# CBSE Class 11th Math 12 – Introduction to Three Dimensional Geometry MCQs

#### The points (p, q, r), (q, r, p) and (r, p, q) are vertices of a(n) ____.

Let the points be A(p, q, r), B(q, r, p) and C(r, p, q). Using distance formula, AB^{2} = (q - p)^{2} + (r - q)^{2} + (p - r)^{2 } Similarly, BC^{2} = (r -q)^{2} + (p - q)^{2} + (q - p)^{2} = AB^{2} (from the above step) AC^{2} = (r - p)^{2} + (p - q)^{2} + (q - r)^{2} = (p - r)^{2} + (q - p)^{2 }+ (r - q)^{2} = A B^{2} So, AB = BC = AC Hence the triangle is equilateral.

#### Name the planes to which the points (0, 1, -2) and (3, 4, 0) belong.

Any point on the Y-Z plane has its x co-ordinate equal to zero. Therefore, the first point is on the Y-Z plane. Similarly, the second point is on the X-Y plane.

#### The length of the median AD of a triangle ABC with vertices A(2, 3, 1), B(5, −5, 1) and C(1, −3, 3) is ____.

The midpoint 'D' of BC is [(5+1/2), (−5−3/2), (1+3/2)] (i.e.) [(6/2), (−8/2), (4/2) or (3, −4, 2). AD = (2, −, 3)^{2} + (3, −, (, −, 4,))^{2 }+ (1, −, 2)^{2} (By distance formula) = 1^{2}+7^{2}+1^{2} = 1 + 49 +1 AD = 51 units

#### The perpendicular distance from the point P(l, m, n) from the co - ordinate axes are ____.

The foot of the perpendicular from the point P in the point A(l, o, o) on the x - axis. AP = l - l^{2}+m^{2}+n^{2}=m^{2}+n^{2} (using distance formula) similarly, the other distances are: n^{2}+l^{2 }and l^{2}+m^{2}.

#### The distance between the points A(a, 3, -5) and B(2, 2a, 0) is 26 units. The values of a are ____.

Using distance formula, AB^{2} = (26)^{2} ⇒ (2 - a)^{2} + (2a - 3)^{2} + ( 0 + 5)^{2}= 26 ⇒ 4 + a^{2 }- 4a + 4a^{2}+ 9 - 12a + 25 = 26 ⇒ 5a^{2} - 16a + 12 = 0 ⇒ 5a^{2} - 10a - 6a + 12 = 0 ⇒ 5a( a - 2) - 6 (a - 2) = 0 ⇒ (5a - 6) (a - 2) = 0 Therefore, a = 6/5 and a = 2

#### The co-ordinates of the foot of the perpendiculars from the point (a, b, c) to the co-ordinate axes are

On the X-axis, the co-ordinates of the foot of the perpendicular from (a, b, c) is (a, o, o), because, on X-axis, the Y & Z co-ordinates are zero. With similar reasoning, we can also find that the co-ordinates of the foot of the perpendiculars on Y & Z axes are: (o, b, o) & (o, o, c) respectively.

#### The equation of the set of points P, the sum of whose distance from A(3, 0, 0) & B(-3, 0, 0) is equal to ____.

Let P(x, y, z) be the point. Therefore, AP = x - 3^{2}+y^{2}+z^{2} BP = x+3^{2}+y^{2}+z^{2} Given, AP + BP = 3 => AP = 3- BP AP^{2 }= 9 + BP^{2} - 6BP (x - 3)^{2} + y^{2} + z^{2} = 9 + (x + 3)^{2} + y^{2} + z^{2} - 6 x + 3^{2}+y^{2}+z^{2} x^{2} + 9 -6x + y^{2} + z^{2} = 9 + x^{2} + 9 + 6x + y^{2} + z^{2} – (6 x+3)^{2}+y^{2}+z^{2} -6x = 9 - 6x+3^{2}+y^{2}+z^{2} -2x = 3 - 2 (x + 3)^{2}+y^{2}+z^{2} 2x+3^{2}+y^{2}+z^{2} = 3 + 2x, Squaring both sides, We get, 4 [ (x + 3)^{2} + y^{2} + z^{2} ] = (3 + 2x)^{2} 4 [x^{2} + y^{2} + z^{2} + 6x + 9] = 9 + 4x^{2 }+12x 4x^{2} + 4y^{2} + 4z^{2} + 24x + 36 - 4x^{2}- 12x - 9 = 0 4y^{2} + 4z^{2 }+ 12x + 27 = 0.

