Which ordered triple represents the point that is 4 units on x-axis only
A (0,4,0)
B (0,0,4)
C (4,0,0)
D (4,4,0)
A point lying only on the x-axis must have y=0 and z=0. Moving 4 units along x gives coordinates (4,0,0), measured from the origin.
The point (3,−2,5) lies in which octant sign pattern
A (−,+,+)
B (+,−,+)
C (+,+,−)
D (−,−,+)
Octants are identified by signs of x, y, z. For (3,−2,5), x is positive, y is negative, and z is positive, so it lies in (+,−,+) octant.
If direction ratios are (1,2,2), the corresponding direction cosines are
A (1/3,2/3,2/3)
B (1/√6,2/√6,2/√6)
C (1/√5,2/√5,2/√5)
D (1/2,1/2,1/2)
Convert DR to DC by dividing each component by √(1²+2²+2²)=√9=3. So direction cosines are (1/3,2/3,2/3), written as (1/√9,2/√9,2/√9).
A line makes equal angles with y- and z-axes when
A l = m
B l = n
C m = n
D l+m+n=1
Direction cosines are (l,m,n) where m=cosβ and n=cosγ. Equal angles with y and z mean β=γ, so cosβ=cosγ, giving m=n (with same sign).
Distance between points (1,2,3) and (4,2,7) equals
A √41 units
B 5 units
C √34 units
D 6 units
Differences are Δx=3, Δy=0, Δz=4. Distance is √(3²+0²+4²)=√(9+16)=√25=5. So √25 units is the exact form.
A point divides segment joining A and B externally when the ratio is
A m:n with same signs
B m=n always
C ratio must be 1:1
D m:n with opposite signs
External division means the dividing point lies outside the segment. In section formula, it is handled by taking one ratio effectively negative, indicating opposite direction weights.
For plane 2x−3y+6z+5=0, a normal vector is
A (5,2,−3)
B (2,3,6)
C (2,−3,6)
D (−2,−3,−6)
In plane ax+by+cz+d=0, the coefficients (a,b,c) form a normal vector. Here a=2, b=−3, c=6, so (2,−3,6) is perpendicular to the plane.
The angle between planes depends mainly on
A their normal vectors
B their d values
C their intercepts only
D their areas only
The angle between two planes is defined as the angle between their normals. Using dot product of normals gives cosθ, independent of plane shifts (d values).
If a line’s direction vector is (a,b,c), then a perpendicular plane must have normal
A perpendicular to (a,b,c)
B parallel to (a,b,c)
C zero normal vector
D (a²,b²,c²)
A line is perpendicular to a plane when it is parallel to the plane’s normal. So if the line direction is (a,b,c), a plane perpendicular to it must have normal proportional to (a,b,c).
Line through (1,1,1) with direction ratios (2,−1,3) in parametric form is
A x=1−2t,y=1−t,z=1−3t
B x=2+t,y=−1+t,z=3+t
C x=1+2t,y=1−t,z=1+3t
D x=1+2t,y=1+t,z=1+3t
Parametric line form is x=x1+at, y=y1+bt, z=z1+ct. Substitute point (1,1,1) and direction (2,−1,3) to get x=1+2t, y=1−t, z=1+3t.
A line intersects plane ax+by+cz+d=0 at one point when it is
A parallel not lying
B lying completely
C parallel to normal
D not parallel plane
If a line is not parallel to a plane, it will cut the plane at exactly one point. If it is parallel, it either never meets or lies entirely in the plane.
Condition for a point P(x1,y1,z1) to lie on plane x+2y−z=4 is
A x1−2y1+z1=4
B x1+2y1−z1=4
C x1+2y1+z1=4
D x1y1z1=4
A point lies on a plane if its coordinates satisfy the plane equation exactly. Substitute x1, y1, z1 into x+2y−z and check it equals 4.
The line of intersection of two planes is generally
A a point only
B a plane itself
C a line
D a sphere
Two distinct non-parallel planes intersect in a straight line. Only special cases are parallel planes (no intersection) or coincident planes (infinite intersections).
If two lines have direction ratios (1,0,1) and (0,1,0), then their dot product is
A 1
B 0
C 2
D −1
Treat direction ratios as vectors u=(1,0,1), v=(0,1,0). Dot product u·v =1·0+0·1+1·0=0, so they are perpendicular directions.
If direction cosines are (l,m,n) and l=m=0, then n must be
A ±1 only
B 0 only
C 1/2 only
D any real number
Since l²+m²+n²=1, if l=0 and m=0, then n²=1. So n must be +1 or −1, meaning the line is along positive or negative z-axis.
Distance from point (1,2,3) to plane x−2y+2z=5 equals
A |1−4+6−5|/3
B |1−4+6−5|/√9
C |1−4+6−5|/√(1+4+4)
D |1−4+6−5|/√(1+2+2)
Use distance formula |ax1+by1+cz1+d|/√(a²+b²+c²). Here plane is x−2y+2z−5=0. Numerator becomes |1−4+6−5| and denominator √(1²+(-2)²+2²)=√9.
