Exercise 6.10-Additional Problems - Chapter 6 - Applications of Vector Algebra - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 6.10-Additional Problems - Chapter 6 - Applications of Vector Algebra - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Additional Problems
Question 1.
If $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$ then $\vec{a} \cdot(\vec{b} \times \vec{c})=$
(a) 6
(b) 10
(c) 12
(d) 24
Solution:
(c) 12
Hint:
$
\begin{aligned}
\vec{a} \cdot(\vec{b} \times \vec{c}) & =\left|\begin{array}{rrr}
2 & 1 & -1 \\
1 & 2 & 1 \\
1 & -1 & 2
\end{array}\right|=2(4+1)-1(2-1)-1(-1-2) \\
& =10-1+3=12
\end{aligned}
$

Question 2.
Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}, \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$, then $[\vec{a} \vec{b} \vec{c}]$ depend on
(a) only $x$
(b) only $y$
(c) Neither x nor y
(d) Both $\mathrm{x}$ and $\mathrm{y}$
Solution:
(c) Neither x nor y
Hint:
$
[\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|=\left|\begin{array}{ccc}
1 & 0 & 0 \\
x & 1 & 1 \\
y & x & 1+x
\end{array}\right|\left[\text { Applying } \mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_1\right]
$
Expanding along $\mathrm{R}_1$, we get
$
\left|\begin{array}{cc}
1 & 1 \\
x & 1+x
\end{array}\right|=1+x-x=1
$
Question 3.
If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{p}, \vec{q}, \vec{r}$ are defined by the relations $\vec{p}=\frac{\vec{b} \times \vec{c}}{[\vec{a} \vec{b} \vec{c}]}, \vec{q}=\frac{\vec{c} \times \vec{a}}{[\vec{a} \vec{b} \vec{c}]}, \vec{r}=\frac{\vec{a} \times \vec{b}}{[\vec{a} \vec{b} \vec{c}]}$, then $\vec{p} \cdot(\vec{a}+\vec{b})+\vec{q} \cdot(\vec{b}+\vec{c})+\vec{r}(\vec{c}+\vec{a})=$
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3

Hint:
$
\begin{aligned}
& \vec{p} \cdot(\vec{a}+\vec{b})=\vec{p} \cdot \vec{a}+\vec{p} \cdot \vec{b} \\
& \frac{(\vec{b} \times \vec{c}) \cdot \vec{a}}{[\vec{a} \vec{b} \vec{c}]}+\frac{[\vec{b} \times \vec{c}] \cdot \vec{b}}{[\vec{a} \vec{b} \vec{c}]}=\frac{[\vec{a} \vec{b} \vec{c}]}{[\vec{a} \vec{b} \vec{c}]}+0=1+0=1 \\
& \text { Similarly } \quad \vec{q} \cdot(\vec{b}+\vec{c})=1 \text { and } \vec{r} \cdot(\vec{c}+\vec{a})=1
\end{aligned}
$
Hence, $\vec{p} \cdot(\vec{a}+\vec{b})+\vec{q}(b+\vec{c})+\vec{r} \cdot(\vec{c}+\vec{a})=1+1+1=3$
Question 4.
The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{k} \times \hat{i})+\hat{k} \cdot(\hat{i} \times \hat{j})=$
(a) 1
(b) 3
(c) $-3$
(d) 0
Solution:
(b) 3
Hint:
$
\begin{gathered}
\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{k} \times \hat{i})+\hat{k} \cdot(\hat{i} \times \hat{j})=\hat{i} \cdot \hat{i}+\hat{j} \cdot \hat{j}+\hat{k} \cdot \hat{k} \quad[\because \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j} \text { and } \hat{i} \times \hat{j}=\hat{k}] \\
\hat{i}^2+\hat{j}^2+\hat{k}^2=1+1+1=3
\end{gathered}
$

Question 5.
Let a, b, c be distinct non-negative numbers. If the vectors $a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$ lie in a plane, then $c$ is
(a) the A.M. of $a$ and $b$
(b) the G.M. of $a$ and $b$
(c) the H.M. of a and b
(d) equal to zero.
Solution:
(b) the G.M. of $a$ and $b$
Hint:
$a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$ are coplanar
$
\begin{array}{lll}
\therefore \quad\left|\begin{array}{lll}
a & a & c \\
1 & 0 & 1 \\
c & c & b
\end{array}\right|=0 & & \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
\end{array}
$

Question 6.
The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+(\hat{i} \times \hat{k}) \cdot \hat{j}$
(a) 1
(b) $-1$
(c) 0
(d) $\hat{j}$
Solution:
(c) 0
Hint:
$
\begin{array}{ll}
\hat{i} \cdot(\hat{j} \times \hat{k})+(\hat{i} \times \hat{k}) \cdot \hat{j} \Rightarrow \hat{i} \cdot \hat{i}+(-\hat{j}) \hat{j} & {[\because \hat{j} \times \hat{k}=\hat{i} \text { and } \hat{i} \times \hat{k}=-\hat{j}]} \\
\Rightarrow & \hat{i}^2-\hat{j}^2=1-1=0
\end{array}
$

