} \times \vec{j})=\vec{a}-a_2 \vec{j}$
(1) $+(2)+(3) \Rightarrow \hat{i} \times(\vec{a} \times \hat{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \hat{k})=3 \vec{a}-\left(a_1 \vec{i}+a_2 \vec{j}+a_3 \vec{k}\right)=3 \vec{a}-\vec{a}=2 \vec{a}$
Question 3.
Prove that $[\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}]=0$

Solution:
$
\begin{aligned}
\text { LHS } & =[\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}] \\
& =(\vec{a}-\vec{b}) \cdot[(\vec{b}-\vec{c}) \times(\vec{c}-\vec{a})] \\
& =(\vec{a}-\vec{b}) \cdot[(\vec{b} \times \vec{c})-(\vec{b} \times \vec{a})-(\vec{c} \times \vec{c})+(\vec{c} \times \vec{a})] \quad \text { [Since } \vec{c} \times \vec{c}=0] \\
& =\vec{a} \cdot(\vec{b} \times \vec{c})-\vec{a} \cdot(\vec{b} \times \vec{a})+\vec{a} \cdot(\vec{c} \times \vec{a})-\vec{b} \cdot(\vec{b} \times \vec{c})+\vec{b} \cdot(\vec{b} \times \vec{c})-\vec{b} \cdot(\vec{c} \times \vec{a}) \\
& =[\vec{a} \vec{b} \vec{c}]-[\vec{b} \vec{c} \vec{a}]=[\vec{a} \vec{b} \vec{c}]-[\vec{a} \vec{b} \vec{c}] \\
& =0=\text { RHS }
\end{aligned}
$

Question 4.
If $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{j}+2 \hat{k}, \vec{c}=-\hat{i}-2 \hat{j}+3 \hat{k}$, verify that
(i) $(\vec{a} \times \vec{b}) \times \vec{c}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}$
(ii) $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$
Solution:
Given $\vec{a}=2 \vec{i}+3 \vec{j}-\vec{k}, \vec{b}=3 \vec{i}+5 \vec{j}-2 \vec{k}, \vec{c}=-\vec{i}-2 \vec{j}+3 \vec{k}$
(i) $(\vec{a} \times \vec{b}) \times \vec{c}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}$
$
\begin{aligned}
\vec{a} \times \vec{b} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 3 & -1 \\
3 & 5 & 2
\end{array}\right|=11 \vec{i}-7 \vec{j}+\vec{k} \\
(\vec{a} \times \vec{b}) \times \vec{c} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
11 & -7 & 1 \\
-1 & -2 & 3
\end{array}\right|=-19 \vec{i}-34 \vec{j}-29 \vec{k} \\
(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a} & =(-2-6-3) \vec{b}-(-3-10+6) \vec{a}=-11 \vec{b}+7 \vec{a} \\
& =-33 \vec{i}-55 \vec{j}-22 \vec{k}+14 \vec{i}+21 \vec{j}-7 \vec{k} \\
& =-19 \vec{i}-34 \vec{j}-29 \vec{k}
\end{aligned}
$
From (1) \& (2), we get
$
(\vec{a} \times \vec{b}) \times \vec{c}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}
$

(ii) $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$
$
\begin{aligned}
\vec{b} \times \vec{c} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & 5 & 2 \\
-1 & -2 & 3
\end{array}\right|=19 \vec{i}-11 \vec{j}-\vec{k} \\
\vec{a} \times(\vec{b} \times \vec{c}) & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 3 & -1 \\
19 & -11 & -1
\end{array}\right|=-14 \vec{i}-17 \vec{j}-79 \vec{k} \\
(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} & =(-2-6-3) \vec{b}-(6+15-2) \vec{c}=-11 \vec{b}-19 \vec{c} \\
& =-33 \vec{i}-55 \vec{j}-22 \vec{k}+19 \vec{i}+38 \vec{j}-57 \vec{k}=-14 \vec{i}-17 \vec{j}-79 \vec{k}
\end{aligned}
$
From (1) \& (2), we get
$
\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}
$
Question 5.
$\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \vec{b}=-\hat{i}+2 \hat{j}-4 \hat{k}, \vec{c}=\hat{i}+\hat{j}+\hat{k}$ then find the value of $(\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{c})$

