Exercise 6.4-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.
Find the vector and cartesian equations of the straight line passing through the point A with position vector $3 \vec{i}-\vec{j}+4 \vec{k}$ and parallel to the vector $-5 \vec{i}+7 \vec{j}+3 \vec{k}$.
Solution:
We know that vector equation of the line through the point with position vector $\vec{a}$ and parallel to $\vec{v}$ is given by $\vec{r}=\vec{a}+t \vec{v}$ where $t$ is a scalar.
Here $\vec{a}=3 \vec{i}-\vec{j}+4 \vec{k}$ and $\vec{v}=-5 \vec{i}+7 \vec{j}+3 \vec{k}$
Vector equation of the line is
$
\vec{r}=(3 \vec{i}-\vec{j}+4 \vec{k})+t(-5 \vec{i}+7 \vec{j}+3 \vec{k})
$
The cartesian equation of the line passing through $(\mathrm{xp} \mathrm{yx}, \mathrm{zx})$ and parallel to a vector whose $\mathrm{d} . \mathrm{r} . \mathrm{s}$ are $1, \mathrm{~m}, \mathrm{n}$ is
$
\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}
$
Here $\quad\left(x_1, y_1, z_1\right)=(3,-1,4)$ and $(l, m, n)=(-5,7,3)$
$\therefore$ The required equation is $\frac{x-3}{-5}=\frac{y+1}{7}=\frac{z-4}{3}$

Question 2.
Find the vector and cartesian equations of the straight line passing through the points $(-5,2,3)$ and $(4,-3$, 6).
Solution:
Vector equation of the straight line passing through two points with position vectors $\vec{a}$ and $\vec{b}$ is given by
$
\vec{r}=\vec{a}+t(\vec{b}-\vec{a})
$
Here
$
\begin{aligned}
\vec{a} & =-5 \vec{i}+2 \vec{j}+3 \vec{k} \text { and } \vec{b}=4 \vec{i}-3 \vec{j}+6 \vec{k} \\
\vec{b}-\vec{a} & =9 \vec{i}-5 \vec{j}+3 \vec{k}
\end{aligned}
$
$\therefore$ Vector equation of the line is
(or) $\quad \begin{aligned} \vec{r} & =-5 \vec{i}+2 \vec{j}+3 \vec{k}+t(9 \vec{i}-5 \vec{j}+3 \vec{k}) \\ \vec{r} & =(1-t)-5 \vec{i}+2 \vec{j}+3 \vec{k}+t(4 \vec{i}-3 \vec{j}+6 \vec{k})\end{aligned}$
Cartesian Form:
The required equation is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
Here $\quad\left(x_1, y_1, z_1\right)=(-5,2,3) ;\left(x_2, y_2, z_2\right)=(4,-3,6)$
$\therefore \quad \frac{x+5}{9}=\frac{y-2}{-5}=\frac{z-3}{3}$ is the cartesian equation of the line.

Question 3.
Find the angle between the lines.
$
\vec{r}=3 \vec{i}+2 \vec{j}-\vec{k}+t(\vec{i}+2 \vec{j}+2 \vec{k}) \text { and } \vec{r}=5 \vec{j}+2 \vec{k}+s(3 \vec{i}+2 \vec{j}+6 \vec{k})
$
Solution:

Let the given lines be in the direction of $\vec{u}$ and $\vec{v}$
Then $\vec{u}=\vec{i}+2 \vec{j}+2 \vec{k}, \vec{v}=3 \vec{i}+2 \vec{j}+6 \vec{k}$
Let $\theta$ be the angle between the given lines
$
\begin{aligned}
\therefore \quad \cos \theta & =\left(\frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}\right) \\
\vec{u} \cdot \vec{v} & =19 ;|\vec{u}|=3 ;|\vec{v}|=7 \\
\therefore \quad \cos \theta & =\frac{19}{21} \quad \Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)
\end{aligned}
$

Question 4.
Find the angle between the following lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-4}{6}$ and $x+1=\frac{y+2}{2}=\frac{z-4}{2}$

Solution:
Angle between two lines is the same as angle between their parallel vectors.
The parallel vectors are: $a=2 \vec{i}+3 \vec{j}+6 \vec{k}$ and $b=\vec{i}+2 \vec{j}+2 \vec{k}$.
If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$.
$
\begin{aligned}
\vec{a} \cdot \vec{b} & =(2)(1)+(3)(2)+(6)(2)=2+6+12=20 \\
|\vec{a}| & =\sqrt{4+9+36}=\sqrt{49}=7 \\
|\vec{b}| & =\sqrt{1+4+4}=\sqrt{9}=3 \\
\text { So, } \cos \theta & =\frac{20}{7 \times 3}=\frac{20}{21} \quad \Rightarrow \theta=\cos ^{-1} \frac{20}{21}
\end{aligned}
$