A plane parallel to x-axis and y-axis must be parallel to
A yz-plane
B xy-plane
C zx-plane
D any plane
If a plane is parallel to both x-axis and y-axis, then it has no tilt along x or y directions. Its normal must be along z-axis, so it is parallel to xy-plane.
If a line has symmetric form (x−2)/1=(y+1)/2=(z−0)/3, then it passes through
A (1,2,3)
B (0,0,0)
C (2,1,0)
D (2,−1,0)
In symmetric form (x−x1)/a=(y−y1)/b=(z−z1)/c, the point on the line is (x1,y1,z1). Here it is (2,−1,0).
A line is in the plane when its direction is
A parallel to normal
B same as d value
C perpendicular to normal
D along intercepts
Any direction lying in the plane must be perpendicular to the plane’s normal vector. So a line in the plane has direction vector whose dot product with normal equals zero.
If u and v are direction vectors of two lines, the lines are skew if they are
A non-parallel, non-coplanar
B parallel and distinct
C intersecting always
D perpendicular always
Skew lines do not intersect and are not parallel, and they lie in different planes. Their directions alone don’t guarantee skew, but non-coplanarity confirms it.
Angle between direction ratios (1,2,2) and (2,1,2) is found using
A distance formula
B dot product
C section formula
D midpoint formula
To find angle between two lines, treat direction ratios as vectors and apply cosθ=(u·v)/(|u||v|). This works in 3D for coplanar or skew directions.
For u=(1,2,2) and v=(2,1,2), u·v equals
A 7
B 6
C 8
D 5
Dot product is 1·2 + 2·1 + 2·2 = 2 + 2 + 4 = 8. This value is then used in the cosθ formula with magnitudes of u and v.
Magnitude of direction vector (1,2,2) is
A √8
B 3
C √5
D √10
Magnitude is √(1²+2²+2²)=√(1+4+4)=√9=3. Writing √9 is fine and helps connect directly to direction cosine conversion.
A line through origin with direction ratios (a,b,c) has parametric form
A x=at,y=bt,z=ct
B x=a+t,y=b+t,z=c+t
C x=a,y=b,z=c
D x=t/a,y=t/b,z=t/c
Through origin means x1=y1=z1=0. Using x=0+at, y=0+bt, z=0+ct gives x=at, y=bt, z=ct, which represents all points on that line.
If direction ratios are (3,−3,0), the line is parallel to
A xz-plane
B yz-plane
C xy-plane
D z-axis
A direction vector with z-component 0 lies parallel to the xy-plane because it has no movement in z direction. So (3,−3,0) indicates direction entirely in xy-plane.
A plane parallel to z-axis must have equation not involving
A z term required
B x must be zero
C y must be zero
D z may be missing
If a plane is parallel to z-axis, moving along z stays within the plane. That happens when z does not appear in its equation, like ax+by+d=0, forming a vertical plane.
A plane perpendicular to z-axis must be of form
A x = constant
B z = constant
C y = constant
D x+y = constant
A plane perpendicular to z-axis has normal along z-axis. That means its equation is z=k. Such a plane is parallel to xy-plane and cuts z-axis at z=k.
The locus of points equidistant from two fixed points in 3D is a
A sphere
B line
C plane
D cone
Points equidistant from two points form the perpendicular bisector plane of the segment joining them. It passes through the midpoint and is perpendicular to the segment direction vector.
If P(1,2,3) and Q(3,4,5), midpoint coordinates are
A (4,6,8)
B (1,2,3)
C (3,4,5)
D (2,3,4)
Midpoint is found by averaging corresponding coordinates: ((1+3)/2, (2+4)/2, (3+5)/2) = (2,3,4). It lies exactly halfway between P and Q.
If a point divides PQ internally in ratio 2:1, it is closer to
A P point
B Q point
C origin always
D midpoint always
In ratio m:n, the dividing point is closer to the endpoint with smaller weight. For 2:1, smaller weight is 1 at Q side, so point lies closer to Q.
If a plane passes through points (1,0,0), (0,1,0), (0,0,1), its equation is
A x+y+z=1
B x+y+z=0
C x−y+z=1
D xyz=1
These are intercept points on x, y, z axes with intercepts 1,1,1. Using intercept form x/1 + y/1 + z/1 = 1 gives x+y+z=1.
A line parallel to plane x+y+z=1 must satisfy for direction vector v
A v×(1,1,1)=0
B |v|=1 always
C v·(1,1,1)=0
D v=(1,1,1)
Plane normal is (1,1,1). Line parallel to plane has direction perpendicular to normal, so dot product with (1,1,1) must be zero.
