Exercise 6.2 - 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

$\mathbf{E x} 6.2$
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 $\vec{a} \cdot(\vec{b} \times \vec{c})$.
Solution:
Given
$
\begin{aligned}
& \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} \\
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 1 & -2 \\
3 & 2 & 1
\end{array}\right| \\
& \vec{b} \times \vec{c}=5 \vec{i}-8 \vec{j}+\vec{k} \\
& \vec{a} \cdot(\vec{b} \times \vec{c})=5+16+3=24 \\
& \text { Aliter: } \quad \vec{a} \cdot(\vec{b} \times \vec{c})=[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}
1 & -2 & 3 \\
2 & 1 & -2 \\
3 & 2 & 1
\end{array}\right|=1(5)+2(8)+3(1)=5+16+3=24 \\
&
\end{aligned}
$
Question 2.
Find the volume of the parallelepiped whose coterminous edges are represented by the vectors $-6 \hat{i}+14 \hat{j}+10 \hat{k}, 14 \hat{i}-10 \hat{j}-6 \hat{k}$ and $2 \hat{i}+4 \hat{j}-2 \hat{k}$.
Solution:
Volume of the parallelepiped $=\| \vec{a}, \vec{b}, \vec{c}]$
$
\begin{aligned}
& {[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}
-6 & 14 & 10 \\
14 & -10 & -6 \\
2 & 4 & -2
\end{array}\right|=-6[20+24]-14[-28+12]+10[56+20]} \\
& =-264+224+760=720 \text { cubic units }
\end{aligned}
$

Question 3.
The volume of the parallelepiped whose coterminus edges are $7 \vec{i}+\lambda \vec{j}-3 \vec{k}, \vec{i}+2 \vec{j}-\vec{k},,-3 \vec{i}+7 \vec{j}+5 \vec{k}$ is 90 cubic units. Find the value of $\lambda$
Solution:
Given, Volume of the parallelepiped $=90$ cubic units
$
\begin{aligned}
& \text { (i.e), } \quad[\vec{a} \vec{b} \vec{c}]=90 \\
& \left|\begin{array}{ccc}
7 & \lambda & -3 \\
1 & 2 & -1 \\
-3 & 7 & 5
\end{array}\right|=90 \\
& \Rightarrow 7[10+7]-\lambda[5-3]-3[7+6]=90 \\
& \Rightarrow 119-2 \lambda-39=90 \\
& \Rightarrow-2 \lambda=10 \quad \Rightarrow \lambda=\frac{-10}{2}=-5 \\
&
\end{aligned}
$

Question 4.
If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors represented by concurrent edges of a parallelepiped of volume 4
cubic units, find the value of $(\vec{a}+\vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b}+\vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c}+\vec{a}) \cdot(\vec{a} \times \vec{b})$
Solution:
Let $\vec{a}, \vec{b}, \vec{c}$ be the concurrent edges of parallelepiped
Given volume of parallelepiped $=4$ cubic units
$
|[\vec{a}, \vec{b}, \vec{c}]|=4
$
But,
$
[\vec{a} \vec{b} \vec{c}]=\pm 4
$
$
\begin{aligned}
(\vec{a}+\vec{b}) \cdot(\vec{b} \times \vec{c}) & =\vec{a} \cdot(\vec{b} \times \vec{c})+\vec{b} \cdot(\vec{b} \times \vec{c})=[\vec{a} \vec{b} \vec{c}]+[\vec{b} \vec{b} \vec{c}]=[\vec{a} \vec{b} \vec{c}]+0 \\
& =[\vec{a} \vec{b} \vec{c}]
\end{aligned}
$
Similarly $(\vec{b}+\vec{c}) \cdot(\vec{c} \times \vec{a})=\vec{b} \cdot(\vec{c} \times \vec{a})=[\vec{b} \vec{c} \vec{a}]=[\vec{a} \vec{b} \vec{c}]$
So, $\quad(1)+(2)+(3)=2[\vec{a} \vec{b} \vec{c}]=3(\pm 4) \Rightarrow \pm 12$

Question 5.
Find the altitude of a parallelepiped determined by the vectors $\vec{a}=-2 \hat{i}+5 \hat{j}+3 \hat{k}, \hat{b}=\hat{i}+3 \hat{j}-2 \hat{k}$ and $\vec{c}=-3 \vec{i}+\vec{j}+4 \vec{k}$ if the base is taken as the parallelogram determined by $\mathrm{b}$ and $\mathrm{c}$.
Solution:
Volume $=$ Base Area $\times$ Height
$\begin{aligned} &|[\vec{a}, \vec{b}, \vec{c}]|=|\vec{b} \times \vec{c}| \times \text { Height } \\ &|[\vec{a}, \vec{b}, \vec{c}]|=\left|\begin{array}{rrr}-2 & 5 & 3 \\ 1 & 3 & -2 \\ -3 & 1 & 4\end{array}\right|=-2(12+2)-5(4-6)+3(1+9) \\ &=-28+10+30=12 \\ & \vec{b} \times \vec{c}=\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 3 & -2 \\ -3 & 1 & 4\end{array}\right|=\vec{i}(14)-\vec{j}(-2)+\vec{k}(10) \\ & \vec{b} \times \vec{c}=14 \vec{i}+2 \vec{j}+10 \vec{k} \\ &|\vec{b} \times \vec{c}|=\sqrt{196+4+100}=\sqrt{300} \cdot \\ & \text { Height }=\frac{12}{\sqrt{300}}=\frac{12}{10 \sqrt{3}}=\frac{6}{5 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{6 \sqrt{3}}{5 \times 3}=\frac{2 \sqrt{3}}{5} \\ &(\sqrt{300}) \times(\text { height })=12 \\ &-12\end{aligned}$

Question 6.
Determine whether the three vectors $2 \hat{i}+3 \hat{j}+\hat{k}, \hat{i}-2 \hat{j}+2 \hat{k}$ and $3 \hat{i}+\hat{j}+3 \hat{k}$ are coplanar.

