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Maths Solve the following Question.(1 Marks)

Introduction To 3D Coordinate Geometry · Maharashtra Board STD 11 Science (MSBSHSE - English Medium). 29 questions.

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Solve the following Question.(1 Marks)

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29 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark
Find the point on x-axis which is equidistant from the points A(3, 2, 2) and B(5, 5, 4).
Answer
We know that the y and z coordinates of the point on the x-axis are 0.
So, let the required point be C(x, y, z)
Now, CA = CB
$\sqrt{(3-\text{x})^2+(2-0)^2+(2-0)^2}=\sqrt{(5-\text{x})^2+(5-0)^2+(4-0)^2}$
$\Rightarrow9-6\text{x}+\text{x}^2+4+4=25-10\text{x}+\text{x}^2+25+16$
$\Rightarrow17-6\text{x}+\text{x}^2=66-10\text{x}+\text{x}^2$
$\Rightarrow4\text{x}=49$
$\Rightarrow\text{x}=\frac{49}{4}$
Hence, the required Point is $\Big(\frac{49}{4},\ 0,\ 0\Big)$
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Question 21 Mark
Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis
Answer
The distance of the point P(3, 5, 12) from x-axis is given by
$\sqrt{5^2+(12)^2}$
$=\sqrt{169}$
$=13$
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Question 41 Mark
Name the octants in which the following points lie:
(-7, 2, -5)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point(-7, 2, -5) are negative, positive and negative, respectively.
Therefore, this point lies in X'OYZ' octant.
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Question 51 Mark
Write the coordinates of third vertex of a triangle having centroid at the origin and two vertices at $(3, -5, 7)$ and $(3, 0, 1)$.
Answer
Let the coordinates of third vertex be $(x_1, y_1, z_1)$
Now,
$\frac{\text{x}_1+3+3}{3}=0,\ \frac{\text{y}_1+5+0}{3}=0\text{ and }\frac{\text{z}_1+7+1}{3}=0$
$\Rightarrow\text{x}_1 = −6,\ \text{y}_1 = 5\text{ and }\text{z}_1 = −8$
Hence, the coordinates of third vertex of a triangle is $(−6, 5, −8)$.
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Question 61 Mark
Name the octants in which the following points lie:
(-5, -4, 7)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point (-5, -4, 7) are negative, negative and positive, respectively.
Therefore, this point lies in X'OY'Z octant .
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Question 71 Mark
Name the octants in which the following points lie:
(5, 2, 3)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point (5, 2, 3) are all positive.
Therefore, this point lies in XOYZ octant.
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Question 81 Mark
Find the ratio in which the line segment joining the points (2, 4, 5) and (3, −5, 4) is divided by the yz-plane.
Answer
Let the yz-plane divide the line sgement joining the points (2, 4, 5) and (3, −5, 4) in m : 1.
Now, we know that on yz-plane the coordinate of x is 0.
$\therefore\frac{\text{m}\times3+1\times2}{\text{m}+1}=0$
$\Rightarrow 3\text{m} + 2 = 0$
$\Rightarrow\text{m}=-\frac{2}{3}$
Hence, yz-plane divide the line sgement joining the points (2, 4, 5) and (3, −5, 4) in 2 : 3 externally.
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Question 91 Mark
Write the distance of the point P(2, 3, 5) from the xy-plane.
Answer
The distance of the point P(2, 3, 5) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 5.
Hence, the distance of the point P(2, 3, 5) from the xy-plane is 5.
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Question 101 Mark
Name the octants in which the following points lie:
(4, -3, 5)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point (4, -3, 5) are positive, negative and positive, respectively.
Therefore, this point lies in XOY'Z octant.
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Question 111 Mark
Write the coordinates of the point P which is five-sixth of the way from A(-2, 0, 6) to B(10, -6, -12).
Answer
Let the coordinates of the point be P(x, y, z)
Now
$\text{PA}=\frac{5}{6}\text{PB}$
$\Rightarrow\frac{\text{PA}}{\text{PB}}=\frac{5}{6}$
$\therefore\ \text{x}=\frac{5\times10-6\times2}{5+6},\ \text{y}=\frac{5\times(-6)+6\times0}{5+6},\ \text{z}=\frac{5\times(-12)+6\times6}{5+6}$
$\therefore\ \text{x}=\frac{38}{11},\ \text{y}=\frac{-24}{11},\ \text{z}=\frac{-34}{11}$
Hence, the coordinates of the point is $\Big(\frac{38}{11},\ \frac{-24}{11},\ \frac{-34}{11}\Big)$
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Question 131 Mark
Name the octants in which the following points lie:
(7, 4, -3)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point (7, 4, -3) are positive, positive and negative, respectively.
