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

The coordinates of a point are (2, -3, 5). The point lies in the __________ octant.

Correct! Wrong!

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Clearly, the y coordinate of the given point is less than zero. So, it should be measured along -ve Y-axis. So, the octant is XOY'Z.

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

Correct! Wrong!

Let the points be A(p, q, r), B(q, r, p) and C(r, p, q). Using distance formula, AB2 = (q - p)2 + (r - q)2 + (p - r)2 Similarly, BC2 = (r -q)2 + (p - q)2 + (q - p)2 = AB2 (from the above step) AC2 = (r - p)2 + (p - q)2 + (q - r)2 = (p - r)2 + (q - p)2 + (r - q)2 = A B2 So, AB = BC = AC Hence the triangle is equilateral.

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

Correct! Wrong!

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 ____.

Correct! Wrong!

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) = 12+72+12 = 1 + 49 +1 AD = 51 units

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

Correct! Wrong!

The foot of the perpendicular from the point P in the point A(l, o, o) on the x - axis. AP = l - l2+m2+n2=m2+n2 (using distance formula) similarly, the other distances are: n2+l2 and l2+m2.

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

Correct! Wrong!

Using distance formula, AB2 = (26)2 ⇒ (2 - a)2 + (2a - 3)2 + ( 0 + 5)2= 26 ⇒ 4 + a2 - 4a + 4a2+ 9 - 12a + 25 = 26 ⇒ 5a2 - 16a + 12 = 0 ⇒ 5a2 - 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

Correct! Wrong!

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 ____.

Correct! Wrong!

Let P(x, y, z) be the point. Therefore, AP = x - 32+y2+z2 BP = x+32+y2+z2 Given, AP + BP = 3 => AP = 3- BP AP2 = 9 + BP2 - 6BP (x - 3)2 + y2 + z2 = 9 + (x + 3)2 + y2 + z2 - 6 x + 32+y2+z2 x2 + 9 -6x + y2 + z2 = 9 + x2 + 9 + 6x + y2 + z2 – (6 x+3)2+y2+z2 -6x = 9 - 6x+32+y2+z2 -2x = 3 - 2 (x + 3)2+y2+z2 2x+32+y2+z2 = 3 + 2x, Squaring both sides, We get, 4 [ (x + 3)2 + y2 + z2 ] = (3 + 2x)2 4 [x2 + y2 + z2 + 6x + 9] = 9 + 4x2 +12x 4x2 + 4y2 + 4z2 + 24x + 36 - 4x2- 12x - 9 = 0 4y2 + 4z2 + 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 ____.

Correct! Wrong!

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

Correct! Wrong!

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 (x1, 0, z1), where x1 ≠ 0 & z1 ≠ 0.

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

Correct! Wrong!

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?

Correct! Wrong!

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 ____.

Correct! Wrong!

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 = 12+32+22 = 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 ____.

Correct! Wrong!

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, AP2 = 52 + 22 + (z - 4)2 = 782 ⇒ 25 + 4 + z2 - 8z + 16 = 78 ⇒ z2 - 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 ____.

Correct! Wrong!

Given, PA/PB = 4/5 25PA2 = 16PB2 25 {(x + 3)2 + (y - 1)2 + (z + 2)2} = 16 { (x - 4)2 + (y - 2)2 + z2 } [ using distance formula] 25(x2 + 6x + 9 + y2 - 2y + 1 + z2 + 4z + 4) = 16 (x2 - 8x + 16 + y2 - 4y + 4 + z2) x2(25 - 16) + y2 (25 - 16) + z2 (25 - 16) + x(150 + 128) +y(-50 + 64) + 100z + 325 - 64 = 0 9x2 + 9y2 + 9z2+ 278x +14y + 100z + 261 = 0

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