Exercise 8.4 - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 8.4
Question 1.
Find the magnitude of $\vec{a} \times \vec{b}$ if $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$. Solution:
$
\begin{aligned}
\vec{a} \times \vec{b} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 3 \\
3 & 5 & -2
\end{array}\right|=\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3) \\
& =-17 \hat{i}+13 \hat{j}+7 \hat{k} \\
|\vec{a} \times \vec{b}| & =\sqrt{17^2+13^2+7^2}=\sqrt{289+169+49}=\sqrt{507}
\end{aligned}
$

Question 2.
Show that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=\overrightarrow{0}$
Solution:
$
\begin{aligned}
& \text { LHS : } \vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b}) \\
& =\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{a}+\vec{c} \times \vec{b} \\
& =\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{b} \times \vec{c}=0 \quad\left(\begin{array}{l}
c \times a=-\vec{a} \times \vec{c} \\
\vec{c} \times \vec{b}=-\vec{b} \times \vec{c}
\end{array}\right) \\
& =\mathrm{RHS} \\
&
\end{aligned}
$
Question 3.
Find the vectors of magnitude $10 \sqrt{3}$ that are perpendicular to the plane which contains $\hat{i}+2 \hat{j}+\hat{k}$ and $\hat{i}+3 \hat{j}+4 \hat{k}$
Solution:

$\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}+4 \hat{k}$
The unit vector $\perp^r$ to $\vec{a}$ and $\vec{b}$ is given by $\hat{n}= \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$
$
\text { Now } \begin{aligned}
\vec{a} \times \vec{b} & =\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
1 & 3 & 4
\end{array}\right|=\hat{i}(8-3)-\hat{j}(4-1)+\hat{k}(3-2)=5 \hat{i}-3 \hat{j}+\hat{k} \\
|\vec{a} \times \vec{b}| & =\sqrt{25+9+1}=\sqrt{35} \\
\therefore \hat{A} & = \pm \frac{5 \hat{i}-3 \hat{j}+\hat{k}}{\sqrt{35}} \\
\therefore 10 \sqrt{3} \hat{n} & =\frac{10 \sqrt{3}(5 \hat{i}-3 \hat{j}+\hat{k})}{\sqrt{35}} \\
& = \pm \frac{10 \sqrt{3}}{\sqrt{35}}(5 \hat{i}-3 \hat{j}+\hat{k})
\end{aligned}
$

Question 4.
Find the unit vectors perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$.
Solution:
$
\begin{aligned}
& \vec{a}=\hat{i}+\hat{j}+\hat{k} \text { and } \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} \\
& \vec{a}+\vec{b}=(\hat{i}+\hat{j}+\hat{k})+(\hat{i}+2 \hat{j}+3 \hat{k})=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
& \vec{a}-\vec{b}=(\hat{i}+\hat{j}+\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})=-\hat{j}-2 \hat{k}
\end{aligned}
$
Let $\vec{a}+\vec{b}=\vec{c}$ and $\vec{a}-\vec{b}=\vec{d}$
Now the unit vectors $\perp^r$ to $\vec{c}$ and $\vec{d}$ is given by $\hat{n}= \pm \frac{\vec{c} \times \vec{d}}{|\vec{c} \times \vec{d}|}$
$
\begin{aligned}
\vec{c} \times \vec{d} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
0 & -1 & -2
\end{array}\right|=\hat{i}(-6+4)-\hat{j}(-4)+\hat{k}(-2)=-2 \hat{i}+4 \hat{j}-2 \hat{k} \\
|\vec{c} \times \bar{d}| & =\sqrt{4+16+4}=\sqrt{24} \\
\hat{n} & = \pm \frac{-2 \hat{i}+4 \hat{j}-2 \hat{k}}{\sqrt{24}}= \pm \frac{2(-\hat{i}+2 \hat{j}-\hat{k})}{\sqrt{4 \times 6}} \\
& = \pm \frac{-2(\hat{i}-2 \hat{j}+\hat{k})}{2 \sqrt{6}}= \pm \frac{(\hat{i}-2 \hat{j}+\hat{k})}{\sqrt{6}}
\end{aligned}
$