Therefore, this point lies in XOYZ' octant.
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Question 141 Mark
If a parallelopiped is formed by the planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, then write the lengths of edges of the parallelopiped and length of the diagonal.
Answer
Lenghs of edges of the parallelopiped are given by
5 - 2, 9 - 3, 7 - 5
= 3, 6, 2
Lenght of the diagonal is given by
$\sqrt{3^2+6^2+2^2}$
$=\sqrt{49}$
$=7\text{ Units}$
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Question 151 Mark
Write the distance of the point P(3, 4, 5) from z-axis.
Answer
The distance of the point P(3, 4, 5) from z-axis is given by
$\sqrt{3^2+4^2}$
$=\sqrt{25}$
$=5$
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Question 161 Mark
The coordinates of the mid-points of sides $AB, BC$ and $CA$ of $\triangle\text{ABC}$ are $D(1, 2, -3), E(3, 0, 1)$ and $F(-1, 1, -4)$ respectively. Write the coordinates of its centroid.
Answer
Let the coordintes of the triangles be $A(x_1, y_2, z_3), B(x_1, y_2, z_3)$ and $C(x_1, y_2, z_3).$
Now,
Mid-point of $AB$ is $D(1, 2, −3)$
$\frac{\text{x}_1+\text{x}_2}{2}=1,\ \frac{\text{y}_1+\text{y}_2}{2}=2\text{ and }\frac{\text{z}_1+\text{z}_2}{2}=-3$
$\Rightarrow x_1 + x_2 = 2, y_1 + y_2 = 4$ and $z_1 + z_2 = -6 ...(i)$
Mid-point of $BC$ is $E(3, 0, 1)$
$\frac{\text{x}_2+\text{x}_3}{2}=3,\ \frac{\text{y}_2+\text{y}_3}{2}=0\text{ and }\frac{\text{z}_2+\text{z}_3}{2}=1$
$\Rightarrow x_2 + x_3 = 6, y_2 + y_3 = 4 = 0$ and $z_2 + z_3 = 2 ...(ii)$
Mid-point of $AC$ is $F(-1, 1, -4)$
$\frac{\text{x}_1+\text{x}_3}{2}=-1,\ \frac{\text{y}_1+\text{y}_3}{2}=1\text{ and }\frac{\text{z}_1+\text{z}_3}{2}=-4$
$\Rightarrow x_1 + x_3 = -2, y_1 + y_3 = 4 = 2$ and $z_1 + z_3 = -8 ...(iii)$
Adding $(1), (2)$ and $(3)$, We get
$2(x_1+ x_2+ x_3) = 6, 2(y_1 + y_2+ y_3) = 6$ and$ 2(z_1 + z_2 + z_3) = -12$
$\Rightarrow x_1+ x_2+ x_3 = 3, y_1 + y_2+ y_3 = 3$ and $z_1 + z_2 + z_3 = -6$
$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}=1,\ \frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}=1\text{ and }\frac{\text{z}_1+\text{z}_2+\text{z}_3}{3}=-2$
Hence, the centroid of the traingle $ABC$ is $(1, 1, -2).$
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Question 171 Mark
Name the octants in which the following points lie:
(-5, -3, -2)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point (-5, -3, -2) are all negative.
Therefore, this point lies in X'OY'Z' octant.
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Question 201 Mark
What is the locus of a point for which y = 0, z = 0?
Answer
We know that on x-axis both y = 0, z = 0.
Hence, the locus of a point for which y = 0, z = 0 is x-axis.
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Question 211 Mark
Name the octants in which the following points lie:
(-5, 4, 3)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point (-5, 4, 3) are negative, positive and positive, respectively.
Therefore, this point lies in X'OYZ octant.
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Question 221 Mark
Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1).
Answer
The coordiante of x point on yz-plane is 0
Let the point be P(0, y, z).
Now, PA = PB
$\sqrt{(2-0)^2+(0-\text{y})^2+(3-\text{z})^2}=\sqrt{(0-0)^2+(3-\text{y})^2+(2-\text{z})^2}$
$\Rightarrow4+\text{y}^2+9-6\text{z+}\text{z}^2=9-6\text{y}+\text{y}^2+4-4\text{z}+\text{z}^2$
$\Rightarrow-6\text{z}=-6\text{y}-4\text{z}$
$\Rightarrow3\text{y}-\text{z}=0\ ...\text{(i)}$
Also, PA = PC
$\sqrt{(2-0)^2+(0-\text{y})^2+(3-\text{z})^2}=\sqrt{(0-0)^2+(0-\text{y})^2+(1-\text{z})^2}$
$\Rightarrow4+\text{y}^2+9-6\text{z+}\text{z}^2=\text{y}^2+1-2\text{z}+\text{z}^2$
$\Rightarrow13-6\text{z}=1-2\text{z}$
$\Rightarrow-4\text{z}=-12$
$\Rightarrow\text{z}=3\ ...\text{(ii)}$
Solving (i) and (ii), we get
y = 1
Hence, the coordinates of the point is (0, 1, 3).
