In 3D Cartesian system, how many coordinate axes are there
A One axis only
B Two axes only
C Three mutually perpendicular
D Four axes total
A 3D coordinate system uses three perpendicular axes: x, y, and z. Any point in space is located using an ordered triple (x, y, z) measured along these axes.
The three coordinate planes in 3D are
A ab, bc, ca
B xy, xz, yz only
C x, y, z planes
D xy, yz, zx
In 3D, each coordinate plane is formed by a pair of axes: xy-plane (z=0), yz-plane (x=0), and zx-plane (y=0). They divide space into different regions.
How many octants are formed by three coordinate planes
A Eight octants
B Four octants
C Six octants
D Ten octants
The three coordinate planes intersect and divide space into 8 parts called octants. Each octant corresponds to a specific sign combination of (x, y, z).
If a point lies on the xy-plane, its z-coordinate is
A Always 1
B Positive only
C Always 0
D Negative only
The xy-plane consists of all points where z=0. So any point lying on the xy-plane must have zero z-coordinate, while x and y can be any real numbers.
Distance between P(x1,y1,z1) and Q(x2,y2,z2) equals
A |x1−x2| only
B √[(Δx)²+(Δy)²+(Δz)²]
C (Δx)+(Δy)+(Δz)
D √[(Δx)+(Δy)+(Δz)]
In 3D, distance uses Pythagoras in three perpendicular directions. Compute differences Δx, Δy, Δz and apply √(Δx²+Δy²+Δz²) for the straight-line distance.
Distance of point P(x,y,z) from origin is
A √(x+y+z)
B x²+y²+z²
C |x|+|y|+|z|
D √(x²+y²+z²)
Origin is (0,0,0). Using 3D distance formula from P to origin gives √[(x−0)²+(y−0)²+(z−0)²] = √(x²+y²+z²).
Midpoint of P(x1,y1,z1) and Q(x2,y2,z2) is
A (x1+x2, y1+y2, z1+z2)
B (x1−x2, y1−y2, z1−z2)
C ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)
D ((x1−x2)/2, …)
The midpoint lies halfway along the segment joining the two points. So each coordinate is the average of corresponding coordinates of the endpoints in 3D.
A point lies on yz-plane if its coordinate is
A x = 0
B y = 0
C z = 0
D x = y
The yz-plane is formed by y- and z-axes, so x-coordinate becomes zero for every point on that plane. y and z may take any real values.
Direction cosines (l, m, n) satisfy
A l+m+n = 1
B l²+m²+n² = 0
C lm+mn+nl = 1
D l²+m²+n² = 1
Direction cosines are cosines of angles made by a line with x, y, z axes. Since they come from a unit direction vector, their squares sum to 1.
Direction ratios (a, b, c) are related to direction cosines by
A Same always
B Must be integers
C Proportional values
D Always between −1 and 1
Direction ratios give a direction vector up to a nonzero scalar multiple. Direction cosines are normalized values. So (a, b, c) ∝ (l, m, n).
If (a,b,c) are direction ratios, direction cosines are
A (a+b+c, …)
B (a/√S, b/√S, c/√S)
C (√a, √b, √c)
D (1/a, 1/b, 1/c)
Convert direction ratios to a unit vector. Here S = a²+b²+c². Dividing each component by √S gives direction cosines that satisfy l²+m²+n²=1.
If a line is parallel to x-axis, its direction ratios can be
A (0,1,0)
B (1,1,0)
C (1,0,0)
D (0,0,1)
A line parallel to x-axis has direction only along x. So y and z components are zero. Any nonzero multiple like (2,0,0) also works.
Two lines are perpendicular if their direction ratios satisfy
A a1/a2 = b1/b2
B a1b2 + b1c2 + c1a2 = 0
C a1+a2 = 0
D a1a2 + b1b2 + c1c2 = 0
Direction ratios act like direction vectors. Two lines are perpendicular when their direction vectors have zero dot product, giving a1a2+b1b2+c1c2 = 0.
Two lines are parallel if their direction ratios are
A Proportional only
B Negative always
C Perpendicular always
D Equal sum only
Parallel lines have the same direction. So their direction ratios (a1,b1,c1) and (a2,b2,c2) must be proportional: a1/a2 = b1/b2 = c1/c2.
Angle between two lines with direction ratios (a1,b1,c1) and (a2,b2,c2) uses
A Cross product only
B Midpoint formula
C Dot product formula
D Section formula
Angle θ between direction vectors is found by cosθ = (a1a2+b1b2+c1c2)/[√(a1²+b1²+c1²)√(a2²+b2²+c2²)]. This works for skew or coplanar lines.
