Miscellaneous Exercise (Revised) - Chapter 11 - Three Dimensional Geometry - Ncert Solutions class 12 - Maths

You can Download the Miscellaneous Exercise (Revised) - Chapter 11 - Three Dimensional Geometry - Ncert Solutions class 12 - Maths with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Ncert Solutions Class 12 Maths: Chapter 11 - Three Dimensional Geometry

Miscellaneous Exercise Question 1.

Find the angle between the lines whose direction ratios are $a, b, c$ and $b-c, c-a, a-b$.

Answer.

Direction ratios of one line are $a, b, c$
$\Rightarrow$ A vector along this line is $\overrightarrow{b_1}=a \hat{i}+b \hat{j}+c \hat{k}$
Direction ratios of second line are $b-c, c-a, a-b$
$\Rightarrow$ A vector along second line is $\overrightarrow{b_2}=(b-c) \hat{i}+(c-a) \hat{j}+(a-b) \hat{k}$
Let $\theta$ be the angle between the two lines, then

$\begin{aligned}
& \cos \theta=\frac{\left|\overrightarrow{b_1} \cdot \overrightarrow{b_2}\right|}{\left|\overrightarrow{b_1}\right| \cdot|\cdot| \overrightarrow{b_2} \mid}=\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2} \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \\
& =\frac{a b-a c+b c-a b+a c-b c}{\sqrt{a^2+b^2+c^2} \sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}=0=\cos 90^{\circ} \\
& \Rightarrow \theta=90^{\circ}
\end{aligned}$

Miscellaneous Exercise Question 2.

Find the equation of the line parallel to $x$-axis and passing through the origin.

Answer.

We know that a unit vector along $x$ - axis is $\hat{i}=\hat{i}+0 \hat{j}+0 \hat{k}$
$\therefore$ Direction cosines of $x$-axis are coefficients of $\hat{i}, \hat{j}, \hat{k}$ in the unit vector
i.e., $1,0,0=l, m, n$
$\therefore$ Equation of the required line passing through the origin $(0,0,0)$ and parallel to $x-$ axis is $\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0} \Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

Vector equation of the required line is $\vec{r}=\vec{a}+\lambda \vec{b}$
$
\begin{aligned}
& \Rightarrow \vec{r}=\overrightarrow{0}+\lambda \hat{i}[\vec{a}=\overrightarrow{0} \text { and } \vec{b}=\hat{i}] \\
& \Rightarrow \vec{r}=\lambda \hat{i} \\
&
\end{aligned}
$

Miscellaneous Exercise Question 3.

If the lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular, find the value of $k$.

Answer.

Given: Equation of one line is $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$
Direction ratios of this line are its denominators, i.e., $-3,2 k, 2=a_1, b_1, c_1$
$\therefore$ A vector along this line is $\overrightarrow{b_1}=-3 \hat{i}+2 k \hat{j}+2 \hat{k}$
Again, equation of second line is $\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}$
Direction ratios of this line are its denominators, i.e., $3 k, 1,-5=a_2, b_2, c_2$

$\therefore$ A vector along this line is $\overrightarrow{b_2}=3 k \hat{i}+\hat{j}-5 \hat{k}$
Since these given lines are perpendicular.
$
\begin{aligned}
& \therefore \overrightarrow{b_1} \cdot \overrightarrow{b_2}=a_1 a_2+b_1 b_2+c_1 c_2=0 \\
& \Rightarrow(-3)(3 k)+(2 k)(1)+2(-5)=0 \Rightarrow-9 k+2 k-10=0 \\
& \Rightarrow-7 k=10 \Rightarrow k=\frac{-10}{7}
\end{aligned}
$

Miscellaneous Exercise Question 4.

Find the shortest distance between lines $\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and
$
\bar{r}=4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})
$

Answer.

Given: Vector equation of one line is $\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$
Comparing with $\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$, we get
$
\overrightarrow{a_1}=6 \hat{i}+2 \hat{j}+2 \hat{k} \text { and } \overrightarrow{b_1}=\hat{i}-2 \hat{j}+2 \hat{k}
$

Again given: Vector equation of another line is $\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})$
Comparing with $\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$, we get
$
\overrightarrow{a_2}=-4 \hat{i}-\hat{k} \text { and } \overrightarrow{b_2}=3 \hat{i}-2 \hat{j}-2 \hat{k}
$

We know that length of shortest distance between two (skew) lines is $\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\right|}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}$..
(i)
$
\begin{aligned}
& \text { Now } \overrightarrow{a_2}-\overrightarrow{a_1}=-4 \hat{i}-\hat{k}-(6 \hat{i}+2 \hat{j}+2 \hat{k})=-4 \hat{i}-\hat{k}-6 \hat{i}-2 \hat{j}-2 \hat{k}=-10 \hat{i}-2 \hat{j}-3 \hat{k} \\
& \text { Again } \overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right|
\end{aligned}
$

Expanding along first row,
$
\begin{aligned}
& =\hat{i}(4+4)-\hat{j}(-2-6)+\hat{k}(-2+6)=8 \hat{i}+8 \hat{j}+4 \hat{k} \\
& \therefore\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=(-10)(8)+(-2)(8)+(-3) 4 \\
& =-80-16-12=-108
\end{aligned}
$

And $\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{(8)^2+(8)^2+(4)^2}=\sqrt{64+64+16}=\sqrt{144}=12$
Putting these values in eq. (i), length of shortest distance $=\frac{|-108|}{12}=\frac{108}{12}=9$

Miscellaneous Exercise Question 5.

Find the vector equation of the line passing through the point $(1,2,-4)$ and perpendicular to the two lines: $\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$.

Answer.

Given: A point on the required line is $\mathrm{A}(1,2,-4)$
$\therefore$ Position vector of point $\mathrm{A}$ is $\vec{a}=(1,2,-4)=\hat{i}+2 \hat{j}-4 \hat{k}$
Also given equations of two lines
$
\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} \text { and } \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}
$
$\therefore$ Direction ratios of given two lines are $\overrightarrow{b_1}=3 \hat{i}-16 \hat{j}+7 \hat{k}$ and $\overrightarrow{b_2}=3 \hat{i}+8 \hat{j}-5 \hat{k}$
$
\text { Now } \vec{b}=\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right|
$

Expanding along first row
$
\begin{aligned}
& =\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48)=24 \hat{i}+36 \hat{j}+72 \hat{k} \\
& \Rightarrow \vec{b}=12(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{aligned}
$
$\therefore$ Equation of the required line is $\vec{r}=\vec{a}+\lambda \vec{b}$

$
\Rightarrow \bar{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(12)(2 \hat{i}+3 \hat{j}+6 \hat{k})
$

Again replacing $12 \lambda$ by $\lambda$,
$
\Rightarrow \vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})
$