Class 11: Three Dimensional Coordinate Geometry – Exercise 28.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Name the octants in which the following points lie:

(i) $\displaystyle (5,2, 3)$ (ii) $\displaystyle (-5, 4, 3)$ (iii) $\displaystyle (4, - 3, 5)$ (iv) $\displaystyle (7 , 4, - 3)$ (v) $\displaystyle (-5, -4,7)$ (vi) $\displaystyle (-5, - 3, -2)$ (vii) $\displaystyle (2, -5, -7)$ (viii) $\displaystyle (-7,2-5)$

Answer:

(i) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (5,2, 3)$ are all positive. Therefore this point lies in $\displaystyle XOYZ$ octant.

(ii) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (-5, 4, 3)$ are negative, positive and positive respectively. Therefore this point lies in $\displaystyle X'OYZ$ octant.

(iii) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (4, - 3, 5)$ are positive, negative and positive respectively. Therefore this point lies in $\displaystyle XOY'Z$ octant.

(iv) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (7 , 4, - 3)$ are positive, positive and negative respectively. Therefore this point lies in $\displaystyle XOYZ'$ octant.

(v) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (-5, -4,7)$ are negative, negative and positive respectively. Therefore this point lies in $\displaystyle X'OY'Z$ octant.

(vi) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (-5, - 3, -2)$ are all negative. Therefore this point lies in $\displaystyle X'OY'Z'$ octant.

(vii) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (2, -5, -7)$ are positive, negative and negative respectively. Therefore this point lies in $\displaystyle XOY'Z'$ octant.

(viii) The x-coordinate, the y-coordinate and the z-coordinate of the point $\displaystyle (-7,2-5)$ are negative, positive and negative respectively. Therefore this point lies in $\displaystyle X'OYZ'$ octant.

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Question 2: Find the image of:

(i) $\displaystyle (-2, 3, 4) \text{ in the YZ-plane. }$

(ii) $\displaystyle (- 5, 4, - 3) \text{ in the XZ-plane.}$

(iii) $\displaystyle (5, 2, -7) \text{ in the XY-plane.}$

(iv) $\displaystyle (- 5, 0, 3) \text{ in the XZ- plane.}$

(v) $\displaystyle (- 4,0, 0) \text{ in the XY-plane.}$

Answer:

(i) $\displaystyle YZ$ plane is the x-axis, hence the sign of $\displaystyle x$ will be changed. Hence the image is $\displaystyle (2, 3, 4)$

(ii) $\displaystyle XZ$ plane is the y-axis, hence the sign of $\displaystyle y$ will be changed. Hence the image is $\displaystyle (-5, -4, -3)$

(iii) $\displaystyle XY$ plane is the z-axis, hence the sign of $\displaystyle z$ will be changed. Hence the image is $\displaystyle (5, 2, 7)$

(iv) $\displaystyle XZ$ plane is the y-axis, hence the sign of $\displaystyle y$ will be changed. Hence the image is $\displaystyle (-5, 0, 3)$

(v) $\displaystyle XY$ plane is the z-axis, hence the sign of $\displaystyle z$ will be changed. Hence the image is $\displaystyle (-4, 0, 0)$

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Question 3: A cube of side 5 has one vertex at the point $\displaystyle (1,0, -1),$ and the three edges from this vertex are, respectively, parallel to the negative $\displaystyle x$ and $\displaystyle y$ axes and positive $\displaystyle z$ axis. Find the coordinates of the other vertices of the cube.

Answer:

Given: A cube has side 5 having one vertex at $\displaystyle (1, 0, -1)$

Side of cube $\displaystyle = 5$

We need to find the coordinates of the other vertices of the cube.

So let the Point $\displaystyle A(1, 0, -1)$ and $\displaystyle AB, AD$ and $\displaystyle AE$ is parallel to negative x-axis, negative y-axis and positive z-axis respectively.

