2 Marks Questions

Question 12 Marks

Using section formula, show that the points A(2, -3, 4), B(-1, 2, 1) and $\text{C}\Big(0, \frac{1}{3}, 2\Big)$ are collinear.

Answer

A(2, -3, 4), B(-1, 2, 1) and $\text{C}\Big(0, \frac{1}{3}, 2\Big)$
DR's of AB are (3, -5, 3)
DR's of BC are $\Big(-1, \frac{5}{3},-1\Big)$
DR's of AC are $\Big(2, \frac{-10}{3},2\Big)$
Its clear that all DR's are proportional.

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Question 22 Marks

Find the coordinates of the points which tisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).

Answer

Tri section points are those which divide line in ratio 1 : 2 or 2 : 1

P(4, 2, -6) and Q(10, -16, 6)

Consider 1 : 2 case, we get

$\Big(\frac{10+8}{3},\frac{-16+4}{3},\frac{6–12}{3}\Big)= (6, -4, -2)$

Consider 2 : 1 case, we get

$\Big(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3}\Big)=(8, -10, 2)$

(6, -4, -2) and (8, -10, 2) are tri section points.

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Question 32 Marks

Given that P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear. Find the ratio in which Q divides PR.

Answer

P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10)
$\text{PQ}=\sqrt{4+4+4}=2\sqrt{3}$
$\text{QR}=\sqrt{16+16 +16}=4\sqrt{3}$
$\text{PQ}:\text{QR}=1:2$

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Question 42 Marks

Find the distance between the following pairs of points:
A(3, 2, -1) and B(-1, -1, -1).

Answer

Distance between points A and B

$\text{AB}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

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

$=\sqrt{(4)^2+(3)^2+(0)^2}$

$=\sqrt{16+9+0}$

$=\sqrt{25}$

$\text{AB = 5 units}$

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Question 52 Marks

Find the distance between the points P and Q having coordinates (-2, 3, 1) and (2, 1, 2).

Answer

Distance between points P and Q

$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

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

$=\sqrt{(-4)^2+(2)^2+(-1)^2}$

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

$\text{PQ}=\sqrt{21}\text{ units}$

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Question 62 Marks

Find the distance between the following pairs of points:
P(1, -1, 0) and Q(2, 1, 2)

Answer

Distance between points P and Q

$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$

$=\sqrt{(1-2)^2+(-1-1)^2+(0-2)^2}$

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

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

$\text{PQ = 3 units}$

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Question 72 Marks

Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.

Answer

Here points are A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7).

$\text{AB}=\sqrt{(3-0)^2+(3-6)^2+(3-3)^2}$

$=\sqrt{9+9}$

$=3\sqrt{2}\text{ units}$

$\text{BC}=\sqrt{(0-1)^2+(6-7)^2+(3-7)^2}$

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

$=3\sqrt{2}\text{ units}$

$\text{AC}=\sqrt{(3-1)^2+(3-7)^2+(3-7)^2}$

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

$=6\text{ units}$

$\text{BD}=\sqrt{(0-4)^2+(6-4)^2+(3-7)^2}$

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

$=6\text{ units}$

$\text{CD}=\sqrt{(1-4)^2+(7-4)^2+(7-7)^2}$

$=\sqrt{9+9}$

$=3\sqrt{2}\text{ units}$

$\text{AD}=\sqrt{(3-4)^2+(3-4)^2+(3-7)^2}$

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

$=3\sqrt{2}\text{ units}$

Since, AB = BC = CD = DA

And AC = BC

So,

A, B, C, D are vertices of a square.

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Question 82 Marks

The coordinates of a point are (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.

Answer

The seven coordinates are as follows:
(-3, -2, -5)
(-3, -2, 5)
(3, -2, -5)
(-3, 2, -5)
(3, 2, 5)
(3, 2, -5)
(-3, 2, 5)

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Question 92 Marks

Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the yz-plane.

Answer

(4, 8, 10) and (6, 10, -8) is divided by the yz-plane.

Equation of yz-plane is x = 0

assume ratio is m : n

Equating x-term, we get

$0=\frac{\text{6m+4n}}{\text{m+n}}$

m : n=-2 : 3

So YZ plane divides the line segment in ratio 2 : 3 externally

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