Class 11 Mathematics Chapter 11 Introduction to Three Dimensional Geometry

Key Concept: Basic Distance Formula

b) $5\sqrt{2}$

[Solution Description] To find the distance between the two given points in three-dimensional space, use the distance formula:

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

Substitute the coordinates of the points into the formula:

$$PQ = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2}$$

Simplify the equation:

$$PQ = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$$

Finally, simplify the square root:

$$PQ = \sqrt{50} = 5\sqrt{2}$$

Key Concept: Plane Intersection

c) X-axis

[Solution Description] The ZX-plane is defined by $y = 0$, and the XY-plane is defined by $z = 0$. The line of intersection occurs where both conditions are satisfied, which means $y = 0$ and $z = 0$. Therefore, the line of intersection is along the x-axis.

Key Concept: Basic Coordinates

a) (0, y, 0)

[Solution Description] Points on the y-axis have coordinates of the form $(0, y, 0)$ because there is no x or z component.

Key Concept: Origin Recognition

a) Yes

[Solution Description] The coordinates of the origin are defined as $(0, 0, 0)$. Therefore, the point $(0, 0, 0)$ is indeed at the origin.

Key Concept: Parallelogram Non-Rectangle

a) Parallelogram with unequal diagonals

[Solution Description] To show that ABCD is a parallelogram, we need to verify that $AB = CD$ and $BC = DA$.

Using the distance formula,

$$AB = \sqrt{(7-3)^2 + (-1-2)^2 + (0-1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26}$$

$$CD = \sqrt{(5-1)^2 + (3-6)^2 + (4-5)^2} = \sqrt{16 + 9 + 1} = \sqrt{26}$$

$$BC = \sqrt{(5-7)^2 + (3+1)^2 + (4-0)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

$$DA = \sqrt{(1-3)^2 + (6-2)^2 + (5-1)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

Since opposite sides are equal, ABCD is a parallelogram. Now check diagonals for inequality:

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

$$BD = \sqrt{(1-7)^2 + (6+1)^2 + (5-0)^2} = \sqrt{36 + 49 + 25} = \sqrt{110}$$

Diagonals are unequal $$(\sqrt{14} \neq \sqrt{110})$$, confirming it is not a rectangle.

Key Concept: Origin Coordinates

b) $(0, 0, 0)$

[Solution Description] The origin in a three-dimensional space is the point where all the coordinate axes intersect. It has coordinates $(0, 0, 0)$ representing zero distance from all planes.

Key Concept: Zero Coordinate

b) 25

[Solution Description] If one or more coordinates are zero, the distance is calculated by considering only non-zero coordinates using the formula:

$$OQ = \sqrt{x_2^2 + y_2^2 + z_2^2}$$

Plugging in the values for the point (7, 0, 24), we have:

$$OQ = \sqrt{7^2 + 0^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$$

Therefore, the distance is 25 units.

Key Concept: Coordinate Calculation

a) (6, 2, 3)

[Solution Description] Translating point $A(1, 2, 3)$ parallel to the x-axis by 5 units means adding 5 to the x-coordinate while keeping the y and z coordinates unchanged. Thus, the new coordinates will be $(1+5, 2, 3)$ which results in $(6, 2, 3)$.

Key Concept: Octant Determination

d) Seventh octant

[Solution Description] In three-dimensional geometry, the sign of each coordinate determines the octant. For the point (-5, 7, -8), the signs are negative, positive, and negative for x, y, and z respectively. This corresponds to the octant where x is negative, y is positive, and z is negative, which is the seventh octant.

