Exercise 8.4-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.
If $\vec{p}=-3 \vec{i}+4 \vec{j}-7 \vec{k}$ and $\vec{q}=6 \vec{i}+2 \vec{j}-3 \vec{k}$ then find $\vec{p} \times \vec{q}$. Verify that $\vec{p}$ and $\vec{p} \times \vec{q}$ are perpendicular to each other and also verify that $\vec{q}$ and $\vec{p} \times \vec{q}$ are perpendicular to each other.
Solution:
$
\begin{aligned}
& \vec{p} \times \vec{q}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-3 & 4 & -7 \\
6 & 2 & -3
\end{array}\right| \\
& =2 \vec{i}-51 \vec{j}-30 \vec{k} \\
& \text { Now } \vec{p} \cdot(\vec{p} \times \vec{q})=(-3 \vec{i}+4 \vec{j}-7 \vec{k}) \cdot(2 \vec{i}-51 \vec{j}-30 \vec{k}) \\
& =-6-204+210=0 \\
&
\end{aligned}
$
Hence $\vec{p}$ and $\vec{p} \times \vec{q}$ are perpendicular to each other.
Now $\quad \vec{q} \cdot(\vec{p} \times \vec{q})=(6 \vec{i}+2 \vec{j}-3 \vec{k}) \cdot(2 \vec{i}-51 \vec{j}-30 \vec{k})$
$
=12-102+90=0
$
Hence $\vec{q}$ and $\vec{p} \times \vec{q}$ are perpendicular to each other.
Question 2.
If $\vec{a}, \vec{b}$ are any two vectors, then prove that $|\vec{a} \times \vec{b}|^2+(\vec{a} \cdot \vec{b})^2=|\vec{a}|^2|\vec{b}|^2$
Solution:
Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$
$
\begin{gathered}
\therefore \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n} \\
|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta \\
|\vec{a} \times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta \\
(\vec{a} \cdot \vec{b})^2=|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta \\
|\vec{a} \times \vec{b}|^2+(\vec{a} \cdot \vec{b})^2=|\vec{a}|^2|\vec{b}|^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=|\vec{a}|^2|\vec{b}|^2
\end{gathered}
$
Question 3.

Find the angle between the vectors $\boldsymbol{2} \vec{i}+\vec{j}-\overrightarrow{\boldsymbol{k}}$ and $\vec{i}+\mathbf{2} \overrightarrow{\boldsymbol{j}}+\overrightarrow{\boldsymbol{k}}$ by using cross product.

Solution:

Let $\vec{a}=2 \vec{i}+\vec{j}-\vec{k} ; \vec{b}=\vec{i}+2 \vec{j}+\vec{k}$
Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$
$
\begin{aligned}
& \therefore \theta=\sin ^{-1}\left(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\right) \\
& \vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 & 1 & -1 \\
1 & 2 & 1
\end{array}\right|=3 \vec{i}-3 \vec{j}+3 \vec{k} \\
&|\vec{a} \times \vec{b}|=\sqrt{3^2+(-3)^2+3^2}=3 \sqrt{3} \\
&|\vec{a}|=\sqrt{2^2+1^2+(-1)^2}=\sqrt{6} \\
&|\vec{b}|=\sqrt{1^2+2^2+1^2}=\sqrt{6} \\
& \therefore \sin \theta=\left(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\right)=\left(\frac{3 \sqrt{3}}{\sqrt{6} \sqrt{6}}\right)=\left(\frac{\sqrt{3}}{2}\right) \\
& \theta=\frac{\pi}{3}
\end{aligned}
$

Question 4.
Find the vector of magnitude 6 which are perpendicular to both the vectors
$
4 \vec{i}-\vec{j}+3 \vec{k} \text { and }-2 \vec{i}+\vec{j}-2 \vec{k}
$
Solution:
Let $\quad \vec{a}=4 \vec{i}-\vec{j}+3 \vec{k} ; \vec{b}=-2 \vec{i}+\vec{j}-2 \vec{k}$
Then
$
\begin{aligned}
|\vec{a} \times \vec{b}| & =\sqrt{(-1)^2+(2)^2+(2)^2}=3 \\
\text { Required vectors } & =6\left[ \pm\left(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right)\right] \\
& = \pm(-2 \vec{i}+4 \vec{j}+4 \vec{k})
\end{aligned}
$
Question 5.
Find the vectors whose length 5 which are perpendicular to the vectors