Since side of cube $\displaystyle = 5$

Point B is $\displaystyle (-4, 0, -1)$

Point D is $\displaystyle (1, -5, -1)$

Point E is $\displaystyle (1, 0, 4)$

Now, EH is parallel to negative y-axis

Point H is $\displaystyle (1, -5, 4)$

HG is parallel to negative x-axis

Point G is $\displaystyle (-4, -5, -1)$

Now, again GC and GF is parallel to negative z-axis and positive y-axis respectively

Point C is $\displaystyle (-4, -5, -1)$

Point F is $\displaystyle (-4, 0, 4)$

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Question 4: Planes are drawn parallel to the coordinate planes through the points $\displaystyle (3,0, -1)$ and $\displaystyle (-2,5,4).$ Find the lengths of the edges of the parallelopiped so formed.

Answer:

$\displaystyle P = ( 3, 0, -1) \text{ and } Q = ( -2, 5, 4)$

284 Clearly, $\displaystyle PA, PB$ and $\displaystyle PC$ are the lengths of the edges of the parallelopiped shown in the figure.

Clearly, $\displaystyle PBEC, QDAF$ are planes parallel to yx-plane such that their distances from yz-plane are $\displaystyle x_1$ and $\displaystyle x_2$ respectively. So,

$\displaystyle PA = ( \text{ distance between the planes } PBEC \text{ and } QDAF) \\ \\ =| x_2 - x_1 |= |-2-3| = |-5 |= 5$

$\displaystyle PB$ is the distance between the planes $\displaystyle PAFC$ and $\displaystyle BDQE$ which are parallel to zx-plane and are at a distance $\displaystyle y_1$ and $\displaystyle y_2$ , respectively, from zx-plane.

$\displaystyle \therefore PB = |y_2 - y_1| = |5-0| = 5$

Similarly, $\displaystyle PC$ is the distance between the planes $\displaystyle PBDA$ and $\displaystyle CEQF$ which are at a distance $\displaystyle z_1$ and $\displaystyle z_2$ respectively, from xy-plane.

$\displaystyle \therefore PC =| z_2- z_1 |= |4 - (-1)| = 5$

Therefore the the lengths of the edges of the parallelopiped are $\displaystyle 5, 5$ and $\displaystyle 5$

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Question 5: Planes are drawn through the points $\displaystyle (5,0,2)$ and $\displaystyle (3, -2,5)$ parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelopiped so formed. 284

Answer:

$\displaystyle P = ( 5, 0, 2) \text{ and } Q = ( 3, -2, 5)$

Clearly, $\displaystyle PA, PB$ and $\displaystyle PC$ are the lengths of the edges of the parallelopiped shown in the figure.

Clearly, $\displaystyle PBEC, QDAF$ are planes parallel to yx-plane such that their distances from yz-plane are $\displaystyle x_1$ and $\displaystyle x_2$ respectively. So,

$\displaystyle PA = ( \text{ distance between the planes } PBEC \text{ and } QDAF) \\ \\ =| x_2 - x_1 |= |3-5| = |-2 |= 2$

$\displaystyle \therefore PB = |y_2 - y_1| = |-2-0| = 2$

$\displaystyle \therefore PC =| z_2- z_1 |= |5-2| = 3$

Therefore the the lengths of the edges of the parallelopiped are $\displaystyle 5, 5$ and $\displaystyle 5$

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Question 6: Find the distances of the point $\displaystyle P (- 4,3,5)$ from the coordinate axes.

Answer:

Given: The point $\displaystyle P (-4, 3, 5)$

The distance of the point from x-axis is given as:

$\displaystyle \text{Distance } = \sqrt{y^2 + z^2} = \sqrt{3^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$

The distance of the point from y-axis is given as:

$\displaystyle \text{Distance } = \sqrt{x^2 + z^2} = \sqrt{(-4)^2 + 5^2} = \sqrt{16+25} = \sqrt{41}$

The distance of the point from z-axis is given as:

$\displaystyle \text{Distance } = \sqrt{x^2 + y^2} = \sqrt{(-4)z^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$

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Question 7: The coordinates of a point are $\displaystyle ( 3, - 2, 5).$ Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.