Exercise 6.10-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 $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ 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. $\quad-(\vec{a} \cdot \vec{b}) \vec{c}=-(\vec{b} \cdot \vec{c}) \vec{a} \quad \Rightarrow t \vec{c}=s \vec{a} \quad \Rightarrow \vec{a} \| \vec{c}$