(Each question 4 marks)

Question 14 Marks

Verify the following: (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.

Answer

Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively. $\text{AB}=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}$ $=\sqrt{(-1)^2+(-1)^2+(-4)^2}$ $=\sqrt{1+1+16}=\sqrt{18}$ $=3\sqrt{2}$ $\text{BC}=\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}$ $=\sqrt{(-3)^2+(3)^2+(0)^2}$ $=\sqrt{9+9}=\sqrt{18}$ $=3\sqrt{2}$ $\text{CA}=\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}$ $=\sqrt{(4)^2+(-2)^2+(4)^2}$ $=\sqrt{16+4+16}$ $=\sqrt{36}$ $=6$ Now, $\text{AB}^2+\text{BC}^2=(3\sqrt{2})^2+(3\sqrt{2})^2=18+18=36=\text{AC}^2$ Therefore, by Pythagoras theorem, ABC is a right triangle. Hence, the given points are the vertices of a right-angled triangle.

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Question 24 Marks

Verify the following: (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Answer

Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C respectively. $\text{AB}=\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2}$ $=\sqrt{(1)^2+(-1)^2+(4)^4}$ $=\sqrt{1+1+16}$ $=\sqrt{18}$ $=3\sqrt{2}$ $\text{BC}=\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2}$ $=\sqrt{(3)^2+(3)^2}$ $=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$ $\text{CA}=\sqrt{(0-4)^2+(7-9)^2+(-10+6)^2}$ $=\sqrt{(-4)^2+(-2)^2+(-4)^2}$ $=\sqrt{16+4+16}=\sqrt{36}=6$ Here, AB = BC ≠ CA Thus, the given points are the vertices of an isosceles triangle.

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Question 34 Marks

Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Answer

Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D respectively. $\text{AB}=\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2}$ $=\sqrt{4+16+16}$ $=\sqrt{36}$ $=6$ $\text{BC}=\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2}$ $=\sqrt{9+25+9}=\sqrt{43}$ $\text{CD}=\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2}$ $=\sqrt{4+16+16}$ $=\sqrt{36}$ $=6$ $\text{DA}=\sqrt{(-1-2)^2+(2+3)^3+(1-4)^2}$ $=\sqrt{9+25+9}=\sqrt{43}$ Here, $\text{AB = CD = 6, BC = AD =}\sqrt{43}$ Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal. Therefore, ABCD is a parallelogram. Hence, the given points are the vertices of a parallelogram.

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Question 44 Marks

A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q(8, 0, 10). Find the coordinates of the point R. [Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by $\Big(\frac{8\text{k}+2}{\text{k}+1},\frac{-3}{\text{k}+1},\frac{10\text{k}+4}{\text{k}+1}\Big)]$

Answer

The coordinates of points P and Q are given as P(2, –3, 4) and Q(8, 0, 10). Let R divide line segment PQ in the ratio k : 1. Hence, by section formula, the coordinates of point R are given by $\Big(\frac{\text{k}(8)+2}{\text{k}+1},\frac{\text{k}(0)-3}{\text{k}+1},\frac{\text{k}(10)+4}{\text{k}+1}\Big)=\Big(\frac{8\text{k}+2}{\text{k}+1},\frac{-3}{\text{k}+1},\frac{10\text{k}+4}{\text{k}+1}\Big)$ It is given that the x-coordinate of point R is 4. $\therefore\frac{8\text{k}+2}{\text{k}+1}=4$ $\Rightarrow8\text{k}+2=4\text{k}+4$ $\Rightarrow4\text{k}=2$ $\Rightarrow\text{k}=\frac{1}{2}$ Therefore, the coordinates of point R are $\begin{pmatrix}4,\frac{-3}{\frac{1}{2}+1},\frac{10\big(\frac{1}{2}\big)+4}{\frac{1}{2}+1}\end{pmatrix}=(4,-2,6)$

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Question 54 Marks

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.

Answer

The three vertices of a parallelogram ABCD are given as A(3, –1, 2), B(1, 2, –4), and C(–1, 1, 2). Let the coordinates of the fourth vertex be D(x, y, z).

We know that the diagonals of a parallelogram bisect each other. Therefore, in parallelogram ABCD, AC and BD bisect each other. $\therefore$ Mid-point of AC = Mid-point of BD $\Rightarrow\Big(\frac{3-1}{2},\frac{-1+1}2{},\frac{2+2}{2}\Big)=\Big(\frac{\text{x}+1}{2},\frac{\text{y}+2}{2},\frac{\text{z}-4}{2}\Big)$ $\Rightarrow(1,0,2)=\Big(\frac{\text{x}+1}{2},\frac{\text{y}+2}{2},\frac{\text{z}-4}{2}\Big)$ $\Rightarrow\frac{\text{x}+1}{2}=1,\frac{\text{y}+2}{2},\text{ and }\frac{\text{z}-4}{2}=2$ $\Rightarrow\text{x}=1,\text{y}=-2\text{ and }\text{z}=8$ Thus, the coordinates of the fourth vertex are (1, –2, 8).

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Question 64 Marks

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

Answer

Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively. Points P, Q, and R are collinear if they lie on a line. $\text{PQ}=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}$ $=\sqrt{(3)^3+(-1)^2+(-2)^2}$ $=\sqrt{9+1+4}$ $=\sqrt{14}$ $\text{QR}=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}$ $=\sqrt{(6)^2+(-2)^2+(-4)^2}$ $=\sqrt{36+4+16}$ $=\sqrt{56}$ $=2\sqrt{14}$ $\text{PR}=\sqrt{(7+2)^2+(0+3)^2+(-1-5)^2}$ $=\sqrt{(9)^2+(-3)^2+(-6)^2}$ $=\sqrt{81+9+36}$ $=\sqrt{126}$ $=3\sqrt{14}$ Here, $\text{PQ + QR}=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=\text{PR}$ Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

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Question 74 Marks

Find the lengths of the medians of the triangle with vertices A(0, 0, 6), B(0,4, 0) and C(6, 0, 0).

Answer

Let AD, BE, and CF be the medians of the given triangle ABC.

Since AD is the median, D is the mid-point of BC. $\therefore$ Coordinates of point D $=\Big(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\Big)=(3,2,0)$ $\text{AD}=\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}=\sqrt{9+4+36}=\sqrt{49}=7$ Since BE is the median, E is the mid-point of AC. $\therefore$ Coordinates of point E $=\Big(\frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2}\Big)=(3,0,3)$ $\text{BE}=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}$ Since CF is the median, F is the mid-point of AB. $\therefore$ Coordinates of point F $=\Big(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\Big)=(0,2,3)$ Length of $\text{CF}=\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}=\sqrt{36+4+9}=\sqrt{49}=7$ Thus, the lengths of the medians of $\triangle\text{ABC}$ are $7,\sqrt{34},$ and $7$

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