For vectors u and v, if u×v is nonzero, the vectors are
A always perpendicular
B not parallel
C always equal
D always unit
Cross product magnitude is |u||v|sinθ. It is zero only when sinθ=0, meaning vectors are parallel or anti-parallel. Nonzero cross product means they are not parallel.
For two planes with normals n1 and n2, they are parallel when
A n1×n2=0
B n1·n2=0
C n1+n2=0 only
D |n1|=|n2|
If normals are parallel, their cross product is zero. That indicates the planes have same orientation, hence the planes are parallel (or coincident).
Distance between two parallel planes 2x+3y−6z+4=0 and 2x+3y−6z−8=0 is
A 4/√49
B 8/√49
C 12/√49
D 16/√49
Distance between parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 is |d1−d2|/√(a²+b²+c²). Here |4−(−8)|=12 and √(4+9+36)=√49.
A point is on zx-plane when which condition holds
A x = 0
B z = 0
C x = y
D y = 0
The zx-plane contains z- and x-axes and is defined by y=0. So any point with y=0 lies on zx-plane regardless of x and z values.
If a line has direction cosines (l,m,n), then the sum l+m+n is
A always 1
B not fixed
C always 0
D always 3
Only the relation l²+m²+n²=1 must hold. The simple sum l+m+n can vary depending on direction and signs, so it is not a constant value.
The projection length of vector a on b equals
A (a×b)/|b|
B |a|/|b|
C (a·b)/|b|
D (a·b)/|a|
Scalar projection (component) of a on b is (a·b)/|b|. It measures how much of vector a lies along the direction of b, with sign showing same or opposite direction.
The vector projection of a on b equals
A [(a×b)/|b|²] b
B [(a·b)/|b|²] b
C [(a·b)/|a|²] a
D [(a·a)/|b|²] b
Vector projection gives a vector along b. Multiply unit direction of b by scalar component: proj_b(a) = [(a·b)/|b|²] b. This is standard in 3D vector geometry.
If a plane’s normal is (0,0,1), then plane is parallel to
A xy-plane
B yz-plane
C zx-plane
D y-axis
Normal (0,0,1) is along z-axis, same as normal of xy-plane. Planes with this normal are of form z=k and are parallel to xy-plane.
If two lines are perpendicular, their direction vectors satisfy
A cross product zero
B ratios equal
C dot product zero
D sums equal
Two lines are perpendicular when their direction vectors have angle 90°. Using dot product u·v=|u||v|cos90°=0, so dot product must be zero.
A line is parallel to z-axis if its direction ratios are
A (k,0,0)
B (0,k,0)
C (k,k,0)
D (0,0,k)
Parallel to z-axis means only z component is nonzero. So direction ratios must be (0,0,k) where k≠0, representing movement purely along z direction.
If line direction ratios are (1,2,3), then a perpendicular direction can be
A (1,2,3)
B (2,−1,0)
C (2,4,6)
D (0,0,1)
Perpendicular vectors have dot product zero. (1,2,3)·(2,−1,0)=2−2+0=0. So (2,−1,0) is a valid direction perpendicular to the given one.
A point-to-line shortest distance in space is measured along
A any segment joining
B parallel to line
C perpendicular from point
D along x-axis only
The shortest distance from a point to a line is the length of the perpendicular segment from the point to the line. Any other joining segment is longer by geometry.
If two planes are coincident, their equations are
A proportional all coefficients
B different normals only
C perpendicular always
D different intercepts only
Coincident planes represent the same set of points, so one equation is a nonzero scalar multiple of the other. That means a:b:c:d are proportional across both equations.
For sphere (x−a)²+(y−b)²+(z−c)²=r², the center is
A (−a,−b,−c)
B (a,b,c)
C (r,r,r)
D (0,0,0)
Standard sphere form shows shifts from x, y, z by constants a, b, c. These represent the coordinates of the center. Radius is r, determined by r² on right side.
If equation x²+y²=16 has no z term, the surface is
A sphere radius 4
B plane parallel xy
C cylinder along z
D cone along z
Since z does not appear, z can take any value. The cross-section in xy-plane is a circle x²+y²=16 of radius 4, extended along z-axis forming a cylinder.
If equation x²+y²=z², the surface represents a
A cylinder along z
B sphere at origin
C plane through origin
D right circular cone
x²+y²=z² is a cone with vertex at origin and axis along z-axis. For fixed z, x²+y² is a circle of radius |z|, showing conical opening.
If a line has direction cosines (1/2,1/2,1/√2), then l²+m²+n² equals
A 2
B 1
C 1/2
D √2
Compute squares: (1/2)²+(1/2)²+(1/√2)² = 1/4+1/4+1/2 = 1. This confirms it is a valid set of direction cosines.
If a plane has equation 3x−4y+0z+7=0, it is parallel to
A z-axis direction
B xy-plane only
C z=constant planes
D x-axis only
Since z term is missing, any change in z does not affect the equation. So the plane extends parallel to z-axis, meaning it is parallel to the z-direction (contains lines parallel to z-axis).