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Three Dimensional Geometry

Multiple Choice Questions

121.

The angle between the curves, y = x and y² - x = 0 at the point (1, 1) is

$\frac{π}{2}$
$\tan^{- 1} (\frac{4}{3})$
$\frac{π}{3}$
$\tan^{- 1} (\frac{3}{4})$

122.

If the distance between (2, 3) and (- 5, 2) is equal to the distance between (x, 2) and (1, 3), then the values of x are

- 6, 8
6, 8
- 8, 6
- 7, 7

123.

The vertices of a triangle are A(3, 7), B (3, 4) and C (5, 4). The equation of the bisector of the angle ABC is

y = x + 1
y = x - 1
y = 3x - 5
y = x

124.

If the angle between a and c is 25°, the angle between b and c is 65° and a + b = c, then the angle between a and b is

40°
115°
25°
90°

125.

The projection of the vector 2i + aj - k on the vector i - 2j + k is $\frac{- 5}{\sqrt{6}}$ . Then, the value of a is equal to

126.
A unit vector in the XOY-plane that makes an angle 30° with the vector i + j and makes an angle 60° with i - j is

$\frac{1}{4} [(\sqrt{6} + \sqrt{2}) i - (\sqrt{6} - \sqrt{2}) j]$

$\frac{1}{2} [(\sqrt{6} - \sqrt{2}) i + (\sqrt{6} + \sqrt{2}) j]$

$\frac{1}{4} [(\sqrt{6} - \sqrt{2}) i + (\sqrt{6} + \sqrt{2}) j]$

$\frac{1}{4} [(\sqrt{6} + \sqrt{2}) i + (\sqrt{6} - \sqrt{2}) j]$

$\frac{1}{4} [(\sqrt{6} + \sqrt{2}) i + (\sqrt{6} - \sqrt{2}) j]$

Let a unit vector in XOY-plane is, $\hat{a} = \frac{a}{|a|}$

$a = a_{1} i + a_{2} j, |a| = \sqrt{{a_{1}}^{2} + {a_{2}}^{2}} Given vectors let b = i + j and c = i - j From question; \hat{a} . b = |\hat{a}| . |b| \cos (30 °) = 1 . \sqrt{2} . \frac{\sqrt{3}}{2} \Rightarrow \frac{\sqrt{6}}{2} \Rightarrow \frac{a . b}{|a|} = \frac{\sqrt{6}}{2} \Rightarrow a_{1} + a_{2} = \frac{\sqrt{6}}{2} |a| . . . (i) and \hat{a} . c = |\hat{a}| . |c| \cos (60 °) = 1 . \sqrt{2} . \frac{1}{2} \Rightarrow \frac{a . c}{|a|} = \frac{\sqrt{2}}{2} |a| . . . (ii) From Eqs . (i) and (ii), 2 a_{1} = (\sqrt{6} + \sqrt{2}) \frac{|a|}{2} a_{1} = \frac{1}{4} (\sqrt{6} + \sqrt{2}) |a| \Rightarrow \frac{a_{1}}{|a|} = \frac{1}{4} (\sqrt{6} + \sqrt{2}) and a_{2} = \frac{1}{4} (\sqrt{6} - \sqrt{2}) |a| \Rightarrow \frac{a_{2}}{|a|} = \frac{1}{4} (\sqrt{6} - \sqrt{2}) So, the unit vector is \hat{a} = \frac{1}{|a|} (a_{1} i + a_{2} j) \hat{a} = \frac{1}{4} \{(\sqrt{6} + \sqrt{2}) i + (\sqrt{6} - \sqrt{2}) j\}$

127.

The angle between the line r = (i + 2j + 3k) + $λ$ (2i + 3j + 4k) and the plane r - (i + j - 2k) = 0 is

0°
60°
30°
90°

128.

The lines r = i + j - k + (3i - j) and r = 4j - k + µ (2i + 3k) intersect at the point

(0, 0, 0)
(0, 0, 1)
(0, - 4, - 1)
(4, 0, - 1)

129.

An equation of the plane through the points (1, 0, 0) and (0, 2, 0) and at a distance $\frac{6}{7}$ units from the origin is

6x + 3y + z - 6 = 0
6x + 3y + 2z - 6= 0
6x + 3y + z + 6 = 0
6x + 3y + 2z + 6 = 0

130.

The projection of a line segment on the axes are 9, 12 and 8. Then, the length of the line segment is

Previous Year Papers

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Multiple Choice Questions

126. A unit vector in the XOY-plane that makes an angle 30° with the vector i + j and makes an angle 60° with i - j is 146 + 2i - 6 - 2j 126 - 2i + 6 + 2j 146 - 2i + 6 + 2j 146 + 2i + 6 - 2j