Class 12 NCERT Solutions – Mathematics Part ii – Chapter 11 – Three Dimensional Geometry – Miscellaneous Exercise

1. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Solution:

The angle θ between the lines with direction cosines a, b, c and b – c, c – a, a – b is given by:

[Tex]cosθ=|\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}}|\\ θ=cos^{-1}|\frac{ab-ac+bc-ab+ac-bc}{\sqrt{a^2+b^2+c^2}.\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}}|\\ =cos^{-1}0\\ [/Tex]

= 90°

Thus, the required angle is 90°.

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

Solution:

The line parallel to x-axis and passing through the origin is x- axis itself.

Let A be any point on the given line.

Thus, coordinates of A are (a,0,0) where a is any real value.

So the direction ratios of OA will be a, 0, 0.

Equation of OA will be:

[Tex]\frac{x-a}0=\frac{y-0}0=\frac{z-0}0\\ ⇒\frac{x}1=\frac{y}0=\frac{z}0=a[/Tex]

Hence, the required equation is [Tex]\frac{x}1=\frac{y}0=\frac{z}0[/Tex].

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

Solution:

Given: a₁ = 3, b₁ = 2k, c₁ = 2 and a₂ = 3k, b₂ = 1, c₂ = -5

If the lines are perpendicular, a₁a₂ + b₁b₂ + c₁c₂ = 0.

⇒ -3(3k) + 2k(1) + 2(-5) = 0

⇒ -9k + 2k -10 = 0

⇒ 7k = -10

⇒ k = -10/7

4. Find the shortest distance between lines [Tex]\vec{r}=6\hat{i}+2\hat{j}+2\hat{k}+λ(\hat{i}-2\hat{j}+2\hat{k})\space and\space\vec{r}=-4\hat{i}-\hat{k}+μ(3\hat{i}-2\hat{j}-2\hat{k}) [/Tex].

Solution:

Shortest distance between two lines [Tex]\vec{r}=\vec{a_1}+λ\vec{b_1}\space and \space \vec{r}=\vec{a_2}+λ\vec{b_2}[/Tex] is given by:

[Tex]d=|\frac{(\vec{b_1}\times\vec{b_2}).(\vec{a_2}-\vec{a_1})}{|\vec{b_1}\times\vec{b_2}|}|[/Tex]

Now,

[Tex]\vec{a_2}-\vec{a_1}=(-4\hat{i}-\hat{k})-(6\hat{i}+2\hat{j}+2\hat{k})\\ =-10\hat{i}-2\hat{j}-3\hat{k}[/Tex]

Also,

[Tex]\vec{b_1}\times\vec{b_2}=(4+4)\hat{i}-(-2-6)\hat{j}+(-2+6)\hat{k}\\ =8\hat{i}+8\hat{j}+4\hat{k}[/Tex]

Substituting these values in the formula, we have:

[Tex]d=|\frac{(8\hat{i}+8\hat{j}+4\hat{k}).(-10\hat{i}-2\hat{j}-3\hat{k})}{|8\hat{i}+8\hat{j}+4\hat{k}|}|\\ =|\frac{-80-16-12}{\sqrt{(8)^2+(8)^2+(4)^2}}|\\ =|\frac{-108}{12}|[/Tex]

= 9

Thus, the shortest distance is 9 units.

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

Solution:

Here, [Tex]\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \space and \space \vec{a}=\hat{i}+2\hat{j}-4\hat{k}[/Tex]

The equation of a line passing through the point (1, 2, – 4) and parallel to [Tex]\vec{b}[/Tex] is given by:

[Tex]\vec{r}=\vec{a}+λ{\vec{b}}\\ ⇒\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})[/Tex]

Since the given two lines are perpendicular, we have:

3b₁ – 16b₂ + 7b₃ = 0

Also, 3b₁ + 8b₂ – 5b₃ = 0

Thus, [Tex]\frac{b_1}{(-16)\times(-5)-8\times7}=\frac{b_2}{7\times3-3\times(-5)}=\frac{b_3}{3\times8-3\times(-16)}\\ ⇒\frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72}\\ ⇒\frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}[/Tex]

So, the direction ratios of \vec{b} are 2, 3 and 6.

Thus, equation of the vector is [Tex]\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+\lambda(2\hat{i}+3\hat{j}+6\hat{k})[/Tex].

Class 12 NCERT Solutions – Chapter 11 – Three Dimensional Geometry

• Exercise 11.1
• Exercise 11.2

Related Articles:

• Angles Between two Lines in 3D Space: Solved Examples
• Equation of a Line in 3D
• Parallel and Perpendicular Lines
• Shortest Distance Between Two Lines in 3D Space | Class 12 Maths