Solution:
Given $\vec{a}=2 \vec{i}+3 \vec{j}-\vec{k}, \vec{b}=-\vec{i}+2 \vec{j}-4 \vec{k}, \vec{c}=\vec{i}+\vec{j}+\vec{k}$
$
\begin{aligned}
(\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{c}) & =\left|\begin{array}{cc}
\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{c} \\
\vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{c}
\end{array}\right|=\left|\begin{array}{cc}
(4+9+1) & (2+3-1) \\
(-2+6+4) & (-1+2-4)
\end{array}\right| \\
& =\left|\begin{array}{cc}
14 & 4 \\
8 & -3
\end{array}\right|=-42-32=-74
\end{aligned}
$

Question 6.
If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar vectors, show that $(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}$

Solution:
Given $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar vectors.
$
\begin{aligned}
(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d}) & =[\vec{a} \vec{b} \vec{d}] \vec{c}-[\vec{a} \vec{b} \vec{c}] \vec{d}=0(\vec{c})-0(\vec{d}) \\
& =\overrightarrow{0}
\end{aligned}
$

Question 7.
$\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}+\hat{k}, \vec{c}=3 \hat{i}+2 \hat{j}+\hat{k}$ and $\vec{a} \times(\vec{b} \times \vec{c})=l \vec{a}+m \vec{b}+n \vec{c}$, find the values of $1, \mathrm{~m}, \mathrm{n}$
Solution:
$
\begin{aligned}
\text { Given } \vec{a} & =\vec{i}+2 \vec{j}+3 \vec{k}, \vec{b}=2 \vec{i}-\vec{j}+\vec{k}, \vec{c}=3 \vec{i}+2 \vec{j}+\vec{k} \\
\vec{a} \times(\vec{b} \times \vec{c}) & =(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=(3+4+3) \vec{b}-(2-2+3) \vec{c} \\
& =10 \vec{b}-3 \vec{c}=20 \vec{i}-10 \vec{j}+10 \vec{k}-9 \vec{i}-6 \vec{j}-3 \vec{k}=11 \hat{i}-16 \hat{j}+7 \hat{k} \\
l \vec{a}+m \vec{b}+n \vec{c} & =l(\vec{i}+2 \vec{j}+3 \vec{k})+m(2 \vec{i}-\vec{j}+\vec{k})+n(3 \vec{i}+2 \vec{j}+\vec{k})
\end{aligned}
$
From (1) \& (2) compare the $\vec{i}, \vec{j}, \vec{k}$ co efficients.
$
\begin{aligned}
l+2 m+3 n & =11 \\
2 l-m+2 n & =-16 \\
3 l+m+n & =7
\end{aligned}
$

Question 8.
If $\hat{a}, \hat{b}, \hat{c}$ are three unit vectors such that $\hat{b}$ and $\hat{c}$ are non-parallel and $\hat{a} \times(\hat{b} \times \hat{c})=\frac{1}{2} \hat{b}$, find the angle between $\hat{a}$ and $\hat{c}$.
Solution:
Given $\quad \vec{a} \times(\vec{b} \times \vec{c})=\frac{1}{2} \vec{b} \quad \Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\frac{1}{2} \vec{b}$
Since $\hat{b}$ and $\hat{c}$ are non collinear vectors. So, equating corresponding coefficients on both sides.
$
\begin{aligned}
& \vec{a} \cdot \vec{c}=\frac{1}{2} \\
& |\vec{a}| \cdot|\vec{c}| \cos \theta=\frac{1}{2} \quad \Rightarrow \cos \theta=\frac{1}{2} \\
& \theta=\cos ^{-1}\left(\frac{1}{2}\right) \quad \Rightarrow \theta=\left(\frac{\pi}{3}\right) \\
&
\end{aligned}
$
$\therefore \hat{a}$ makes an angle with $\hat{c}$ is $\frac{\pi}{3}$