Anand Classes provides comprehensive and well-structured NCERT Solutions for Three Dimensional Geometry Exercise 11.2 of Class 12 Chapter 11, designed to help students understand direction ratios, direction cosines, and concepts of 3D geometry with clarity. These notes and solutions are crafted to simplify complex problems, strengthen conceptual understanding, and support CBSE board exam preparation with accuracy and precision. Click the print button to download study material and notes.
Access NCERT Solutions of Chapter -11 Three Dimensional Geometry Exercise 11.2 Class 12 Math
NCERT Question 11 : Show that the lines
$$\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} \quad and \\[1em]
\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$$
are perpendicular to each other.
Solution
The given symmetric equations are:
$$\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$$
and
$$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}.$$
Thus, the direction ratios are
$$\vec b_1=(7,-5,1),$$
$$\vec b_2=(1,2,3).$$
To check perpendicularity, use the condition
$$a_1a_2 + b_1b_2 + c_1c_2 = 0.$$
Compute:
$$7(1) + (-5)(2) + 1(3) = 7 – 10 + 3 = 0.$$
Hence, the lines are perpendicular.
Final Result
$$\boxed{\text{The given lines are perpendicular.}}$$
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NCERT Question.12 : Find the shortest distance between the lines
$$\vec r = \hat i + 2\hat j + \hat k + \lambda(\hat i – \hat j + \hat k)\\[1em]
\vec r = 2\hat i – \hat j – \hat k + \mu(2\hat i + \hat j + 2\hat k)$$
Solution
The shortest distance between two skew lines
$$\vec r = \vec a_1 + \lambda \vec b_1 \quad \text{and} \quad \vec r = \vec a_2 + \mu \vec b_2$$
is
$$d = \frac{\left|(\vec a_2 – \vec a_1)\cdot (\vec b_1 \times \vec b_2)\right|}{|\vec b_1 \times \vec b_2|}$$
Step 1: Identify vectors
$$\vec a_1 = \hat i + 2\hat j + \hat k$$
$$\vec b_1 = \hat i – \hat j + \hat k$$
$$\vec a_2 = 2\hat i – \hat j – \hat k$$
$$\vec b_2 = 2\hat i + \hat j + 2\hat k$$
Step 2: Compute
$$\vec a_2 – \vec a_1 = (2\hat i – \hat j – \hat k) – (\hat i + 2\hat j + \hat k)$$
$$\vec a_2 – \vec a_1= \hat i – 3\hat j – 2\hat k$$
Step 3: Cross product
$$
\vec b_1 \times \vec b_2=
\begin{vmatrix}
\hat i & \hat j & \hat k \\
1 & -1 & 1 \\
2 & 1 & 2
\end{vmatrix}
$$
Computing the determinant:
$$
\vec b_1 \times \vec b_2
= (-2 – 1)\hat i – (2 – 2)\hat j + (1 + 3)\hat k
$$
$$
\vec b_1 \times \vec b_2= -3\hat i + 0\hat j + 4\hat k
$$
Magnitude:
$$
|\vec b_1 \times \vec b_2| = \sqrt{(-3)^2 + 0^2 + 4^2}
= \sqrt{9 + 16}
= \sqrt{25}
= 5
$$
Step 4: Dot product
$$
(\vec a_2 – \vec a_1)\cdot (\vec b_1 \times \vec b_2)
= (\hat i – 3\hat j – 2\hat k)\cdot(-3\hat i + 4\hat k)
$$
$$
(\vec a_2 – \vec a_1)\cdot (\vec b_1 \times \vec b_2)= -3(1) + 4(-2)
$$
$$
(\vec a_2 – \vec a_1)\cdot (\vec b_1 \times \vec b_2)= -3 – 8 = -11
$$
Step 5: Distance
$$d = \frac{\left|(\vec a_2 – \vec a_1)\cdot (\vec b_1 \times \vec b_2)\right|}{|\vec b_1 \times \vec b_2|}$$
$$
d = \frac{| -11 |}{5}
= \frac{11}{5}
$$