Solution:
$
\text { Let } \begin{aligned}
\vec{a} & =2 \vec{i}+3 \vec{j}+\vec{k} \\
\vec{b} & =\vec{i}-2 \vec{j}+2 \vec{k} \\
\vec{c} & =3 \vec{i}+\vec{j}+3 \vec{k}
\end{aligned}
$
we know that $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if $[\vec{a} \vec{b} \vec{c}]=0$
$
[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}
2 & 3 & 1 \\
1 & -2 & 2 \\
3 & 1 & 3
\end{array}\right|=2(-6-2)-3(3-6)+(1+6)=-16+9+7=0
$
$\therefore$ Gives vectors are coplanar

Question 7.
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}$ and $\vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$. If $c_1=1$ and $c_2=2$, find $c_3$ such that $\vec{a}, \vec{b}$ and $\vec{c}$ and $c$ are coplanar.
Solution:
Given
$
\begin{aligned}
\vec{a} & =\vec{i}+\vec{j}+\vec{k} \\
\vec{b} & =\vec{i} \\
\vec{c} & =c_1 \vec{i}+c_2 \vec{j}+c_3 \vec{k} \text { are coplanar. }
\end{aligned}
$
But $c_1=1$ and $c_2=2$
So
$
\vec{c}=\vec{i}+2 \vec{j}+c_3 \vec{k}
$
We know that $[\vec{a} \vec{b} \vec{c}]=0$
$
\begin{aligned}
\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 0 \\
1 & 2 & c_3
\end{array}\right| & =0 \\
1[0]-1\left[c_3\right]+1[2] & =0 \quad \Rightarrow-c_3+2=0 \quad \Rightarrow c_3=2
\end{aligned}
$

Question 8.
If $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}, \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$, , show that $[\vec{a} \vec{b} \vec{c}]$ depends neither $\mathrm{x}$ nor $\mathrm{y}$.
Solution:
$
\begin{aligned}
& \text { Given } \vec{a}=\vec{i}-\vec{k} \\
& \vec{b}=x \vec{i}+\vec{j}+(1-x) \vec{k} \\
& \vec{c}=y \vec{i}+x \vec{j}+(1+x-y) \vec{k} \\
& {[\vec{a} \vec{b} \vec{c}]=\left|\begin{array}{ccc}
1 & 0 & -1 \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|=1(1+x-y)-x(1-x)-1\left(x^2-y\right)} \\
& =1+x-y-x+x^2-x^2+y \\
& {[\vec{a} \vec{b} \vec{c}]=1} \\
&
\end{aligned}
$
$\therefore$ Clearly $[\vec{a} \vec{b} \vec{c}]$ depends on neither $x$ nor $y$

Question 9.
If the vectors
$a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$
are coplanar, prove that $\mathrm{c}$ is the geometric mean of $\mathrm{a}$ and $\mathrm{b}$.
Solution:
Let $\vec{a}_1=a \vec{i}+a \vec{j}+c \vec{k}$
$
\begin{aligned}
& \vec{a}_2=\vec{i}+\vec{k} \\
& \vec{a}_3=c \vec{i}+c \vec{j}+b \vec{k}
\end{aligned}
$
But $\vec{a}_1, \vec{a}_2, \vec{a}_3$ are coplanar (Given)
So, $\left[\begin{array}{lll}\vec{a}_1 & \vec{a}_2 & \vec{a}_3\end{array}\right]=0$
$
\begin{aligned}
\left|\begin{array}{lll}
a & a & c \\
1 & 0 & 1 \\
c & c & b
\end{array}\right|=0 \quad & \Rightarrow a[0-c]-a[b-c]+c[c-0]=0 \\
& \Rightarrow-a c-a b+a c+c^2=0 \\
& \Rightarrow c^2=a b \quad \Rightarrow c=\sqrt{a b}
\end{aligned}
$
$\therefore \mathrm{c}$ is the geometric means of ' $\mathrm{a}$ ' and ' $\mathrm{b}$ '.

Question 10.
Let $\vec{a}, \vec{b}, \vec{c}$ be three non-zero vectors such that $\vec{c}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$ show that $[\vec{a}, \vec{b}, \vec{c}]^2=\frac{1}{4}|\vec{a}|^2|\vec{b}|^2$.
Solution:
Given $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$. So $\vec{c}$ is parallel to $\vec{a} \times \vec{b}$
$
\begin{aligned}
{[\vec{a} \vec{b} \vec{c}] } & =\vec{a} \cdot(\vec{b} \times \vec{c}) \\
|[\vec{a}, \vec{b}, \vec{c}]| & =|\vec{a}||\vec{b} \times \vec{c}|=|\vec{a}||\vec{b}||\vec{c}| \sin \left(\frac{\pi}{6}\right) \\
|[\vec{a}, \vec{b}, \vec{c}]| & =|\vec{a}||\vec{b}||\vec{c}|\left(\frac{1}{2}\right)
\end{aligned}
$
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
$