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Question 231 Mark
Name the octants in which the following points lie:
(2, -5, -7)
Answer
The x-coordinate, the y-coordinate and the z-coordinate of the point (2, -5, -7) are positive, negative and negative, respectively.
Therefore, this point lies in XOY'Z' octant.
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Question 241 Mark
If the distance between the points P(a, 2, 1) and Q(1, −1, 1) is 5 units, find the value of a.
Answer
$\text{PQ}=5$
$\Rightarrow\sqrt{(1-\text{a})^2+(-1-2)^2+(1-1)^2}=5$
$\Rightarrow(1-\text{a})^2+(-3)^2=25$
$\Rightarrow1-2\text{a+}\text{a}^2+9-25=0$
$\Rightarrow\text{a}^2-2\text{a}-15=0$
$\Rightarrow(\text{a}-5)(\text{a}+3)=0$
$\Rightarrow\text{a}=5,\ -3$
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Question 251 Mark
Find the point on y-axis which is at a distance of $\sqrt{10}$ units from the point (1, 2, 3).
Answer
We know that the x and z coordinates of the point on the y-axis are 0.
So, let the required point be (0, y, 0)
Now,
$\sqrt{(1-0)^2+(2-\text{y})^2+(3-0)^2}=\sqrt{10}$
$\Rightarrow\ 1+4-4\text{y}+\text{y}^2+9=10$
$\Rightarrow\ \text{y}^2-4\text{y}+4=0$
$\Rightarrow\ (\text{y}-2)^2=0$
$\text{y}=2,\ 2$
Hence, the required point is (0, 2, 0)
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Question 261 Mark
Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.
Answer
We know that the x and z coordinates on y-axis are 0
The coordinates of the foot of the perpendicular from a point (1, 2, 3) on y-axis are (0, 2, 0)
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Question 271 Mark
If the origin is the centroid of a triangle ABC having vertices A(a, 1, 3), B(-2, b -5) and C(4, 7, c) find the values of a, b, c.
Answer
We have A(a, 1, 3), B(-2, b, -5) and C(4, 7, c)
Now,
$\frac{\text{a}-2+4}{3}=0,\ \frac{1+\text{b}+7}{3}=0\text{ and }\frac{3-5+\text{c}}{3}=0$
$\Rightarrow\text{a}=-2,\ \text{b}=-8\text{ and }\text{c}=2$
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Question 291 Mark
Find the coordinates of a point equidistant from the origin and points A(a, 0, 0), B(0, b, 0) and C(0, 0, c).
Answer
Let the point be P(x, y, z).
Now, PO = PA
$\sqrt{(0-\text{x})^2+(0-\text{y})^2+(0-\text{z})^2}=\sqrt{(\text{a}-\text{x})^2+(0-\text{y})^2+(0-\text{z})^2}$
$\Rightarrow\text{x}^2+\text{y}^2+\text{z}^2=\text{a}^2-2\text{ax}+\text{x}^2+\text{y}^2+\text{z}^2$
$\Rightarrow0=\text{a}^2-2\text{ax}$
$\Rightarrow\text{x}=\frac{\text{a}}{2}$
Also, PO = PB
$\sqrt{(0-\text{x})^2+(0-\text{y})^2+(0-\text{z})^2}=\sqrt{(0-\text{x})^2+(\text{b}-\text{y})^2+(0-\text{z})^2}$
$\Rightarrow\text{x}^2+\text{y}^2+\text{z}^2=\text{x}^2-2\text{by}+\text{y}^2+\text{z}^2$
$\Rightarrow0=\text{b}^2-2\text{by}$
$\Rightarrow\text{y}=\frac{\text{b}}{2}$
Again, PO = PC
$\sqrt{(0-\text{x})^2+(0-\text{y})^2+(0-\text{z})^2}=\sqrt{(0-\text{x})^2+(\text{b}-\text{y})^2+(\text{c}-\text{z})^2}$
$\Rightarrow\text{x}^2+\text{y}^2+\text{z}^2=\text{x}^2+\text{y}^2+\text{c}^2-2\text{cz}+\text{z}^2$
$\Rightarrow0=\text{c}^2-2\text{cz}$
$\Rightarrow\text{z}=\frac{\text{c}}{2}$
Hence, the coordinates of the point is $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2},\frac{\text{c}}{2}\Big)$
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