Key Concept: Distance Formula

b) $5\sqrt{2}$

[Solution Description] To find the distance between the points $A(1, 2, 3)$ and $B(4, 6, 8)$, we use the distance formula in 3D:

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

Substitute the coordinates of the points into the formula:

$$AB = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2}$$

Calculate each squared difference:

$$(4 – 1)^2 = 9, \quad (6 – 2)^2 = 16, \quad (8 – 3)^2 = 25$$

Sum these results:

$$9 + 16 + 25 = 50$$

Finally, take the square root:

$$AB = \sqrt{50} = 5\sqrt{2}$$

Key Concept: Origin Distance

c) $\sqrt{50}$

[Solution Description] To find the distance from the origin $(0, 0, 0)$ to a point $Q(x_2, y_2, z_2)$, use the formula:

$$OQ = \sqrt{x_2^2 + y_2^2 + z_2^2}.$$

Substituting $x_2 = -3$, $y_2 = 4$, and $z_2 = 5$ into the formula, we have:

$$OQ = \sqrt{(-3)^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}.$$

Hence, the distance is $\sqrt{50}$.

Key Concept: Origin Coordinates

b) (0,0,0)

[Solution Description] The point where all three coordinate axes intersect in a 3D system is called the origin, which has the coordinates (0, 0, 0).

Key Concept: Distance from Origin

c) $10\sqrt{2}$

[Solution Description] To find the distance of a point $(x, y, z)$ from the origin $(0, 0, 0)$, we use the distance formula in 3D:

$$d = \sqrt{x^2 + y^2 + z^2}$$

Substituting $x = 6$, $y = -8$, and $z = 10$ into the equation, we have:

$$d = \sqrt{6^2 + (-8)^2 + 10^2} = \sqrt{36 + 64 + 100} = \sqrt{200} = 10\sqrt{2}$$

Therefore, the distance is $10\sqrt{2}$.

Key Concept: Centroid Calculation

b) (2, 2, 2)

[Solution Description] The centroid formula is given by $$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$$.

Let the coordinates of R be $(x, y, z)$.

Given that the centroid is at (2, 3, 4):

$$\frac{1 + 3 + x}{3} = 2 \Rightarrow 4 + x = 6 \Rightarrow x = 2$$

$$\frac{2 + 5 + y}{3} = 3 \Rightarrow 7 + y = 9 \Rightarrow y = 2$$

$$\frac{3 + 7 + z}{3} = 4 \Rightarrow 10 + z = 12 \Rightarrow z = 2$$

The coordinates of vertex R are $(2, 2, 2)$.

Key Concept: Midpoint Calculation

a) (5, 6, 7)

[Solution Description] To find the midpoint, compute $$\left(\frac{3 + 7}{2}, \frac{4 + 8}{2}, \frac{5 + 9}{2}\right)$$. Calculating each component:

– For the x-coordinate:

$$\frac{3 + 7}{2} = \frac{10}{2} = 5$$

– For the y-coordinate:

$$\frac{4 + 8}{2} = \frac{12}{2} = 6$$

– For the z-coordinate:

$$\frac{5 + 9}{2} = \frac{14}{2} = 7$$

So, the midpoint is $(5, 6, 7)$.

Key Concept: Centroid Calculation

b) $G(1, \frac{10}{3}, \frac{4}{3})$

[Solution Description] The formula for the centroid is:

$$G\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$$

Substituting the given vertices:

$$G_x = \frac{2 + (-4) + 5}{3} = \frac{3}{3} = 1$$,

$$G_y = \frac{3 + (-1) + 8}{3} = \frac{10}{3}$$,

$$G_z = \frac{5 + 6 + (-7)}{3} = \frac{4}{3}$$.

Therefore, the centroid is $$G(1, \frac{10}{3}, \frac{4}{3})$$.

Key Concept: Identify Collinear Points

b) No, they are not collinear

[Solution Description] We need to check if $PQ + QR = PR$ for collinearity using point distances.

