ID: 978 Type: Subjective Source: NCERT Class 11 Year: null

Question 1: Find the octant in which the points $(-3,1,2)$ and $(-3,1,-2)$ lie.

ID: 979 Type: Subjective Source: NCERT Class 11 Year: null

Question 2: Find the distance between the points $P(1,-3,4)$ and $Q(-4,1,2)$

ID: 980 Type: Subjective Source: NCERT Class 11 Year: null

Question 3: Show that the points $P(-2,3,5)$, $Q(1,2,3)$ and $R(7, 0, -1)$ are collinear.

ID: 981 Type: Subjective Source: NCERT Class 11 Year: null

Question 4: Are the points $A(3,6,9)$, $B(10, 20, 30)$ and $C(25,-41,5)$, the vertices of a right angled triangle?

ID: 982 Type: Subjective Source: NCERT Class 11 Year: null

Question 5: Find the equation of set of points P such that $PA^2+PB^2=2k^2$, where A and B are the points $(3, 4, 5)$ and $(-1,3,-7)$, respectively.

ID: 983 Type: Subjective Source: NCERT Class 11 Year: null

Question 6: Show that the points $A(1,2,3)$, $B(-1,-2,-1)$, $C(2,3,2)$ and $D(4,7,6)$ are the vertices of a parallelogram ABCD, but it is not a rectangle.

ID: 984 Type: Subjective Source: NCERT Class 11 Year: null

Question 7: Find the equation of the set of the points P such that its distances from the points $A(3, 4, -5)$ and $B(-2, 1, 4)$ are equal.

ID: 985 Type: Subjective Source: NCERT Class 11 Year: null

Question 8: The centroid of a triangle ABC is at the point $(1, 1, 1)$. If the coordinates of A and B are $(3, –5, 7)$ and $(-1, 7, – 6)$, respectively, find the coordinates of the point C.

ID: 986 Type: Subjective Source: NCERT Class 11 Year: null

Question 9: A point is on the $x$-axis. What are its $y$-coordinate and $z$-coordinates?

ID: 987 Type: Subjective Source: NCERT Class 11 Year: null

Question 10: A point is in the XZ-plane. What can you say about its $y$-coordinate?

ID: 988 Type: Subjective Source: NCERT Class 11 Year: null

Question 11: Name the octants in which the following points lie: $(1, 2, 3)$, $(4, -2, 3)$, $(4, -2, -5)$, $(4, 2, 5)$, $(–4, 2, -5)$, $(–4, 2, 5)$, $(-3, -1, 6)$, $(-2, -4, -7)$ .

ID: 989 Type: Subjective Source: NCERT Class 11 Year: null

Question 12: Fill in the blanks: (i) The $x$-axis and $y$-axis taken together determine a plane known as

ID: 990 Type: Subjective Source: NCERT Class 11 Year: null

Question 13: Fill in the blanks: (ii) The coordinates of points in the XY-plane are of the form

ID: 991 Type: Subjective Source: NCERT Class 11 Year: null

Question 14: Fill in the blanks: (iii) Coordinate planes divide the

ID: 992 Type: Subjective Source: NCERT Class 11 Year: null

Question 15: Find the distance between the following pairs of points: (i) $(2, 3, 5)$ and $(4, 3, 1)$

ID: 993 Type: Subjective Source: NCERT Class 11 Year: null

Question 16: Find the distance between the following pairs of points: (ii) $(-3, 7, 2)$ and $(2,4,-1)$

ID: 994 Type: Subjective Source: NCERT Class 11 Year: null

Question 17: Find the distance between the following pairs of points: (iii) $(-1,3,-4)$ and $(1,-3,4)$

ID: 995 Type: Subjective Source: NCERT Class 11 Year: null

Question 18: Find the distance between the following pairs of points: (iv) $(2,-1,3)$ and $(-2,1,3)$

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🤖 AI Generated Solution:

**Correct Option if MCQ:** N/A **Reasoning:** * $d = \sqrt{(-2-2)^2 + (1-(-1))^2 + (3-3)^2}$ * $d = \sqrt{(-4)^2 + (2)^2 + (0)^2}$ * $d = \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5}$

ID: 996 Type: Subjective Source: NCERT Class 11 Year: null

Question 19: Show that the points $(-2, 3, 5)$, $(1, 2, 3)$ and $(7, 0, -1)$ are collinear.

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🤖 AI Generated Solution:

**Reasoning:** * Vector AB = (1 - (-2), 2 - 3, 3 - 5) = (3, -1, -2) * Vector BC = (7 - 1, 0 - 2, -1 - 3) = (6, -2, -4) * Vector BC = 2 * Vector AB, hence collinear.

ID: 997 Type: Subjective Source: NCERT Class 11 Year: null

Question 20: Verify the following: (i) $(0,7,-10)$, $(1, 6, -6)$ and $(4,9,-6)$ are the vertices of an isosceles triangle.

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🤖 AI Generated Solution:

**Reasoning:** * Calculate the squared distance between pairs of vertices: * $d^2((0,7,-10),(1,6,-6)) = (1-0)^2 + (6-7)^2 + (-6 - (-10))^2 = 1 + 1 + 16 = 18$ * $d^2((1,6,-6),(4,9,-6)) = (4-1)^2 + (9-6)^2 + (-6 - (-6))^2 = 9 + 9 + 0 = 18$ * $d^2((0,7,-10),(4,9,-6)) = (4-0)^2 + (9-7)^2 + (-6 - (-10))^2 = 16 + 4 + 16 = 36$ * Two sides have equal length ($\sqrt{18}$). * Therefore, the triangle is isosceles.

