Exercise 6.4 - 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

Ex $6.4$
Question 1.
Find the non-parametric form of vector equation and Cartesian equations of the straight line passing through the point with position vector $4 \hat{i}+3 \hat{j}-7 \hat{k}$ and parallel to the vector $2 \hat{i}-6 \hat{j}+7 \hat{k}$

Solution:
Given
$
\vec{a}=4 \vec{i}+3 \vec{j}-7 \vec{k}, \vec{b}=2 \vec{i}-6 \vec{j}+7 \vec{k}
$
$\therefore$ Non- Parametric vector equation of the straight line is $(\vec{r}-\vec{a}) \times \vec{b}=\overrightarrow{0}$
$
(\vec{r}-(4 \vec{i}+3 \vec{j}-7 \vec{k})) \times(2 \vec{i}-6 \vec{j}+7 \vec{k})=\overrightarrow{0}
$
Cartesian equation:
$
\begin{aligned}
\frac{x-x_1}{b_1}=\frac{y-y_1}{b_2}=\frac{z-z_1}{b_3} & \left(\begin{array}{rrr}
x_1 & y_1 & z_1 \\
4 & 3 & -7
\end{array}\right) \\
\frac{x-4}{2}=\frac{y-3}{-6}=\frac{z+7}{7} & \left(\begin{array}{rrr}
b_1 & b_2 & b_3 \\
2 & -6 & 7
\end{array}\right)
\end{aligned}
$
Question 2.
Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point $(-2,3,4)$ and parallel to the straight line $\frac{x-1}{-4}=\frac{y+3}{5}=\frac{8-z}{6}$
Solution:
Given $\quad \vec{a}=-2 \vec{i}+3 \vec{j}+4 \vec{k}, \vec{b}=-4 \vec{i}+5 \vec{j}+6 \vec{k}$
$\therefore$ Parametric vector equation of the straight line is
$
\begin{aligned}
\vec{r} & =\vec{a}+t \vec{b} \\
\vec{r} & =(-2 \vec{i}+3 \vec{j}+4 \vec{k})+t(-4 \vec{i}+5 \vec{j}+6 \vec{k}), t \in \mathbb{R} .
\end{aligned}
$
Cartesian equation:
$
\begin{aligned}
& \frac{x-x_1}{b_1}=\frac{y-y_1}{b_2}=\frac{z-z_1}{b_3} \\
& \frac{x+2}{-4}=\frac{y-3}{5}=\frac{z-4}{6}\left(\begin{array}{rrr}
x_1 & y_1 & z_1 \\
-2 & 3 & 4
\end{array}\right)
\end{aligned}
$

Question 3.
Find the points where the straight line passes through $(6,7,4)$ and $(8,4,9)$ cuts the $x z$ and $y z$ planes.
Solution:
Given straight line passing through the points $(6,7,4)$ and $(8,4,9)$.
Direction ratio of the straight line joining these two points $2,-3,-5$.
Cartesian equation:
$
\begin{aligned}
& \frac{x-6}{2}=\frac{y-7}{-3}=\frac{z-4}{5} \quad \Rightarrow \frac{x-6}{2}=\frac{y-7}{-3}=\frac{z-4}{5}=t \text { (say) } \\
& (2 t+6,-3 t+7,5 t+4)
\end{aligned}
$
$\therefore$ (i) The straight line cuts the $x z$-plane.
So we get
$
\begin{aligned}
y & =0 \\
-3 t+7 & =0 \\
-3 t & =-7 \quad \Rightarrow t=\frac{7}{3} \\
5 t+4 & =5\left(\frac{7}{3}\right)+4=\frac{35+12}{3}=\frac{47}{3} \\
2 t+6 & =2\left(\frac{7}{3}\right)+6=\frac{14+18}{3}=\frac{32}{3}
\end{aligned}
$
The required point $\left(\frac{32}{3}, 0, \frac{47}{3}\right)$
(ii) The straight Line cuts yz-plane
So we get $x=0$
$
\begin{aligned}
& 2 t+6=0 \Rightarrow 2 t=-6 \\
& t=-3 \\
& -3 t+7=-3(-3)+7=9+7=16 \\
& 5 t+4=5(-3)+4=-15+4=-11
\end{aligned}
$
The required point $(0,16,-11)$.