Final Answer
$$\boxed{\frac{11}{5}}$$
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NCERT Question.13 : Find the shortest distance between the lines
$$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \quad \text{and} \quad \\[1em] \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$$
Solution
For two lines in symmetric form
$$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$$
and
$$\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$$
the shortest distance is
$$
d=\frac{
\left|
\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix}
\right|
}{
\sqrt{(b_1c_2 – b_2c_1)^2 + (c_1a_2 – c_2a_1)^2 + (a_1b_2 – a_2b_1)^2}
}
$$
Step 1: Identify parameters
From the first line:
$$x_1=-1, y_1=-1, z_1=-1$$
$$a_1=7, b_1=-6, c_1=1$$
From the second line:
$$x_2=3, y_2=5, z_2=7$$
$$a_2=1, b_2=-2, c_2=1$$
Step 2: Evaluate determinant
$$
\begin{vmatrix}
4 & 6 & 8 \\
7 & -6 & 1 \\
1 & -2 & 1
\end{vmatrix}
=4(-6+2)-6(7-1)+8(-14+6)
$$
$$
=-16 – 36 – 64 = -116
$$
Step 3: Denominator
$$
(b_1c_2 – b_2c_1)^2 + (c_1a_2 – c_2a_1)^2 + (a_1b_2 – a_2b_1)^2
$$
Substitute:
$$
(-6\cdot 1 – (-2)\cdot 1)^2 + (1\cdot 1 – 1\cdot 7)^2 + (7(-2) – 1(-6))^2
$$
$$
=(-6+2)^2 + (1-7)^2 + (-14+6)^2
$$
$$
=16 + 36 + 64 = 116
$$
Thus denominator:
$$\sqrt{116} = 2\sqrt{29}$$
Step 4: Distance
$$
d=\frac{
\left|
\begin{vmatrix}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix}
\right|
}{
\sqrt{(b_1c_2 – b_2c_1)^2 + (c_1a_2 – c_2a_1)^2 + (a_1b_2 – a_2b_1)^2}
}
$$
$$
d=\frac{| -116 |}{2\sqrt{29}}=\frac{116}{2\sqrt{29}}=\frac{58}{\sqrt{29}}
$$
Rationalizing:
$$
d=\frac{58}{\sqrt{29}}\cdot \frac{\sqrt{29}}{\sqrt{29}}=\frac{58\sqrt{29}}{29}=2\sqrt{29}
$$
Final Answer
$$\boxed{2\sqrt{29}}$$
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NCERT Question.14 : Find the shortest distance between the lines
$$\vec r=\hat i+2\hat j+3\hat k+\lambda(\hat i-3\hat j+2\hat k) \\[1em]\vec r=4\hat i+5\hat j+6\hat k+\mu(2\hat i+3\hat j+\hat k)$$
Solution :
Position vectors
$$\vec a_1=(1,2,3),\qquad \vec a_2=(4,5,6)$$
Direction vectors
$$\vec b_1=(1,-3,2),\qquad \vec b_2=(2,3,1)$$
Formula (vector method)
For skew lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$ the shortest distance is
$$
d=\frac{\left|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\right|}{\big|\vec b_1\times\vec b_2\big|}
$$
1. Difference of position vectors
$$\vec a_2-\vec a_1=(4-1,5-2,6-3)=(3,3,3)$$
2. Cross product
$$
\vec b_1\times\vec b_2=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
1 & -3 & 2\\
2 & 3 & 1
\end{vmatrix}
= -9\hat i +3\hat j +9\hat k
$$
3. Magnitude of cross product
$$
\big|\vec b_1\times\vec b_2\big|=\sqrt{(-9)^2+3^2+9^2}
=\sqrt{81+9+81}
=\sqrt{171}
=3\sqrt{19}
$$
4. Dot product
$$
(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)
=(3\hat i+3\hat j+3\hat k)\cdot(-9\hat i+3\hat j+9\hat k)$$
$$(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2) =3(-9+3+9)=9
$$
Distance
Substitute into the formula:
$$
d=\frac{|9|}{3\sqrt{19}}=\frac{3}{\sqrt{19}}
$$
Final Result
$$\boxed{\frac{3}{\sqrt{19}}}$$