Question 5.
Find the area of the parallelogram whose two adjacent sides are determined by the vectors $\hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$
Solution:
The area of the parallelogram with $\vec{a}$ and $\vec{b}$ as adjacent sides is $|\vec{a} \times \vec{b}|$
Here $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}$
$
\begin{aligned}
|\vec{a} \times \vec{b}| & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
3 & -2 & 1
\end{array}\right|=\hat{i}(2+6)-\hat{j}(1-9)+\hat{k}(-2-6)=8 \hat{i}+8 \hat{j}-8 \hat{k} \\
\therefore|\vec{a} \times \vec{b}| & =\sqrt{8^2+8^2+8^2}=8 \sqrt{3} \text {. So, area of the parallelogram }=8 \sqrt{3} \text { sq. unit . }
\end{aligned}
$
Question 6.
Find the area of the triangle whose vertices are $\mathrm{A}(3,-1,2), \mathrm{B}(1,-1,-3)$ and $\mathrm{C}(4,-3,1)$
Solution:
$
\begin{aligned}
& \mathrm{A}=(3,-1,2) ; \mathrm{B}=(1,-1,-3) \text { and } \mathrm{C}=(4,-3,1) \\
& \therefore \overrightarrow{\mathrm{OA}}=3 \hat{i}-\hat{j}+2 \hat{k} ; \overrightarrow{\mathrm{OB}}=\hat{i}-\hat{j}-3 \hat{k} \text { and } \overrightarrow{\mathrm{OC}}=4 \hat{i}-3 \hat{j}+\hat{k} \\
& \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\frac{1}{2}|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|=\frac{1}{2}|\overrightarrow{\mathrm{CA}} \times \overrightarrow{\mathrm{CB}}| \\
& \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(\hat{i}-\hat{j}-3 \hat{k})-(3 \hat{i}-\hat{j}+2 \hat{k})=\hat{i}-\hat{j}-3 \hat{k}-3 \hat{i}+\hat{j}-2 \hat{k} \\
& =-2 \hat{i}-5 \hat{k} \\
& \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=4 \hat{i}-3 \hat{j}+\hat{k}-3 \hat{i}+\hat{j}-2 \hat{k} \\
& =\hat{i}-2 \hat{j}-\hat{k} \\
& \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 0 & -5 \\
1 & -2 & -1
\end{array}\right| \\
& =\hat{i}(-10)-\hat{j}(2+5)+\hat{k}(4-0) \\
& =-10 \hat{i}-7 \hat{j}+4 \hat{k} \\
& |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{100+49+16}=\sqrt{165} \\
& \text { Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\frac{\sqrt{165}}{2} \text { sq. units } \\
&
\end{aligned}
$

Question 7.
If $\vec{a}, \vec{b}, \vec{c}$ are position vectors of the vertices $\mathrm{A}, \mathrm{B}, \mathrm{C}$ of a triangle $\mathrm{ABC}$, show that the area of the triangle $\mathrm{ABC}$ is $\frac{1}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|$. Also deduce the condition for collinearity of the points A, B, C

Solution:

Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$
Now
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\vec{b}-\vec{a} \\
\overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=\vec{c}-\vec{a} \\
\text { Hence, area of } \triangle \mathrm{ABC} & =\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\frac{1}{2}|(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})| \\
& =\frac{1}{2}|\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{a} \times \vec{c}+\vec{a} \times \vec{a}| \\
& =\frac{1}{2}|\vec{b} \times \vec{c}+\vec{a} \times \vec{b}+\vec{c} \times \vec{a}| \\
\text { Area of } \triangle \mathrm{ABC} & =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|
\end{aligned}
$
If the points $A, B, C$ are collinear, then the area of $\triangle \mathrm{ABC}=0$.
$
\begin{aligned}
& =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|=0 \\
& =|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|=0 \\
& \text { (or) } \vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=\overrightarrow{0}
\end{aligned}
$
Thus $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=\overrightarrow{0}$ is the required condition of collinearity of the points with positions $\vec{a}, \vec{b}, \vec{c}$.
Question 8.
For any vector $\vec{a}$ prove that $|\vec{a} \times \hat{i}|^2+|\vec{a} \times \hat{\boldsymbol{j}}|^2+|\vec{a} \times \hat{\boldsymbol{k}}|^2=\mathbf{2}|\vec{a}|^2$.
Solution:

$
\begin{aligned}
& \vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \\
& |\vec{a}|=\sqrt{\vec{a}_1^2+\vec{a}_2^2+\vec{a}_3^2} \\
& \vec{a}^2=\vec{a}_1^2+\vec{a}_2^2+\vec{a}_3^2 \\
& \vec{a} \times \hat{i}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
1 & 0 & 0
\end{array}\right|=\left[\vec{a}_3 \hat{j}-\vec{a}_2 \hat{k}\right] \\
& |\vec{a} \times \hat{i}|^2=\left(\vec{a}_3 \hat{j}-\vec{a}_2 \hat{k}\right) \cdot\left(\vec{a}_3 \hat{j}-\vec{a}_2 \hat{k}\right)=\vec{a}_3{ }^2+\vec{a}_2^2 \\
& \vec{a} \times \hat{j}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
0 & 1 & 0
\end{array}\right|=(-1)\left[\vec{a}_3 \hat{i}-\vec{a}_1 \hat{k}\right] \\
& |\vec{a} \times \hat{j}|^2=\vec{a}_3^2+\vec{a}_1^2 \\
& \text { and } \vec{a} \times \hat{k}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
a_1 & a_2 & a_3 \\
0 & 0 & 1
\end{array}\right|=\vec{a}_2 \hat{i}-\vec{a}_1 \hat{j} \\
& |\vec{a} \times \hat{k}|^2=a_2^2+a_1^2 \\
& \therefore \quad \text { LHS }=|\vec{a} \times \hat{i}|^2+|\vec{a} \times \hat{j}|^2+|\vec{a} \times \hat{k}|^2 \\
& =\vec{a}_3^2+\vec{a}_2^2+\vec{a}_3^2+\vec{a}_1^2+\vec{a}_2^2+\vec{a}_1^2 \\
& =2\left(\vec{a}_1^2+\vec{a}_2^2+\vec{a}_3^2\right)=2 \vec{a}^2=\text { RHS. } \\
&
\end{aligned}
$
Question 9.
Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{b}}=\overrightarrow{\boldsymbol{a}} \cdot \overrightarrow{\boldsymbol{c}}=\mathbf{0}$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$. Prove that $\vec{a}= \pm \frac{2}{\sqrt{3}}(\vec{b} \times \vec{c})$
Solution:

Given $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
Also $\vec{a} \cdot \vec{b}=0 \Rightarrow \vec{a} \perp^r$ to $\vec{b}$
$
\vec{a} \cdot \vec{c}=0 \Rightarrow \vec{a} \perp^r \text { to } \vec{c}
$
from (i) and (ii) $\vec{a} \perp^r$ to $\vec{b}$ and $\vec{c}$ (but $\vec{b} \times \vec{c}$ is $\perp^r$ to $\vec{b}$ and $\vec{c}$ )
$
\begin{aligned}
& \Rightarrow \vec{a} \text { is } \| l \text { to } \vec{b} \times \vec{c} \\
& \Rightarrow \quad \vec{a}=t(\vec{b} \times \vec{c}) \\
& \text { But }|\vec{a}|=1 \Rightarrow|t(\vec{b} \times \vec{c})|=1 \\
& \text { (i.e.) } t\left[|\vec{b}||\vec{c}| \sin \left(\frac{\pi}{3}\right)\right]=1 \\
& \Rightarrow t\left(\frac{\sqrt{3}}{2}\right)=1 \Rightarrow t=\frac{2}{\sqrt{3}}
\end{aligned}
$
So, $\quad \vec{a}= \pm \frac{2}{\sqrt{3}}(\vec{b} \times \vec{c})$

Question 10.
Find the angle between the vectors $2 \hat{i}+\hat{j}-\hat{k}$ and $\hat{i}+2 \hat{j}+\hat{k}$ using vector product

Solution:
The angle between $\vec{a}$ and $\vec{b}$ using vector product is given by
$
\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}
$
Here $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$
$
\begin{aligned}
\vec{a} \times \vec{b} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & -1 \\
1 & 2 & 1
\end{array}\right|=\hat{i}(1+2)-\hat{j}(2+1)+\hat{k}(4-1) \\
& =3 \hat{i}-3 \hat{j}+3 \hat{k} \\
|\vec{a} \times \vec{b}| & =\sqrt{9+9+9}=\sqrt{9 \times 3}=3 \sqrt{3} \\
|\vec{a}| & =\sqrt{4+1+1}=\sqrt{6} \\
|\vec{b}| & =\sqrt{1+4+1}=\sqrt{6} \\
\sin \theta & =\frac{3 \sqrt{3}}{\sqrt{6} \sqrt{6}}=\frac{3 \sqrt{3}}{6}=\frac{\sqrt{3}}{2} \\
\therefore \theta & =\pi / 3 .
\end{aligned}
$