Question 4.
Find the direction cosines of the straight line passing through the points $(5,6,7)$ and $(7,9,13)$. Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points.
Solution:
Given straight line passing through the points $(5,6,7)$ and $(7,9,13)$
$
\therefore \text { d.r.s: } 2,3,6
$
$
\begin{aligned}
& \text { d.c.s }=\left(\frac{2}{\sqrt{2^2+3^2+6^2}}, \frac{3}{\sqrt{2^2+3^2+6^2}}, \frac{6}{\sqrt{2^2+3^2+6^2}}\right) \\
& \text { d.c.s }=\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right)
\end{aligned}
$
$\therefore$ Parametric form of vector equation passing through give two points.
Given
$
\begin{aligned}
\vec{r} & =\vec{a}+t(\vec{b}-\vec{a}) \\
\vec{a} & =5 \vec{i}+6 \vec{j}+7 \vec{k}, \vec{b}=7 \vec{i}+9 \vec{j}+13 \vec{k} \\
\vec{r} & =(5 \vec{i}+6 \vec{j}+7 \vec{k})+t(2 \vec{i}+3 \vec{j}+6 \vec{k})
\end{aligned}
$
Cartesian equation:
$
\begin{aligned}
\frac{x-x_1}{x_2-x_1} & =\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \\
\frac{x-5}{2} & =\frac{y-6}{3}=\frac{z-7}{6}
\end{aligned}
$
Note: Selection of $\vec{a}$ and $\vec{b}$ is your choice.

Question 5.
Find the acute angle between the following lines.
(i) $\vec{r}=(4 \hat{i}-\hat{j})+t(\hat{i}+2 \hat{j}-2 \hat{k}), \vec{r}=(\hat{i}-2 \hat{j}+4 \hat{k})+s(-\hat{i}-2 \hat{j}+2 \hat{k})$
(ii) $\frac{x+4}{3}=\frac{y-7}{4}=\frac{z+5}{5}, \vec{r}=4 \hat{k}+t(2 \hat{i}+\hat{j}+\hat{k})$.
(iii) $2 x=3 y=-z$ and $6 x=-y=-4 z$
Solution:

(i)
$
\begin{aligned}
& \text { Given } \vec{b}=\vec{i}+2 \vec{j}-2 \vec{k}, \vec{d}=-\vec{i}-2 \vec{j}+2 \vec{k} \\
& \cos \theta=\frac{|\vec{b} \cdot \vec{d}|}{|\vec{b}||\vec{d}|}=\frac{|(-1-4-4)|}{\sqrt{1+4+4} \sqrt{1+4+4}}=\frac{9}{3 \times 3} \\
& \cos \theta=1 \\
& \theta=\cos ^{-1}(1) \quad \Rightarrow \theta=0^{\circ} \\
&
\end{aligned}
$

(ii)
$
\text { Given } \begin{aligned}
\vec{b} & =3 \vec{i}+4 \vec{j}+5 \vec{k}, \vec{d}=2 \vec{i}+\vec{j}+\vec{k} \\
\cos \theta & =\frac{|\vec{b} \cdot \vec{d}|}{|\vec{b}||\vec{d}|} \\
\vec{b} \cdot \vec{d} & =6+4+5=15 \\
|\vec{b}| & =\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2} \\
|\vec{d}| & =\sqrt{4+1+1}=\sqrt{6}
\end{aligned}
$
$(1) \Rightarrow \quad \cos \theta=\frac{15}{(5 \sqrt{2})(\sqrt{6})}=\frac{3}{2 \sqrt{3}} \Rightarrow \theta=\cos ^{-1}\left(\frac{3}{2 \sqrt{3}}\right)$

(iii)

$\begin{aligned}
& \vec{b} \cdot \vec{d}=\frac{1}{12}-\frac{1}{3}+\frac{1}{4}=\frac{1-4+3}{12}=0 \\
& \cos \theta=\frac{|\vec{b} \cdot \vec{d}|}{|\vec{b}||\vec{d}|}=\frac{0}{\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+(-1)^2} \sqrt{\left(\frac{1}{6}\right)^2+(-1)^2+\left(\frac{-1}{4}\right)^2}} \\
& \cos \theta=0 \quad \Rightarrow \theta=\cos ^{-1}(0) \\
& \theta=\frac{\pi}{2} \\
&
\end{aligned}$

Question 6.
The vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(7,2,1), \mathrm{B}(6,0,3)$, and $\mathrm{C}(4,2,4)$. Find $\angle \mathrm{ABC}$.
Solution:
Let
$
\begin{aligned}
\overrightarrow{\mathrm{OA}} & =7 \vec{i}+2 \vec{j}+\vec{k}, \overrightarrow{\mathrm{OB}}=6 \vec{i}+0 \vec{j}+3 \vec{k}, \overrightarrow{\mathrm{OC}}=4 \vec{i}+2 \vec{j}+4 \vec{k} \\
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=-\vec{i}-2 \vec{j}+2 \vec{k} \\
\overrightarrow{\mathrm{BC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=-2 \vec{i}+2 \vec{j}+\vec{k} \\
\cos \theta & =\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{AB}}| \cdot|\overrightarrow{\mathrm{BC}}|} \quad \Rightarrow \cos \theta=\frac{2-4+2}{\sqrt{1+4+4} \sqrt{4+4+1}} \\
\cos \theta & =0 \quad \Rightarrow \theta=\frac{\pi}{2} \\
\theta & =\cos ^{-1}(0) \quad \Rightarrow
\end{aligned}
$