#### The ratio in which the line joining A(4, −2, 6) and B(2, −3, 2) is divided by the Y-axis is ____.

Let the ratio be k: 1 any point on the Y−axis is (0, y1, 0) Let P be the point of division. By section formula, P is [(2k−4)/(k+1), (−3k−2/k+1), (4k−8/k+1)] Equating the x or the z co−ordinate to 0, we get, k = 2 Therefore, the ratio is 2 : 1.

#### The co- ordinates on any point in the X-Z plane is of the form

Any point on the X-Z plane has its y co-ordinate equal to zero. The co-ordinates of any point, hence, takes the form (x_{1}, 0, z_{1}), where x1 ≠ 0 & z_{1} ≠ 0.

#### The co-ordinate planes divide the space into _______________ parts known as ____________

The co-ordinate planes divide the space into 8 parts known as octants

#### What is the image of (-5, 2, 3) in the Y-Z plane?

The image of the given point P(-5, 2, 3) in the Y-Z plane will be as much below the Y-Z plane as P is above it. So, the image is (5, 2, 3).

#### The points A(5, -8, 3), B(4, -5, 1) and C(3, -2, -1) are ____.

Using distance formula, AB = (4-5)^{2 }+ (-5+8)^{2 }+(1-3)^{2} = 1 + 9 + 4 =14 units BC = (3-4)^{2 }+ (-2+5)^{2 }+ (-1-1)^{2 }= 1^{2}+3^{2}+2^{2} = 14 units AC = (3-5)^{2 }+ (-2+8)^{2 }+ (-1-3)^{2}= 4 + 36 +16 = 56 AC = 214 units So, we find that, AB + BC = AC Therefore, A, B & C are collinear.

#### The co-ordinates of a point on the Z axis which is at a distance of 78 units from the point (5, -2, 4) are ____.

Any point on the Z axis can be taken as A(0, 0, z). Let P be the point (5, -2, 4) Therefore, using distance formula, AP^{2} = 5^{2} + 2^{2} + (z - 4)^{2} = 78^{2} ⇒ 25 + 4 + z^{2} - 8z + 16 = 78 ⇒ z^{2} - 8z -33 = 0 ⇒ (z - 11) (z + 3) = 0 ⇒ z = 11, z = -3 Therefore, the points are (0, 0, 11) and (0, 0, -3)

#### The locus of a point P which moves such that PA: PB = 4: 5, where, A and B are the points (-3, 1, -2) and (4, 2, 0) respectively, is ____.

Given, PA/PB = 4/5 25PA^{2} = 16PB^{2} 25 {(x + 3)^{2 }+ (y - 1)^{2} + (z + 2)^{2}} = 16 { (x - 4)^{2} + (y - 2)^{2} + z^{2} } [ using distance formula] 25(x^{2} + 6x + 9 + y^{2} - 2y + 1 + z^{2} + 4z + 4) = 16 (x^{2} - 8x + 16 + y^{2} - 4y + 4 + z^{2}) x^{2}(25 - 16) + y^{2} (25 - 16) + z^{2} (25 - 16) + x(150 + 128) +y(-50 + 64) + 100z + 325 - 64 = 0 9x^{2 }+ 9y^{2} + 9z^{2}+ 278x +14y + 100z + 261 = 0