ID: 998 Type: Subjective Source: NCERT Class 11 Year: null

Question 21: Verify the following: (ii) $(0, 7, 10)$, $(-1, 6, 6)$ and $(-4,9,6)$ are the vertices of a right angled triangle.

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🤖 AI Generated Solution:

**Reasoning:** * Calculate the squared lengths of the sides: $PQ^2 = 1^2 + 1^2 + 4^2 = 18$, $QR^2 = 3^2 + (-3)^2 + 0^2 = 18$, $PR^2 = 4^2 + (-2)^2 + 4^2 = 36$. * Check if the Pythagorean theorem holds: $18 + 18 = 36$. * $PQ^2 + QR^2 = PR^2$, confirming it's a right-angled triangle.

ID: 999 Type: Subjective Source: NCERT Class 11 Year: null

Question 22: Verify the following: (iii) $(-1, 2, 1)$, $(1, -2, 5)$, $(4, -7, 8)$ and $(2, -3, 4)$ are the vertices of a parallelogram.

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🤖 AI Generated Solution:

**Correct Option if MCQ:** (iii) **Reasoning:** * $\vec{AB} = (2, -4, 4)$ and $\vec{DC} = (2, -4, 4)$ * $\vec{AD} = (3, -5, 3)$ and $\vec{BC} = (3, -5, 3)$ * Opposite sides are equal and parallel.

ID: 1000 Type: Subjective Source: NCERT Class 11 Year: null

Question 23: Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

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🤖 AI Generated Solution:

**Correct Option if MCQ:** Not Applicable **Reasoning:** * Let $(x, y, z)$ be a point equidistant from $(1, 2, 3)$ and $(3, 2, -1)$. * $(x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y-2)^2 + (z+1)^2$ * $x^2 - 2x + 1 = x^2 - 6x + 9 \Rightarrow 4x = 8 \Rightarrow x = 2$

ID: 1001 Type: Subjective Source: NCERT Class 11 Year: null

Question 24: 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 $10$.

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🤖 AI Generated Solution:

**Correct Option if MCQ:** C **Reasoning:** * The set of points P forms an ellipse with foci at $A(4,0,0)$ and $B(-4,0,0)$. * The sum of the distances $PA + PB = 10$, so $2a = 10$, which means $a = 5$. * The distance between foci is $2c = 8$, so $c = 4$. For an ellipse, $b^2 = a^2 - c^2 = 5^2 - 4^2 = 25 - 16 = 9$. The equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

ID: 1002 Type: Subjective Source: NCERT Class 11 Year: null

Question 25: Three vertices of a parallelogram ABCD are $A(3,-1,2)$, $B(1,2,-4)$ and $C(-1,1,2)$. Find the coordinates of the fourth vertex.

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🤖 AI Generated Solution:

**Correct Option if MCQ:** D **Reasoning:** * $\vec{AB} = B - A = (-2, 3, -6)$ * $\vec{DC} = C - D$ * $B - A = C - D \implies D = C - (B - A) = C - B + A$ * $D = (-1, 1, 2) - (1, 2, -4) + (3, -1, 2) = (1, -2, 6)$

ID: 1003 Type: Subjective Source: NCERT Class 11 Year: null

Question 26: Find the lengths of the medians of the triangle with vertices $A(0,0,6)$, $B(0,4,0)$ and $(6, 0, 0)$.

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🤖 AI Generated Solution:

**Correct Option if MCQ:** N/A **Reasoning:** * $m_a = \sqrt{(\frac{0+6}{2}-0)^2 + (\frac{4+0}{2}-0)^2 + (\frac{0+0}{2}-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7$ * $m_b = \sqrt{(\frac{0+6}{2}-0)^2 + (\frac{0+0}{2}-4)^2 + (\frac{6+0}{2}-0)^2} = \sqrt{9+16+9} = \sqrt{34}$ * $m_c = \sqrt{(\frac{0+0}{2}-6)^2 + (\frac{0+4}{2}-0)^2 + (\frac{6+0}{2}-0)^2} = \sqrt{36+4+9} = \sqrt{49} = 7$

ID: 1004 Type: Subjective Source: NCERT Class 11 Year: null

Question 27: If the origin is the centroid of the triangle PQR with vertices $P(2a,2,6)$, $Q(-4,3b,-10)$ and $R(8,14,2c)$, then find the values of $a, b$ and $c$.

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🤖 AI Generated Solution:

**Correct Option if MCQ:** Not applicable (this is a calculation question) **Reasoning:** * Centroid formula: $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$ * $(\frac{2a-4+8}{3}, \frac{2+3b+14}{3}, \frac{6-10+2c}{3}) = (0,0,0)$ * $a = -2, b = -16/3, c = 2$

ID: 1005 Type: Subjective Source: NCERT Class 11 Year: null

Question 28: If A and B be the points $(3, 4, 5)$ and $(-1, 3, -7)$, respectively, find the equation of the set of points P such that $PA^2+PB^2=k^2$, where $k$ is a constant.