Question 7.
If the straight line joining the points $(2,1,4)$ and $(a-1,4,-1)$ is parallel to the line joining the points $(0,2$, $b-1)$ and $(5,3,-2)$, find the values of $a$ and $b$.
Solution:
Let $\overrightarrow{\mathrm{OA}}=2 \hat{i}+\hat{j}+4 \vec{k}, \overrightarrow{\mathrm{OB}}=(a-1) \vec{i}+4 \vec{j}-\vec{k}$
So,
$
\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(a-3) \vec{i}+3 \vec{j}-5 \vec{k}
$
Let $\overrightarrow{\mathrm{OC}}=2 \vec{i}+(b-1) \vec{k}$
$
\begin{aligned}
\overrightarrow{\mathrm{OD}} & =5 \vec{i}+3 \vec{j}-2 \vec{k} \\
\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OC}} & =5 i+j+(-b-1) \vec{k}
\end{aligned}
$
Given $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ are parallel.
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =m \overrightarrow{\mathrm{CD}} \\
(a-3) \vec{i}+3 \vec{j}-5 \vec{k} & =m(5 \vec{i}+\vec{j}+(-b-1) \vec{k})
\end{aligned}
$
Equating $\vec{i}, \vec{j}, \vec{k}$ coefficient on both side
$
\begin{aligned}
& \begin{array}{l|l|l}
a-3 & =5 m \\
a-3 & =5(3)
\end{array} \mid 3=m \quad \begin{aligned}
m(-b-1) & =-5 \\
3(-b-1) & =-5
\end{aligned} \quad \Rightarrow-b-1=\frac{-5}{3} \\
& a-3=15 \\
& -b=\frac{-5}{3}+1=\frac{-5+3}{3}=\frac{-2}{3} \Rightarrow b=\frac{2}{3} \\
& \therefore a=18 ; b=\frac{2}{3} \\
&
\end{aligned}
$

Question 8.
If the straight lines $\frac{x-5}{5 m+2}=\frac{2-y}{5}=\frac{1-z}{-1}$ and $x=\frac{2 y+1}{4 m}=\frac{1-z}{-3}$ are perpendicular to each other, find the value of $m$.
Solution:
$
\frac{x-5}{5 m+2}=\frac{2-y}{5}=\frac{1-z}{-1} \text { and } x=\frac{2 y+1}{4 m}=\frac{1-z}{-3} \text { are perpendicular. }
$
Rewrite the above equations.
$
\frac{x-5}{5 m+2}=\frac{y-2}{-5}=\frac{z-1}{1} \text { and } x=\frac{y+\frac{1}{2}}{2 m}=\frac{z-1}{3}
$
So, we get,
$
\begin{aligned}
\vec{d} & =\vec{i}+2 m \vec{j}+3 \vec{k} & & \\
\vec{b} \cdot \vec{d} & =0 \text { (Given) } & & \\
5 m+2-10 m+3 & =0 & & \Rightarrow-5 m+5=0 \\
-5 m & =-5 & & \Rightarrow m=1
\end{aligned}
$
$
\vec{b}=(5 m+2) \vec{i}-5 \vec{j}+\vec{k}
$

$\begin{aligned}
\vec{d} & =\vec{i}+2 m \vec{j}+3 \vec{k} & & \\
\vec{b} \cdot \vec{d} & =0 \text { (Given) } & & \\
5 m+2-10 m+3 & =0 & & \Rightarrow-5 m+5=0 \\
-5 m & =-5 & & \Rightarrow m=1
\end{aligned}$

Question 9.
Show that the points $(2,3,4),(-1,4,5)$ and $(8,1,2)$ are collinear.
Solution:
Given points are (2, 3,4), $(-1,4,5)$ and $(8,1,2)$ Equation of the line joining of the first and second point is $\frac{x-2}{-3}=\frac{y-3}{1}=\frac{z-4}{1}=\mathrm{m}$ (say)
$(-3 m+2, m+3, m+4)$
On putting $m=-2$, we get the third point is $(8,1,2)$
$\therefore$ Given points are collinear.