Exercise 8.2-Additional Questions - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 8.2-Additional Questions - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions 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.
Show that the points whose position vectors given by
$(i)-2 \hat{i}+3 \hat{j}+5 \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k}, 7 \hat{i}-\hat{k}$
(ii) $\hat{i}-2 \hat{j}+3 \hat{k}, 2 \hat{i}+3 \hat{j}-4 \hat{k}$ and $-7 \vec{j}+10 \vec{k}$
Solution:

(i) Let the given point be $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. To prove $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are collinear we have to prove that
$
\overrightarrow{\mathrm{AB}}=t \overrightarrow{\mathrm{AC}}
$
Now $\overrightarrow{\mathrm{OA}}=-2 \hat{i}+3 \hat{j}+5 \hat{k}, \overrightarrow{\mathrm{OB}}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\overrightarrow{\mathrm{OC}}=7 \hat{i}-\hat{k}$
$
\begin{aligned}
& \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\hat{i}+2 \hat{j}+3 \hat{k})-(-2 \hat{i}+3 \hat{j}+5 \hat{k}) \\
&=\hat{i}(1+2)+\hat{j}(2-3)+\hat{k}(3-5) \\
&=3 \hat{i}-\hat{j}-2 \hat{k} \\
&-\ldots \quad . \quad \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(7 \hat{i}-\hat{k})-(-2 \hat{i}+3 \hat{j}+5 \hat{k}) \\
&=\hat{i}(7+2)+\hat{j}(0-3)+\hat{k}(-1-5) \\
&=9 \hat{i}-3 \hat{j}-6 \hat{k}=3(3 \hat{i}-\hat{j}-2 \hat{k}) \\
& \overrightarrow{\mathrm{AC}}=3 \overrightarrow{\mathrm{AB}} \\
& \Rightarrow \quad \text { the point } \mathrm{A}, \mathrm{B} \text { and } \mathrm{C} \text { are collinear. }
\end{aligned}
$

(ii) Let the given point be $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$
$\overrightarrow{\mathrm{OA}}=\hat{i}-2 \hat{j}+3 \hat{k}, \overrightarrow{\mathrm{OB}}=2 \hat{i}+3 \hat{j}-4 \hat{k}$ and $\overrightarrow{\mathrm{OC}}=-7 \hat{j}+10 \hat{k}$
Now
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(2 \vec{i}+3 \vec{j}-4 \vec{k})-(\vec{i}-2 \vec{j}+3 \vec{k}) \\
& =\hat{i}(2-1)+\hat{j}(3+2)+\hat{k}(-4-3) \\
& =\hat{i}+5 \hat{j}-7 \hat{k} \\
\overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(-7 \hat{j}+10 \hat{k})-(\hat{i}-2 \hat{j}+3 \hat{k}) \\
& =\hat{i}(0-1)+\hat{j}(-7+2)+\hat{k}(10-3) \\
& =-\hat{i}-5 \hat{j}+7 \hat{k}=-(\hat{i}+5 \hat{j}-7 \hat{k}) \\
\overrightarrow{\mathrm{AC}} & =-\overrightarrow{\mathrm{AB}}
\end{aligned}
$
$
\Rightarrow \quad \overrightarrow{\mathrm{AC}}=-\overrightarrow{\mathrm{AB}}
$
So, the point $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ are collinear.
Question 2.
Find the unit vectors parallel to the sum of $3 \hat{i}-5 \hat{j}+8 \hat{k}$ and $-2 \hat{j}-2 \hat{k}$ Solution:
Let the given vectors be $\vec{a}=3 \hat{i}-5 \hat{j}+8 \hat{k}$ and $\vec{b}=-2 \hat{j}-2 \hat{k}$
Now
$
\begin{aligned}
\vec{a}+\vec{b} & =(3 \hat{i}-5 \hat{j}+8 \hat{k})+(-2 \hat{j}-2 \hat{k}) \\
& =\hat{i}(3)+\hat{j}(-5-2)+\hat{k}(8-2) \\
& =3 \hat{i}-7 \hat{j}+6 \hat{k} \\
|\vec{a}+\vec{b}| & =\sqrt{9+49+36}=\sqrt{94} \text { units }
\end{aligned}
$
The unit vectors parallel to $\vec{a}+\vec{b}$ are $\pm \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}= \pm \frac{3 \hat{i}-7 \hat{j}+6 \hat{k}}{\sqrt{94}}$.

