11th Grade > Mathematics
INTRODUCTION TO THREE DIMENSIONAL GEOMETRY MCQs
Introduction To Three Dimensional Geometry
Total Questions : 49
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Answer: Option A. -> x = 2z
:
A
Let P(x,y,z) represent the set of points whichequidistant from A(1,2,3) and B(3,2,-1).
Equating AP=BP using the distance formula, we have
(x−1)2+(y−2)2+(z−3)2=(x−3)2+(y−2)2+(z+1)2⇒−2x−4y−6z=−6x−4y+2z⇒4x−8z=0⇒x=2z
:
A
Let P(x,y,z) represent the set of points whichequidistant from A(1,2,3) and B(3,2,-1).
Equating AP=BP using the distance formula, we have
(x−1)2+(y−2)2+(z−3)2=(x−3)2+(y−2)2+(z+1)2⇒−2x−4y−6z=−6x−4y+2z⇒4x−8z=0⇒x=2z
Answer: Option B. -> P=(95,25,−15)
:
B
Let P(x,y,z) divide the line segment joining A(1,-2,3) and B(3,4,-5) internally in the ratio 2:3. Therefore,
x=2×3+3×12+3=95y=2×4+3×(−2)2+3=25z=2×(−5)+3×32+3=−15
P=(95,25,−15)
:
B
Let P(x,y,z) divide the line segment joining A(1,-2,3) and B(3,4,-5) internally in the ratio 2:3. Therefore,
x=2×3+3×12+3=95y=2×4+3×(−2)2+3=25z=2×(−5)+3×32+3=−15
P=(95,25,−15)
Answer: Option A. -> - ,+, -
:
A
For a point in octantVI, the x and z coordinates are negative and the y coordinate is positive.
:
A
For a point in octantVI, the x and z coordinates are negative and the y coordinate is positive.
Answer: Option A. -> True
:
A
Let A=(4,-2,3), B=(5,1,2) and C=(7,7,0).
AB=√(5−4)2+(1−(−2))2+(2−3)2=√12+32+12=√11
BC=√(7−5)2+(7−1)2+(0−2)2=√22+62+22=√44=2√11
AC=√(7−4)2+(7−(−2))2+(0−3)2=√32+92+32=√99=3√11
AC=AB+BC. Hence, A, B and C are collinear. The statement is true.
:
A
Let A=(4,-2,3), B=(5,1,2) and C=(7,7,0).
AB=√(5−4)2+(1−(−2))2+(2−3)2=√12+32+12=√11
BC=√(7−5)2+(7−1)2+(0−2)2=√22+62+22=√44=2√11
AC=√(7−4)2+(7−(−2))2+(0−3)2=√32+92+32=√99=3√11
AC=AB+BC. Hence, A, B and C are collinear. The statement is true.
Answer: Option A. -> (1,2,2)
:
A
Let C(x,y,z) be the centroid of the triangle whose vertices are (0,3,5), (1,4,-2) and (2,-1,3)
x=0+1+23=33=1y=3+4−13=63=2z=5−2+32=63=2
The centroid of the triangle is (1,2,2)
:
A
Let C(x,y,z) be the centroid of the triangle whose vertices are (0,3,5), (1,4,-2) and (2,-1,3)
x=0+1+23=33=1y=3+4−13=63=2z=5−2+32=63=2
The centroid of the triangle is (1,2,2)
Answer: Option B. -> 5 units
:
B
Using the distance formula, the distance between the points (3,0,4) and (-1,3,4) is
d=√(−1−3)2+(3−0)2+(4−4)2=√42+52+02=√25=5units
:
B
Using the distance formula, the distance between the points (3,0,4) and (-1,3,4) is
d=√(−1−3)2+(3−0)2+(4−4)2=√42+52+02=√25=5units
Answer: Option A. -> 1:2
:
A
Let Qdivide PR in the ratio k:1. Equate the x-coordinateof the point of division to 5.
∴3+9kk+1=53+9k=5k+54k=2k=12
Since it is given that P, Q and Rare collinear, we need to calculate k onlyfor any one of thecoordinates.
:
A
Let Qdivide PR in the ratio k:1. Equate the x-coordinateof the point of division to 5.
∴3+9kk+1=53+9k=5k+54k=2k=12
Since it is given that P, Q and Rare collinear, we need to calculate k onlyfor any one of thecoordinates.
Answer: Option D. -> II, III, VI, VII
:
D
The x-coordinate of a point is negative in the octants II, III, VI, VII
:
D
The x-coordinate of a point is negative in the octants II, III, VI, VII
Answer: Option D. -> None
:
D
The point (2,0,3) lies on the x-z plane as its y-coordinate is 0. Hence it is part of no octant.
:
D
The point (2,0,3) lies on the x-z plane as its y-coordinate is 0. Hence it is part of no octant.
Answer: Option C. -> √43 units
:
C
The distance between the points (-3,7,2) and (2,4,-1),
d=√(−3−2)2+(7−4)2+(−1−2)2d=√43
:
C
The distance between the points (-3,7,2) and (2,4,-1),
d=√(−3−2)2+(7−4)2+(−1−2)2d=√43