Exercise 6.3-Additional Problems - 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

Additional Problems
Question 1.
If $\vec{a}=3 \vec{i}+2 \vec{j}-4 \vec{k}, \vec{b}=5 \vec{i}-3 \vec{j}+6 \vec{k}, \vec{c}=5 \vec{i}-\vec{j}+2 \vec{k}$, find (i) $\vec{a} \times(\vec{b} \times \vec{c})($ (ii) $(\vec{a} \times \vec{b}) \times \vec{c}$ and show that they are not equal.
Solution:
$(i)$
$
\begin{aligned}
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
5 & -3 & 6 \\
5 & -1 & 2
\end{array}\right|=20 \vec{j}+10 \vec{k} \\
& \therefore \quad \vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
3 & 2 & -4 \\
0 & 20 & 10
\end{array}\right|=100 \vec{i}-30 \vec{j}+60 \vec{k} \\
&
\end{aligned}
$
(ii)
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 3 & 2 & -4 \\ 5 & -3 & 6\end{array}\right|=-38 \vec{j}-19 \vec{k}$
$(\vec{a} \times \vec{b}) \times \vec{c}=\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 0 & -38 & -19 \\ 5 & -1 & 2\end{array}\right|=-95 \vec{i}-95 \vec{j}+190 \vec{k}$

Question 2 .
. If $\vec{a}=2 \vec{i}+3 \vec{j}-\vec{k}, \vec{b}=-2 \vec{i}+5 \vec{k}, \vec{c}=\vec{j}-3 \vec{k}$ verify that $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$.
Solution:
$
\begin{aligned}
\vec{b} \times \vec{c} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-2 & 0 & 5 \\
0 & 1 & -3
\end{array}\right|=\vec{i}(-5)-\vec{j}(6)+\vec{k}(-2)=-5 \vec{i}-6 \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 & -1 \\
-5 & -6 & -2
\end{array}\right|=\vec{i}(-6-6)-\vec{j}(-4-5)+\vec{k}(-12+15) \\
& =-12 \vec{i}+9 \vec{j}+3 \vec{k} \\
\vec{a} \cdot \vec{c} & =(2 \vec{i}+3 \vec{j}-\vec{k}) \cdot(\vec{j}-3 \vec{k})=(2)(0)+3(1)+(-1)(-3)=3+3=6 \\
\vec{a} \cdot \vec{b} & =(2 \vec{i}+3 \vec{j}-\vec{k}) \cdot(-2 \vec{i}+5 \vec{k}) \\
& =(2)(-2)+(3)(0)+(-1)(5)=-4+0-5=-9
\end{aligned}
$
So, $(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=6[-2 \vec{i}+5 \vec{k}]-(-9)[\vec{j}-3 \vec{k}]=6[-2 \vec{i}+5 \vec{k}]+9[\vec{j}-3 \vec{k}]$
$
=-12 \vec{i}+30 \vec{k}+9 \vec{j}-27 \vec{k}=-12 \vec{i}+9 \vec{j}+3 \vec{k}
$
$
\text { (1) }=(2) \Rightarrow \quad \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}
$