Exercise 8.2-Additional Questions - Chapter 8 Vector Algebra–I 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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.