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NCERT Question.15 : Find the shortest distance between the lines
$$\vec r=(1-t)\hat i+(t-2)\hat j+(3-2t)\hat k \quad and \\[1em] \vec r=(s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k$$
Solution :
Step 1 — Write both lines in standard vector form
Rewrite the first line as $\vec r=\vec a_1+t\vec b_1$ (note the parameter sign change):
$$\vec r=(\hat i-2\hat j+3\hat k)+t(-\hat i+\hat j-2\hat k)$$
so
$$\vec a_1=(1,-2,3),\qquad \vec b_1=(-1,1,-2).$$
Rewrite the second line as $\vec r=\vec a_2+s\vec b_2$:
$$\vec r=(\hat i-\hat j-\hat k)+s(\hat i+2\hat j-2\hat k)$$
so
$$\vec a_2=(1,-1,-1),\qquad \vec b_2=(1,2,-2).$$
Step 2 — Vector formula for shortest distance
For skew lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$ the shortest distance is
$$
d=\frac{\big|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\big|}{\lVert \vec b_1\times\vec b_2\rVert}
$$
Step 3 — Compute the pieces
Difference of position vectors:
$$\vec a_2-\vec a_1=(1-1, -1-(-2), -1-3)=(0,1,-4).$$
Cross product:
$$
\vec b_1\times\vec b_2=
\begin{vmatrix}
\hat i & \hat j & \hat k\\
-1 & 1 & -2\\
1 & 2 & -2
\end{vmatrix}
=(2,-4,-3).
$$
Magnitude of the cross product:
$$
\lVert \vec b_1\times\vec b_2\rVert=\sqrt{2^2+(-4)^2+(-3)^2}=\sqrt{4+16+9}=\sqrt{29}.
$$
Dot product:
$$
(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=(0,1,-4)\cdot(2,-4,-3)=0-4+12=8.
$$
Step 4 — Distance
Substitute into the formula:
$$
d=\frac{|8|}{\sqrt{29}}=\frac{8}{\sqrt{29}}.
$$
Final Result
$$\boxed{\frac{8}{\sqrt{29}}}$$
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NCERT Question 16 : Find the vector and Cartesian equation of the line passing through the origin and the point (5, -2, 3)
Solution
The line passes through the origin, so its position vector is
$$\vec{a} = \vec{0}$$
The direction ratios of the line through the origin $(0, 0, 0)$ and the point $(5, -2, 3)$ are
$$(5-0, -2-0, 3-0) = (5, -2, 3)$$
Hence, the line is parallel to the vector
$$\vec{b} = 5\hat{i} – 2\hat{j} + 3\hat{k}$$
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is
$$\vec{r} = \vec{a} + \lambda \vec{b}, ;\lambda \in \mathbb{R}$$
Substituting the values, we get
$$\vec{r} = \vec{0} + \lambda (5\hat{i} – 2\hat{j} + 3\hat{k})$$
$$\vec{r} = \lambda (5\hat{i} – 2\hat{j} + 3\hat{k})$$
The Cartesian form of a line passing through point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is
$$\frac{x – x_1}{a} = \frac{y – y_1}{b} = \frac{z – z_1}{c}$$
For this line, we have $(x_1, y_1, z_1) = (0, 0, 0)$ and $(a, b, c) = (5, -2, 3)$, so the Cartesian equation is
$$\frac{x – 0}{5} = \frac{y – 0}{-2} = \frac{z – 0}{3}$$
$$\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}$$
Final Result
Vector form:
$$\boxed{\vec{r} = \lambda (5\hat{i} – 2\hat{j} + 3\hat{k}), ;\lambda \in \mathbb{R}}$$
Cartesian form:
$$\boxed{\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}}$$
Perfect study solution for NCERT 3D geometry exercises, ideal for JEE and CBSE preparation.
