Anand Classes provides comprehensive and exam-oriented NCERT Solutions for Three Dimensional Geometry Miscellaneous Exercise, Chapter-11, Class 12 Maths, helping students strengthen spatial understanding and improve problem-solving accuracy. These notes cover all important concepts, formulas, solved examples, and step-by-step solutions designed to support CBSE board preparation. Students can rely on these well-structured notes for quick revision and deeper conceptual clarity. Click the print button to download study material and notes.
NCERT Question.1 : Find the angle between the lines whose direction ratios are
$(a,b,c)$ and $(b-c,c-a,a-b)$
Solution
The angle between two lines whose direction ratios are
$(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ is
$$
\cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{,2}+b_1^{,2}+c_1^{,2}}\sqrt{a_2^{,2}+b_2^{,2}+c_2^{,2}}}
$$
Given
$$a_1=a,\quad b_1=b,\quad c_1=c$$
$$a_2=b-c,\quad b_2=c-a,\quad c_2=a-b$$
Substitute in the formula:
$$
a(b-c)+b(c-a)+c(a-b)
$$
Expand:
$$
ab-ac+bc-ab+ca-bc=0
$$
Thus,
$$
\cos\theta=\frac{0}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}=0
$$
Therefore,
$$
\theta=90^\circ
$$
Final Result
$$\boxed{90^\circ}$$
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NCERT Question.2 : Find the equation of a line parallel to the $x$–axis and passing through the origin
Solution
A line passing through a point $(x_1, y_1 , z_1)$ and having direction ratios $(a , b , c)$ is written as
$$
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}
$$
The given line passes through the origin:
$$
x_1=0 ; y_1=0 ; z_1=0
$$
Since the line is parallel to the x–axis, its direction ratios are:
$$
a=1 ; b=0 ; c=0
$$
Thus, the equation of the required line becomes:
$$
\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}
$$
Which simplifies to:
$$
x=\lambda \quad y=0 \quad z=0
$$
Final Result
$$
\boxed{x=\lambda ; y=0 ; z=0}
$$
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NCERT Question.3 : If the lines
$$\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\quad and \\[1em] \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$$
are perpendicular, find the value of $k$.
Solution
Direction ratios of the first line are
$$a_1=-3 \quad b_1=2k \quad c_1=2$$
Direction ratios of the second line are
$$a_2=3k \quad b_2=1 \quad c_2=-5$$
For two lines to be perpendicular,
$$
a_1a_2+b_1b_2+c_1c_2=0
$$
Substituting the values:
$$
(-3)(3k)+(2k)(1)+(2)(-5)=0
$$
$$
-9k+2k-10=0
$$
$$
-7k-10=0
$$
$$
-7k=10
$$
$$
k=-\frac{10}{7}
$$
Final Result
$$\boxed{k=-\frac{10}{7}}$$
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NCERT Question.4 : Find the shortest distance between the lines
$$\vec r=6\hat i+2\hat j+2\hat k+\lambda(\hat i-2\hat j+2\hat k) \quad and \\[1em]\vec r=-4\hat i+0\hat j-\hat k+\mu(3\hat i-2\hat j-2\hat k)$$
Solution
Write both lines in the form $\vec r=\vec a+\lambda\vec b$ and identify vectors
$$\vec a_1=6\hat i+2\hat j+2\hat k,\qquad \vec b_1=\hat i-2\hat j+2\hat k$$
$$\vec a_2=-4\hat i+0\hat j-\hat k,\qquad \vec b_2=3\hat i-2\hat j-2\hat k$$
Compute the difference of position vectors
$$\vec a_2-\vec a_1=(-4-6)\hat i+(0-2)\hat j+(-1-2)\hat k=-10\hat i-2\hat j-3\hat k$$
Compute the cross product
$$
\vec b_1\times\vec b_2=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
1 & -2 & 2\\
3 & -2 & -2
\end{vmatrix}
$$
$$
\vec b_1\times\vec b_2=\hat i((-2)(-2)-2(-2))-\hat j(1(-2)-2\cdot 3)+\hat k(1(-2)-(-2)\cdot 3)
$$
$$
\vec b_1\times\vec b_2=\hat i(4+4)-\hat j(-2-6)+\hat k(-2+6)=8\hat i+8\hat j+4\hat k
$$
Magnitude of the cross product
$$\big\lVert\vec b_1\times\vec b_2\big\rVert=\sqrt{8^2+8^2+4^2}=\sqrt{64+64+16}=\sqrt{144}=12$$
Dot product of the cross with $\vec a_2-\vec a_1$
$$
(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=(8\hat i+8\hat j+4\hat k)\cdot(-10\hat i-2\hat j-3\hat k)
$$
$$
(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=8(-10)+8(-2)+4(-3)=-80-16-12=-108
$$
Shortest distance formula for skew lines
$$
d=\left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)}{\big\lVert\vec b_1\times\vec b_2\big\rVert}\right|
$$
Substitute values
$$
d=\left|\frac{-108}{12}\right|=9
$$
Final Result
$$\boxed{9\ \text{units}}$$
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NCERT Question.5 : Find the vector equation of the line passing through $(1,2,-4)$ and perpendicular to the two lines
$$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} \qquad\text{and}\\[1em] \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$$
Solution
Step 1: Direction vectors of the two given lines
From the symmetric forms:
$$\vec b_1 = 3\hat i – 16\hat j + 7\hat k$$
$$\vec b_2 = 3\hat i + 8\hat j – 5\hat k$$
Step 2: Find the direction vector of the required line
The required line is perpendicular to both lines, so its direction vector is
$$\vec b = \vec b_1 \times \vec b_2$$
Compute the cross product:
$$
\vec b =
\begin{vmatrix}
\hat i & \hat j & \hat k \\
3 & -16 & 7 \\
3 & 8 & -5
\end{vmatrix}
$$
$$
\vec b= \hat i((-16)(-5) – 7\cdot8)-\hat j(3(-5) – 7\cdot3)+\hat k(3\cdot8 – (-16)(3))
$$
$$
\vec b= \hat i(80 – 56) -\hat j(-15 – 21)+\hat k(24 + 48)
$$
$$
\vec b= 24\hat i + 36\hat j + 72\hat k
$$
Factor out (12):
$$\vec b’ = 2\hat i + 3\hat j + 6\hat k$$
Step 3: Position vector of the given point
Point $(1,2,-4)$:
$$\vec a = \hat i + 2\hat j – 4\hat k$$
Step 4: Vector equation of the required line
$$
\vec r = \vec a + \lambda \vec b’
$$
$$
\vec r = (\hat i + 2\hat j – 4\hat k) + \lambda(2\hat i + 3\hat j + 6\hat k)
$$
Final Result
$$
\boxed{\vec r =(\hat i + 2\hat j – 4\hat k) + \lambda(2\hat i + 3\hat j + 6\hat k)}
$$
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