Calculate $PQ$:

$$PQ = \sqrt{(2 – 1)^2 + (3 – 1)^2 + (5 – 1)^2}$$

$$= \sqrt{1 + 4 + 16}$$

$$= \sqrt{21}$$

Calculate $QR$:

$$QR = \sqrt{(3 – 2)^2 + (5 – 3)^2 + (9 – 5)^2}$$

$$= \sqrt{1 + 4 + 16}$$

$$= \sqrt{21}$$

Calculate $PR$:

$$PR = \sqrt{(3 – 1)^2 + (5 – 1)^2 + (9 – 1)^2}$$

$$= \sqrt{4 + 16 + 64}$$

$$= \sqrt{84}$$

Since $$\sqrt{21} + \sqrt{21} \neq \sqrt{84}$$, the points are not collinear.

Key Concept: Distance Formula

c) $5\sqrt{2}$

[Solution Description] The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ can be calculated using the formula:

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

Substitute the given values:

$$d = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$$

So, the distance is $5\sqrt{2}$ units.

Key Concept: Distance Calculation

c) $5\sqrt{3}$

[Solution Description] The distance formula between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is given by

$$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}.$$

Substituting the coordinates of $P(2, -1, 3)$ and $Q(-3, 4, -2)$:

$$PQ = \sqrt{((-3) – 2)^2 + (4 – (-1))^2 + ((-2) – 3)^2}.$$

Simplifying each term inside the square root:

$$(-3) – 2 = -5, \quad \therefore (-5)^2 = 25$$,

$$4 – (-1) = 4 + 1 = 5, \quad \therefore 5^2 = 25$$,

$$(-2) – 3 = -5, \quad \therefore (-5)^2 = 25$$.

Therefore,

$$PQ = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}.$$

Thus, the distance is $5\sqrt{3}$.

Key Concept: Distance Formula

c) 7.07

[Solution Description] Using the distance formula $$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}$$:

$$PQ = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2}$$

Calculate each term individually:

$$(x_2 – x_1)^2 = (4 – 1)^2 = 3^2 = 9$$

$$(y_2 – y_1)^2 = (6 – 2)^2 = 4^2 = 16$$

$$(z_2 – z_1)^2 = (8 – 3)^2 = 5^2 = 25$$

Sum these values:

$$PQ = \sqrt{9 + 16 + 25} = \sqrt{50} \approx 7.07$$

Thus, the distance is approximately 7.07 units.

Key Concept: Octant Naming

b) Second octant

[Solution Description] To determine the octant where a point lies, check the signs of its coordinates:

– x-coordinate: negative

– y-coordinate: positive

– z-coordinate: positive

These sign combinations indicate that the point is in the second octant.

Key Concept: Distance Formula

c) $\sqrt{61}$

[Solution Description] To find the distance between two points $$A(x_1, y_1, z_1) = (1, 2, 3)$$ and $$B(x_2, y_2, z_2) = (4, 6, 9)$$, we use the 3D distance formula:

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

Substituting the values:

$$AB = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (9 – 3)^2} = \sqrt{3^2 + 4^2 + 6^2} = \sqrt{9 + 16 + 36} = \sqrt{61}$$

Hence, the distance is $\sqrt{61}$.

Key Concept: Distance Calculation

c) $3\sqrt{3}$

[Solution Description] To find the distance between the two points in 3D space, we use the formula $$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}$$. Substituting the given coordinates:

$$PQ = \sqrt{(4 – 1)^2 + (5 – 2)^2 + (6 – 3)^2}$$

$$= \sqrt{3^2 + 3^2 + 3^2}$$

$$= \sqrt{9 + 9 + 9}$$

$$= \sqrt{27}$$

$$= 3\sqrt{3}$$

Therefore, the distance is $$3\sqrt{3}$$.

Key Concept: Origin Coordinates

a) (0, 0, 0)

[Solution Description] The origin in a three-dimensional coordinate system is the point where all axes intersect. It is represented by the coordinates (0, 0, 0).

Key Concept: Basic Formula Application

c) (7, 9)

[Solution Description] To find the coordinates of point P dividing AB externally in the ratio 2:1, we use the formula for external division:

$$\left( \frac{mx_2 – nx_1}{m-n}, \frac{my_2 – ny_1}{m-n} \right)$$

where $m = 2$, $n = 1$, $x_1 = 1$, $y_1 = 3$, $x_2 = 4$, $y_2 = 6$.