Question 3.
The vertices of a triangle have position vectors $4 \hat{i}+5 \hat{j}+6 \hat{k}, 5 \hat{i}+6 \hat{j}+4 \hat{k}, 6 \hat{i}+4 \hat{j}+5 \hat{k}$ Prove that the triangle is equilateral.
Solution:
Let $\mathrm{ABC}$ be the triangles with position vectors $\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$
(i.e.,) $\overrightarrow{\mathrm{OA}}=4 \hat{i}+5 \hat{j}+6 \hat{k}, \overrightarrow{\mathrm{OB}}=5 \hat{i}+6 \hat{j}+4 \hat{k}$ and $\overrightarrow{\mathrm{OC}}=6 \hat{i}+4 \hat{j}+5 \hat{k}$
Now
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(5 \hat{i}+6 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+6 \hat{k}) \\
& =\hat{i}+\hat{j}-2 \hat{k} \\
|\overrightarrow{\mathrm{AB}}| & =\sqrt{1+1+4}=\sqrt{6} \text { units } \\
\overrightarrow{\mathrm{BC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=(6 \hat{i}+4 \hat{j}+5 \hat{k})-(5 \hat{i}+6 \hat{j}+4 \hat{k}) \\
& =\hat{i}-2 \hat{j}+\hat{k} \\
|\overrightarrow{\mathrm{BC}}| & =\sqrt{1+4+1}=\sqrt{6} \text { units } \\
\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} & =(6 \hat{i}+4 \hat{j}+5 \hat{k})-(4 \hat{i}+5 \hat{j}+6 \hat{k}) \\
& =2 \hat{i}-\hat{j}-\hat{k} \\
|\overrightarrow{\mathrm{AC}}| & =\sqrt{4+1+1}=\sqrt{6} \text { units }
\end{aligned}
$
Now $|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}|=\sqrt{6}$ units
$\Rightarrow \triangle \mathrm{ABC}$ is an equilateral triangle.
Question 4.
Prove that the points $2 \hat{i}+3 \hat{j}+4 \hat{k}, 3 \hat{i}+4 \hat{j}+2 \hat{k}, 4 \hat{i}+2 \hat{j}+3 \hat{\boldsymbol{k}}$ form an equilateral triangle.

Solution:
Let $\mathrm{ABC}$ be the given triangle with vertices $\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$.
Now $\overrightarrow{\mathrm{OA}}=2 \hat{i}+3 \hat{j}+4 \hat{k} ; \overrightarrow{\mathrm{OB}}=3 \hat{i}+4 \hat{j}+2 \hat{k} ; \overrightarrow{\mathrm{OC}}=4 \hat{i}+2 \hat{j}+3 \hat{k}$
$
\begin{aligned}
& \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(3 \hat{i}+4 \hat{j}+2 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\
& =\hat{i}(3-2)+\hat{j}(4-3)+\hat{k}(2-4) \\
& =\hat{i}+\hat{j}-2 \hat{k} \\
& \therefore \quad|\overrightarrow{\mathrm{AB}}|=\sqrt{1+1+4}=\sqrt{6} \\
& \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=(4 \hat{i}+2 \hat{j}+3 \hat{k})-(3 \hat{i}+4 \hat{j}+2 \hat{k}) \\
& =\hat{i}(4-3)+\hat{j}(2-4)+\hat{k}(3-2) \\
& =\hat{i}-2 \hat{j}+\hat{k} \\
& \therefore \quad|\overrightarrow{B C}|=\sqrt{1+4+1}=\sqrt{6} \\
& \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(4 \hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\
& =\hat{i}(4-2)+\hat{j}(2-3)+\hat{k}(3-4) \\
& =2 \hat{i}-\hat{j}-\hat{k} \\
& \therefore \quad|\overrightarrow{\mathrm{AC}}|=\sqrt{4+1+1}=\sqrt{6} \\
&
\end{aligned}
$
So
Now $|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}| \Rightarrow \mathrm{ABC}$ is an equilateral triangle.
(i.e.,) the given points from an equilateral triangle.

Question 5.
Examine whether the vectors $\hat{\boldsymbol{i}}+\mathbf{3} \hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}, 2 \hat{\boldsymbol{i}}-\hat{\boldsymbol{j}}-\hat{\boldsymbol{k}}$ and $\mathbf{7} \hat{\boldsymbol{j}}+5 \hat{\boldsymbol{k}}$ are coplanar

Solution:
Let the given vectors be
$
\vec{a}=\hat{i}+3 \hat{j}+\hat{k} ; \vec{b}=2 \hat{i}-\hat{j}-\hat{k} \text { and } \vec{c}=7 \hat{j}+5 \hat{k}
$
To prove that the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar we have to prove that
$
\begin{aligned}
\vec{a} & =m \vec{b}+n \vec{c} \text { where } m \text { and } n \text { are scalars. } \\
\hat{i}+3 \hat{j}+\hat{k} & =m(2 \hat{i}-\hat{j}-\hat{k})+n(7 \hat{j}+5 \hat{k})
\end{aligned}
$
Equating $\hat{i}, \hat{j}$ and $\hat{k}$ components we get,
(i.e.,)
$
\begin{aligned}
1 & =2 m+0 \\
2 m & =1 \Rightarrow m=\frac{1}{2} \\
3 & =-m+7 n
\end{aligned}
$
(i.e.)
$
\begin{aligned}
-m+7 n & =3 \\
1 & =-m+5 n
\end{aligned}
$

$
-m+5 n=1
$
Substituting $m=1 / 2$ in (ii) we get
$
-1 / 2+7 n=3
$
(i.e.,)
$
\therefore \quad n=\frac{7}{2 \times 7}=\frac{1}{2}
$
Substituting $m=\frac{1}{2}$ and $n=\frac{1}{2}$ in (iii) we get
$\mathrm{LHS}=1$
$
\begin{array}{rlr}
\text { RHS } & =-\left(\frac{1}{2}\right)+5\left(\frac{1}{2}\right)=-\frac{1}{2}+\frac{5}{2} & \\
& =\frac{-1+5}{2}=\frac{4}{2}=2 & \text { LHS } \neq \text { RHS }
\end{array}
$
$\Rightarrow$ We are not able to write one vector as a linear combination of the other two vectors
$\Rightarrow$ the given vectors are not coplanar.