NCERT Solutions For Class 11 Maths Chapter 11 Introduction To Three Dimensional Geometry Exercise 11.2 - 2025-26

Exercise 11.2 

1. Find the distance between the following pairs of points:

1. $(2, 3, 5)$ and $(4, 3, 1)$

2. $(-3, 7, 2)$ and $(2, 4, -1)$

3. $(-1, 3, -4)$ and $(1, -3, 4)$                

4. $(2, -1, 3)$ and $(-2, 1, 3)$                              

Ans: (i) $(2, 3, 5)$ and $(4, 3, 1)$:

We use the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

$d = \sqrt{(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2} = \sqrt{(2)^2 + (0)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$

(ii) $(-3, 7, 2)$ and $(2, 4, -1)$:

$d = \sqrt{(2 + 3)^2 + (4 - 7)^2 + (-1 - 2)^2} = \sqrt{(5)^2 + (-3)^2 + (-3)^2} = \sqrt{25 + 9 + 9} = \sqrt{43}$

(iii) $(-1, 3, -4)$ and $(1, -3, 4)$:

$d = \sqrt{(1 + 1)^2 + (-3 - 3)^2 + (4 + 4)^2} = \sqrt{(2)^2 + (-6)^2 + (8)^2} = \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26}$

(iv) $(2, -1, 3)$ and $(-2, 1, 3)$:

$d = \sqrt{(-2 - 2)^2 + (1 + 1)^2 + (3 - 3)^2} = \sqrt{(-4)^2 + (2)^2 + (0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$

2. Show that the points $(-2, 3, 5)$, $(1, 2, 3)$, and $(7, 0, -1)$ are collinear.

Ans: To prove collinearity, we check if the distances between the points satisfy $AB + BC = AC$. 

- Distance between $A(-2, 3, 5)$ and $B(1, 2, 3)$:

$AB = \sqrt{(1 + 2)^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{(3)^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

- Distance between $B(1, 2, 3)$ and $C(7, 0, -1)$:

$BC = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{(6)^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56}$

- Distance between $A(-2, 3, 5)$ and $C(7, 0, -1)$:

$AC = \sqrt{(7 + 2)^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{(9)^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126}$

Since $AB + BC \neq AC$, the points are not collinear.

3. Verify the following:

(i) Show that $(0, 7, -10)$, $(1, 6, -6)$, and $(4, 9, -6)$ are vertices of an isosceles triangle.

Ans:

We check if two sides of the triangle are equal:

- Distance between $(0, 7, -10)$ and $(1, 6, -6)$:

$d = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 + 10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$

- Distance between $(1, 6, -6)$ and $(4, 9, -6)$:

$d = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 + 6)^2} = \sqrt{9 + 9 + 0} = \sqrt{18}$

- Distance between $(0, 7, -10)$ and $(4, 9, -6)$:

$d = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (-6 + 10)^2} = \sqrt{16 + 4 + 16} = \sqrt{36}$

Since two sides are equal, the triangle is isosceles.

(ii) Show that $(0, 7, 10)$, $(-1, 6, 6)$, and $(4, 9, 6)$ are the vertices of a right-angled triangle.

Ans: To verify if the triangle is right-angled, we check if the square of the longest side equals the sum of the squares of the other two sides.

- Distance between $(0, 7, 10)$ and $(-1, 6, 6)$:

$d_1 = \sqrt{(-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$

- Distance between $(-1, 6, 6)$ and $(4, 9, 6)$:

$d_2 = \sqrt{(4 + 1)^2 + (9 - 6)^2 + (6 - 6)^2} = \sqrt{(5)^2 + (3)^2 + (0)^2} = \sqrt{25 + 9 + 0} = \sqrt{34}$

- Distance between $(0, 7, 10)$ and $(4, 9, 6)$:

$d_3 = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2} = \sqrt{(4)^2 + (2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Now check if the Pythagorean theorem holds: 

$d_1^2 + d_2^2 = 18 + 34 = 52$

$d_3^2 = 6^2 = 36$

Since $ d_1^2 + d_2^2 \neq d_3^2 $, the triangle is not right-angled based on this configuration.

(iii) Show that $(-1, 2, 1)$, $(1, -2, 5)$, $(4, -7, 8)$, and $(2, -3, 4)$ are the vertices of a parallelogram.

Ans: To verify if the points form a parallelogram, we need to show that the diagonals bisect each other, i.e., the midpoints of both diagonals should coincide.

- Midpoint of diagonal $(-1, 2, 1)$ and $(4, -7, 8)$:

$M_1 = \left(\frac{-1 + 4}{2}, \frac{2 - 7}{2}, \frac{1 + 8}{2}\right) = \left(\frac{3}{2}, \frac{-5}{2}, \frac{9}{2}\right)$

- Midpoint of diagonal $(1, -2, 5)$ and $(2, -3, 4)$:

$M_2 = \left(\frac{1 + 2}{2}, \frac{-2 - 3}{2}, \frac{5 + 4}{2}\right) = \left(\frac{3}{2}, \frac{-5}{2}, \frac{9}{2}\right)$

Since the midpoints are the same, the diagonals bisect each other, proving that the points form a parallelogram.

4. Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

Ans: Let the coordinates of the point $P(x, y, z)$ be equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$. Then,

$\text{Distance from } P(x, y, z) \text{ to } (1, 2, 3) = \text{Distance from } P(x, y, z) \text{ to } (3, 2, -1)$

Using the distance formula:

$\sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2} = \sqrt{(x - 3)^2 + (y - 2)^2 + (z + 1)^2}$

Squaring both sides:

$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$

Expanding both sides:

$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9)  = (x^2 - 6x + 9) + (y^2 - 4y + 4) + (z^2 + 2z + 1)$

Simplifying:

$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1$

Cancel the common terms:

$-2x + 1 - 6z + 9 = -6x + 9 + 2z + 1$

Simplifying further:

$-2x - 6z + 10 = -6x + 2z + 10$

Bringing all terms to one side:

$4x - 8z = 0$

Thus, the equation of the set of points is:

$x = 2z$

5. Find the equation of the set of points $P$, the sum of whose distances from $A(4, 0, 0)$ and $B(-4, 0, 0)$ is equal to 10.

Ans:This is the equation of an ellipse with foci at points $A(4, 0, 0)$ and $B(-4, 0, 0)$, and the sum of the distances from these points is equal to the constant 10. The standard equation of an ellipse with foci at $(a, 0, 0)$ and $(-a, 0, 0)$, and the sum of the distances equal to $2c$, is:

$2a = 10, \quad \Rightarrow a = 5, \quad b^2 = a^2 - f^2$

Here, the distance between the foci is 8, so $f = 4$. Thus:

$b^2 = 5^2 - 4^2 = 25 - 16 = 9, \quad b = 3$

Therefore, the equation of the ellipse is:

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

Conclusion

The NCERT Solutions for Class 11 Maths Chapter 11: Introduction to Three Dimensional Geometry, Exercise 11.2 offer a clear and structured approach to mastering key concepts like distance and section formulas in 3D space. By working through these questions, students will solidify their understanding of spatial relationships and improve their problem-solving skills. This exercise is fundamental in building a foundation for advanced topics in geometry and vector analysis. Using Vedantu’s solutions, students can easily grasp these concepts and be better prepared for their exams.

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