$\vec{a}=3 \vec{i}+\vec{j}-4 \vec{k}$ and $\vec{b}=6 \vec{i}+5 \vec{j}-2 \vec{k}$
Solution:
The unit vector perpendicular to $\vec{a}$ and $\vec{b}$ is $\hat{n}= \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
$
\begin{aligned}
\vec{a} \times \vec{b} & =\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -4 \\
6 & 5 & -2
\end{array}\right| \\
& =\hat{i}(-2+20)-\hat{j}(-6+24)+\hat{k}(15-6) \\
& =18 \hat{i}-18 \hat{j}+9 \hat{k} \\
|\vec{a} \times \vec{b}| & =\sqrt{324+324+81}=\sqrt{729}=\sqrt{81 \times 9}=27 \\
\hat{n} & = \pm \frac{18 \hat{i}-18 \hat{j}+9 \hat{k}}{27}= \pm \frac{9(2 \hat{i}-2 \hat{j}+\hat{k})}{27} \\
& = \pm \frac{2 \hat{i}-2 \hat{j}+\hat{k}}{3} \\
\therefore 5 \hat{n} & = \pm \frac{5(2 \hat{i}-2 \hat{j}+\hat{k})}{3}
\end{aligned}
$

Question 6.
If $\vec{a} \times \vec{b}=\vec{c} \times \vec{d}$ and $\vec{a} \times \vec{c}=\vec{b} \times \vec{d}$ show that $\vec{a}-\vec{d}$ and $\vec{b}-\vec{d}$ are parallel.
Solution:
Given $\vec{a} \times \vec{b}=\vec{c} \times \vec{d}$ and $\vec{a} \times \vec{c}=\vec{b} \times \vec{d}$
Now $(\vec{a}-\vec{d}) \times(\vec{b}-\vec{c})$
$
\begin{aligned}
& =\vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{d} \times \vec{b}+\vec{d} \times \vec{c} \\
& =\vec{a} \times \vec{b}-\vec{a} \times \vec{c}+\vec{b} \times \vec{d}-\vec{c} \times \vec{d} \\
& =(\vec{a} \times \vec{b}-\vec{c} \times \vec{d})-(\vec{a} \times \vec{c}-\vec{b} \times \vec{d})=0
\end{aligned}
$
$\Rightarrow \vec{a}-\vec{d}$ and $\vec{b}-\vec{c}$ are parallel.
Question 7.
Find the angle between two vectors $\vec{a}$ and $\vec{b}$ if $|\overrightarrow{\boldsymbol{a}} \times \overrightarrow{\boldsymbol{b}}|=\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{b}}$

Solution:
Let $\theta$ be the angle between the vector $\vec{a}$ and $\vec{b}$.
$
\begin{aligned}
& \therefore \vec{a} \times \vec{b}=a b \sin \theta \hat{n} \\
& \text { So, }|\vec{a} \times \vec{b}|=|a b \sin \theta \hat{n}|=a b \sin \theta ; \vec{a} \cdot \vec{b}=a b \cos \theta \\
& \text { Given }|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b} \\
& \Rightarrow \quad a b \sin \theta=a b \cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=1 \\
& \text { i.e., } \tan \theta=1 \Rightarrow \theta=\frac{\pi}{4}
\end{aligned}
$
$\therefore$ The angle between the vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$

Question 8.
If $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}=0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\pi / 6$.
Prove that $\vec{a}= \pm 2(\vec{b} \times \vec{c})$
Solution:
$
\begin{aligned}
\vec{b} \times \vec{c} & =|\vec{b}| \times|\vec{c}| \sin \theta \hat{n} \\
& =(1)(1) \sin (\pi / 6)(\hat{n})=\frac{1}{2}(\hat{n}) \\
\vec{a} \cdot \vec{b} & =0 \Rightarrow \vec{a} \perp^{\mathrm{r}} \text { to } \vec{b} ; \vec{a} \cdot \vec{c}=0 \Rightarrow \vec{a} \perp^r \text { to } \vec{c}
\end{aligned}
$
$\therefore \quad \vec{a}$ is $\perp^{\mathrm{r}}$ to $\vec{b}$ and $\vec{c} \Rightarrow \vec{a}$ parallel to $\vec{b} \times \vec{c}$.
From (i) and (ii), $\vec{a}= \pm t(\vec{b} \times \vec{c})$.
$
\Rightarrow|\vec{a}|=|t(\vec{b} \times \vec{c})| \text {. But }|\vec{a}|=1 \Rightarrow t|\vec{b} \times \vec{c}|=1
$
(i.e.) $t\left[(1)(1) \sin \frac{\pi}{6}\right]=1 \Rightarrow t / 2=1 \Rightarrow t=2$
So $\quad \vec{a}= \pm 2(\vec{b} \times \vec{c})$
Question 9.
If $|\vec{a}|=2,|\vec{b}|=7$ and $\vec{a} \times \vec{b}=3 \hat{\boldsymbol{i}}-2 \hat{\boldsymbol{j}}+6 \hat{\boldsymbol{k}}$ find the angle between $\vec{a}$ and $\vec{b}$

Solution:
If $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then
$
\begin{aligned}
\sin \theta & =\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|} ; \text { Here } \vec{a} \times \vec{b}=3 \hat{i}-2 \hat{j}+6 \hat{k} \\
\therefore|\vec{a} \times \vec{b}| & =\sqrt{9+4+36}=7 \\
\text { So, } \sin \theta & =\frac{7}{2 \times 7}=\frac{1}{2} ; \therefore \theta=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} .
\end{aligned}
$