If direction cosines are (l,m,n), angle with x-axis is
A sin⁻¹(l)
B cos⁻¹(l)
C tan⁻¹(l)
D cos⁻¹(m)
By definition, l = cosα where α is angle between line and x-axis. So α = cos⁻¹(l). Similarly, β = cos⁻¹(m), γ = cos⁻¹(n).
A sphere with center at origin and radius r has equation
A x+y+z = r
B x²+y² = r²
C x²+z² = r²
D x²+y²+z² = r²
All points at fixed distance r from origin satisfy distance formula √(x²+y²+z²)=r. Squaring gives the standard sphere equation x²+y²+z²=r².
General equation of a plane in 3D is
A ax²+by²+cz² = 0
B xy+yz+zx = 0
C ax+by+cz+d = 0
D ax+by = 0 only
A plane is a first-degree surface in x, y, z. The coefficients a, b, c represent components of a normal vector to the plane, and d shifts the plane.
Normal vector to plane ax+by+cz+d=0 is
A (a, b, c)
B (a+b+c)
C (d, a, b)
D (ab, bc, ca)
For plane ax+by+cz+d=0, vector (a,b,c) is perpendicular to the plane. Any line along this vector is normal to the plane and makes right angle with it.
Distance of point (x1,y1,z1) from plane ax+by+cz+d=0 is
A |ax1+by1+cz1+d|
B √(a²+b²+c²)
C |d|/√(x1²+y1²+z1²)
D |ax1+by1+cz1+d|/√(a²+b²+c²)
Substitute the point into plane equation to get signed value. Divide its absolute value by magnitude of normal vector √(a²+b²+c²) to obtain perpendicular distance.
Intercept form of plane cutting axes at a, b, c is
A ax+by+cz = 1
B x+y+z = abc
C x/a + y/b + z/c = 1
D x/a + y/b = 1
If a plane meets x-, y-, z-axes at (a,0,0), (0,b,0), (0,0,c), its intercept form becomes x/a + y/b + z/c = 1.
Equation of line through point (x1,y1,z1) with direction ratios (a,b,c) is
A x+y+z = 0
B (x−x1)/a = (y−y1)/b = (z−z1)/c
C ax+by+cz = 0
D x/x1 = y/y1 = z/z1
A line in symmetric form uses a point and a direction vector. Each ratio equals a parameter t, representing movement along the direction (a,b,c) from the given point.
Parametric form of the same line can be written as
A x = x1+at, y = y1+bt, z = z1+ct
B x = x1−at only
C ax+by+cz = 1
D x²+y²+z² = t
Taking the common ratio as parameter t, we get x−x1 = at, y−y1 = bt, z−z1 = ct. This gives parametric equations for the line.
Direction ratios of line joining P(x1,y1,z1) and Q(x2,y2,z2) are
A (x1+x2, y1+y2, z1+z2)
B (x1x2, y1y2, z1z2)
C (x1−y1, y1−z1, …)
D (x2−x1, y2−y1, z2−z1)
The direction vector from P to Q is obtained by subtracting coordinates: Q−P. This gives direction ratios, and any proportional triple represents the same direction.
If two planes are perpendicular, their normals satisfy
A n1 × n2 = 0
B n1 = n2
C n1 · n2 = 0
D |n1| = |n2|
Angle between planes equals angle between their normal vectors. Two planes are perpendicular when their normals are perpendicular, i.e., dot product of normals is zero.
Angle between planes ax+by+cz+d=0 and a’x+b’y+c’z+d’=0 depends on
A normals (a,b,c) and (a’,b’,c’)
B d and d’ only
C intercepts only
D plane areas
The angle between planes is the angle between their normal vectors. Use cosθ = (aa’+bb’+cc’)/[√(a²+b²+c²)√(a’²+b’²+c’²)].
A line is perpendicular to a plane if line direction is
A Parallel to plane
B Perpendicular to normal
C Same as intercepts
D Along plane normal
A plane’s normal vector is perpendicular to the plane. If a line’s direction vector is parallel to the normal, then the line makes a 90° angle with the plane.
A line is parallel to a plane if line direction vector is
A Perpendicular to plane
B Equal to normal
C Perpendicular to normal
D Zero vector
A line is parallel to a plane when it has no component along the plane’s normal direction. So the line direction vector must be perpendicular to the normal vector.
If l, m, n are direction cosines, then each lies between
A 0 and 1 only
B −1 and 1
C 1 and 2
D −2 and 2
Direction cosines are cosines of angles with axes. Cosine values always lie between −1 and 1. Their signs depend on the octant and direction of the line.