Question 7.
The value of $(\hat{i}-\hat{j}, \hat{j}-\hat{k}, \hat{k}-\hat{i})$ is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(a) 0
Hint:
$
\begin{gathered}
(\hat{i}-\hat{j}) \cdot[(\hat{j}-\hat{k} \times(\hat{k}-\hat{i}))] \\
(\hat{i}-\hat{j}) \cdot[\hat{j} \times \hat{k}-\hat{j} \times \hat{i}-\hat{k} \times \hat{k}+\hat{k} \times \hat{i}]=\hat{i} \cdot \hat{i}+\hat{i} \cdot \hat{k}+\hat{i} \cdot \hat{j}-\hat{j} \cdot \hat{i}-\hat{j} \cdot \hat{k}-\hat{j} \cdot \hat{j} \\
=\hat{i}^2+0+0-0-0-\hat{j}^2=1-1=0
\end{gathered}
$

Question 8.
If $\vec{u}=\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})$, then
(a) $\vec{u}$ is a unit vector (b) $\vec{u}=\vec{a}+\vec{b}+\vec{c}$
(c) $\vec{u}=\overrightarrow{0}$
(d) $\vec{u} \neq \overrightarrow{0}$
Solution:
(c) $\vec{u}=\overrightarrow{0}$
Hint:
$
\begin{aligned}
& \vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})=\overrightarrow{0} \\
& \Rightarrow \quad \vec{u}=\overrightarrow{0} \\
&
\end{aligned}
$

Question 9.
The area of the parallelogram having a diagonal $3 \vec{i}+\vec{j}-\vec{k}$ and a side $\vec{i}-3 \vec{j}+4 \vec{k}$ is ...............
(a) $10 \sqrt{3}$
(b) $6 \sqrt{30}$
(c) $\frac{3}{2} \sqrt{30}$
(d) $3 \sqrt{30}$
Solution:
(d) $3 \sqrt{30}$
Solution:

$\begin{aligned} \text { Area of } \triangle \mathrm{ABC} & =\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ \therefore \quad \text { Area of parallelogram } & =2 \times \text { Area of } \triangle \mathrm{ABC} \\ & =2 \times \frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} & =\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & -3 & 4 \\ 3 & 1 & -1\end{array}\right|=-\vec{i}+13 \vec{j}+10 \vec{k} \\ \therefore \quad|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| & =\sqrt{1+69+100}=\sqrt{270}=3 \sqrt{30}\end{aligned}$

Question 10.
If $\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(a \times \vec{b})=\vec{x} \times \vec{y}$, then
(a) $\vec{x}=\overrightarrow{0}$
(b) $\vec{y}=\overrightarrow{0}$
(c) $\vec{x}$ and $\vec{y}$ are parallel
(d) $\vec{x}=\overrightarrow{0}$ or $\vec{y}=\overrightarrow{0}$ or $\vec{x}$ and $\vec{y}$ are parallel
Solution:
(d) $\vec{x}=\overrightarrow{0}$ or $\vec{y}=\overrightarrow{0}$ or $\vec{x}$ and $\vec{y}$ are parallel
Hint:
$\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})=\overrightarrow{0}$
So
$
\vec{x} \times \vec{y}=\overrightarrow{0} \quad \Rightarrow \vec{x}=\overrightarrow{0} \quad \text { or } \vec{y}=\overrightarrow{0}
$
or $\quad \vec{x} \| \vec{y}$

Question 11.
If $\overrightarrow{\mathrm{PR}}=2 \vec{i}+\vec{j}+\vec{k}, \overrightarrow{\text { Question }}=-\vec{i}+3 \vec{j}+2 \vec{k}$, then the area of the quadrilateral PQRS is
(a) $5 \sqrt{3}$
(b) $10 \sqrt{3}$
(c) $\frac{5 \sqrt{3}}{2}$
(d) $\frac{3}{2}$
Solution:
(c) $\frac{5 \sqrt{3}}{2}$

Hint:
$
\begin{aligned}
\text { Area of } \mathrm{PQRS} & =\frac{1}{2}|\overrightarrow{\mathrm{PR}} \times \overrightarrow{\mathrm{QS}}| \\
\overrightarrow{\mathrm{PR}} \times \overrightarrow{\mathrm{QS}} & =\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 1 & 1 \\
-1 & 3 & 2
\end{array}\right|=\vec{i}(2-3)-\vec{j}(4+1)+\vec{k}(6+1) \\
& =-\vec{i}-5 \vec{j}+7 \vec{k} \\
|\overrightarrow{\mathrm{PR}} \times \overrightarrow{\mathrm{QS}}| & =\sqrt{1+25+49}=\sqrt{75}=5 \sqrt{3} \\
\text { Area of } \mathrm{PQRS} & =\frac{1}{2}(5 \sqrt{3})=\frac{5 \sqrt{3}}{2}
\end{aligned}
$
Question 12.
If $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}$ for non-coplanar vectors $\vec{a}, \vec{b}, \vec{c}$ then
(a) $\vec{a}$ parallel to $\vec{b}$
(b) $\vec{b}$ parallel to $\vec{c}$
(c) $\vec{c}$ parallel to $\vec{a}$
(d) $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$
Solution:
(c) $\vec{c}$ parallel to $\vec{a}$
Hint:
Given $\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}$
$
\Rightarrow(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{b} \cdot \vec{c}) \vec{a}
$
i.e. $\quad-(\vec{a} \cdot \vec{b}) \vec{c}=-(\vec{b} \cdot \vec{c}) \vec{a} \quad \Rightarrow t \vec{c}=s \vec{a} \quad \Rightarrow \vec{a} \| \vec{c}$