Calculating the x-coordinate:

$$\frac{2 \times 4 – 1 \times 1}{2-1} = \frac{8 – 1}{1} = 7$$

Calculating the y-coordinate:

$$\frac{2 \times 6 – 1 \times 3}{2-1} = \frac{12 – 3}{1} = 9$$

Therefore, the coordinates of P are (7, 9).

Key Concept: Distance Calculation

c) $\sqrt{57}$

[Solution Description] To calculate the distance between two points in space, we use the distance formula:

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

Here, $x_1 = 3$, $y_1 = 4$, $z_1 = 5$, $x_2 = -1$, $y_2 = -1$, and $z_2 = 1$. Substituting these values into the formula, we get:

$$PQ = \sqrt{((-1) – 3)^2 + ((-1) – 4)^2 + (1 – 5)^2}$$

Simplifying further:

$$PQ = \sqrt{(-4)^2 + (-5)^2 + (-4)^2} = \sqrt{16 + 25 + 16} = \sqrt{57}$$

Therefore, the distance is $\sqrt{57}$.

Key Concept: Perpendicular Distance Calculation

b) 7

[Solution Description] To find the perpendicular distance of a point $(x, y, z)$ from the YZ-plane, we use the formula $|x|$. With $x = 7$, the distance is $|7| = 7$.

Key Concept: Origin Distance

c) $\sqrt{14}$

[Solution Description] Use the distance formula from the origin:

$$OQ = \sqrt{x^2 + y^2 + z^2}$$

where $x = 2$, $y = -1$, $z = 3$. Substitute these values:

$$OQ = \sqrt{2^2 + (-1)^2 + 3^2}$$

$$= \sqrt{4 + 1 + 9}$$

$$= \sqrt{14}$$

Thus, the distance from the origin to the point is $\sqrt{14}$.

Key Concept: Coordinate Plane Division

c) 8

[Solution Description] In a 3D Cartesian coordinate system, the three perpendicular planes (XY, YZ, and ZX) divide the space into eight distinct regions known as octants. Each octant represents a unique combination of positive or negative values for the $x$, $y$, and $z$ coordinates.

Key Concept: Collinearity Proof

a) Yes

[Solution Description] Calculate the distance between each pair of points:

Distance $d_{12}$ between $(1, 2, 3)$ and $(4, 6, 8)$:

$$d_{12} = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$$

Distance $d_{23}$ between $(4, 6, 8)$ and $(7, 10, 13)$:

$$d_{23} = \sqrt{(7 – 4)^2 + (10 – 6)^2 + (13 – 8)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}$$

Distance $d_{13}$ between $(1, 2, 3)$ and $(7, 10, 13)$:

$$d_{13} = \sqrt{(7 – 1)^2 + (10 – 2)^2 + (13 – 3)^2} = \sqrt{6^2 + 8^2 + 10^2} = \sqrt{36 + 64 + 100} = \sqrt{200}$$

Check collinearity: $d_{12} + d_{23} = d_{13}$

$$\sqrt{50} + \sqrt{50} = \sqrt{200}$$

Since the equality holds, the points are collinear.

Key Concept: Origin Coordinates

b) (0, 0, 0)

[Solution Description] The origin in a three-dimensional coordinate system is the point where all the axes intersect, having coordinates (0, 0, 0).

Key Concept: Octant Identification

b) Second Octant

[Solution Description] The sign of the coordinates determines the octant in 3D space. If:

-$x < 0$, $y > 0$, and $z < 0$, then the point lies in the second octant. Thus, the point $(-5, 3, -7)$ has: - a negative $x$-coordinate, - a positive $y$-coordinate, - a negative $z$-coordinate, indicating it lies in the second octant.

Key Concept: Octant Naming

b) Second octant

[Solution Description] The point (-3, 4, -5) has a negative x-coordinate, positive y-coordinate, and negative z-coordinate. This describes the octant where x is negative, y is positive, and z is negative.

Key Concept: Coordinate Calculation

b) $(2, \frac{10}{3}, 5)$

[Solution Description] To find the coordinates of point M which divides the line segment AB in the ratio 1:2, we use the section formula:

$$x = \frac{mx_2 + nx_1}{m+n}, \; y = \frac{my_2 + ny_1}{m+n}, \; z = \frac{mz_2 + nz_1}{m+n}$$

Here, $x_1 = 1$, $y_1 = 2$, $z_1 = 3$, $x_2 = 4$, $y_2 = 6$, $z_2 = 9$, $m = 1$, $n = 2$.

Calculate each coordinate:

$$x = \frac{(1)(4) + (2)(1)}{1+2} = \frac{4 + 2}{3} = 2$$

$$y = \frac{(1)(6) + (2)(2)}{1+2} = \frac{6 + 4}{3} = \frac{10}{3}$$

$$z = \frac{(1)(9) + (2)(3)}{1+2} = \frac{9 + 6}{3} = 5$$

Thus, the coordinates of M are $$(2, \frac{10}{3}, 5)$$.

Key Concept: Origin Recognition

c) (0, 0, 0)

[Solution Description] The origin in three-dimensional space is defined by the point where all three axes intersect, resulting in coordinates where x, y, and z are all equal to zero. Therefore, the coordinates of the origin are (0, 0, 0).

Key Concept: Collinearity Check

a) Yes, they are collinear.

[Solution Description] To check for collinearity, calculate distances $PQ$, $QR$, and $PR$ and verify if $PQ + QR = PR$.

$$PQ = \sqrt{(2 – 1)^2 + (3 – 2)^2 + (4 – 3)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$$

$$QR = \sqrt{(3 – 2)^2 + (4 – 3)^2 + (5 – 4)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$$

$$PR = \sqrt{(3 – 1)^2 + (4 – 2)^2 + (5 – 3)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$$

Since $$PQ + QR = \sqrt{3} + \sqrt{3} = 2\sqrt{3} = PR$$, the points are collinear.

Key Concept: Coordinate Calculation

c) $(4, -3, 8)$

[Solution Description] To transform the given point $(4, -3, 5)$ by shifting 3 units along the z-axis, we add 3 to the z-coordinate while keeping the x and y coordinates unchanged. Therefore, the new coordinates will be:

$$(4, -3, 5 + 3) = (4, -3, 8)$$

Key Concept: Midpoint Calculation

a) (1, -0.5, 5)

[Solution Description] To find the midpoint $M(x, y, z)$ of a line segment connecting points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, use the formula:

$$x = \frac{x_1 + x_2}{2}$$, $$y = \frac{y_1 + y_2}{2}$$, $$z = \frac{z_1 + z_2}{2}$$.

Substituting the given coordinates:

$$x = \frac{3 + (-1)}{2} = \frac{2}{2} = 1$$,

$$y = \frac{-5 + 4}{2} = \frac{-1}{2}$$,

$$z = \frac{7 + 3}{2} = \frac{10}{2} = 5$$.

So, the midpoint is $(1, -0.5, 5)$.

Key Concept: Collinearity Check

b) No

[Solution Description] To determine if the points are collinear, we need to check if the sum of distances $AB + BC$ equals the distance $AC$. Use the distance formula:

$$AB = \sqrt{(4 – 1)^2 + (6 – 2)^2 + (8 – 3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}$$

$$BC = \sqrt{(7 – 4)^2 + (10 – 6)^2 + (13 – 8)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}$$

$$AC = \sqrt{(7 – 1)^2 + (10 – 2)^2 + (13 – 3)^2} = \sqrt{6^2 + 8^2 + 10^2} = \sqrt{200}$$

Since $$AB + BC = \sqrt{50} + \sqrt{50} = \sqrt{100} = 10$$ and $AC = \sqrt{200}$, the points are not collinear.