If a point divides PQ internally in ratio m:n, its coordinates are
A (mx1+nx2)/(m+n) etc
B (x1+x2)/2 always
C (mx1−nx2)/(m−n) etc
D (mx2+nx1)/(m+n) etc
Internal section formula in 3D gives weighted average closer to the larger weight. If P is (x1,y1,z1) and Q is (x2,y2,z2), the dividing point is (mx2+nx1)/(m+n), similarly for y and z.
If m=n in internal division of PQ, the point becomes
A Origin always
B Same as P
C Midpoint of PQ
D Same as Q
When m=n, the section formula gives equal weights to P and Q. So coordinates become averages of corresponding coordinates, which is exactly the midpoint of the segment PQ.
Condition for collinearity of points P, Q, R in 3D is
A PQ vector ∥ PR vector
B PQ = QR always
C x1+x2+x3 = 0
D y1=y2=y3
Three points are collinear if the vectors PQ and PR are parallel, meaning one is a scalar multiple of the other. This ensures all three lie on one straight line.
A basic locus of points at fixed distance r from (a,b,c) is
A Plane
B Line
C Cone
D Sphere
All points whose distance from a fixed point (a,b,c) equals r form a sphere. Using 3D distance formula and squaring gives (x−a)²+(y−b)²+(z−c)²=r².
Direction cosines of line through points P and Q are obtained by
A Adding coordinates
B Using plane equation
C Normalizing PQ vector
D Taking midpoint only
Direction ratios are (x2−x1, y2−y1, z2−z1). Divide each by the vector magnitude √(Δx²+Δy²+Δz²) to get direction cosines.
If two lines have direction ratios (1,2,3) and (2,4,6), they are
A Perpendicular
B Parallel
C Intersecting always
D Skew always
Since (2,4,6) is exactly 2 times (1,2,3), direction ratios are proportional. That means directions are the same, so the lines are parallel (may be distinct or coincident).
If direction ratios (a,b,c) satisfy a²+b²+c² = 0, then
A Line is x-axis
B Line is plane
C Angles are acute
D Vector is zero
Sum of squares equals zero only when a=b=c=0. That gives a zero vector, which cannot represent a valid direction for a line. Direction ratios must not all be zero.
Angle between lines is acute when their dot product is
A Negative
B Zero
C Positive
D Undefined
For direction vectors u and v, u·v = |u||v|cosθ. If dot product is positive, cosθ>0 so θ is acute. If negative, θ is obtuse; if zero, θ is right.
A plane parallel to xy-plane has equation form
A z = constant
B ax+by+cz = 0
C x = constant
D y = constant
Any plane parallel to xy-plane has the same normal as xy-plane, i.e., along z-axis. So it is given by z = k, where k is a constant.
A plane parallel to yz-plane has equation form
A x = constant
B x = constant
C z = constant
D x+y = constant
The yz-plane has equation x=0. Any plane parallel to it keeps the same normal along x-axis, so it becomes x = k for some constant k.
Vector form of line through position vector a with direction b is
A r = a × b
B r · b = 0
C r = a + tb
D r = t(a+b)
In vector form, any point r on the line is obtained by starting from a fixed point with position vector a and moving t units along direction vector b.
If a line has direction cosines (0,1,0), it is parallel to
A x-axis
B z-axis
C xy-plane
D y-axis
Direction cosines indicate direction components along axes. (0,1,0) means no x or z component and full y component, so the line is parallel to the y-axis.
If a line has equal angles with all three axes, its direction cosines are
A (1,0,0)
B (1/2, 1/2, 0)
C (1/√3, 1/√3, 1/√3)
D (0,1/√2,1/√2)
Equal angles imply l=m=n. Using l²+m²+n²=1 gives 3l²=1, so l=1/√3. Same for m and n (taking same sign for same octant).
The intersection of two non-parallel planes is generally a
A Line
B Point
C Plane
D Sphere
Two planes in 3D usually intersect in a straight line, unless they are parallel (no intersection) or coincident (infinite intersections). Non-parallel distinct planes meet in a line.
A line lies in a plane if it satisfies the plane equation for
A One point only
B Three points only
C No points
D Two points on line
If two distinct points of a line satisfy the plane equation, the entire line lies in the plane because a plane containing two points of a line contains the whole line segment and its extension.
The shortest distance between skew lines is measured along
A Any joining segment
B Common perpendicular
C Parallel direction only
D Coordinate axes only
Skew lines do not intersect and are not parallel. The shortest distance between them is the length of the segment perpendicular to both lines, called the common perpendicular.
Distance between two parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 is
A |d1−d2|
B √(a²+b²+c²)
C |d1−d2|/√(a²+b²+c²)
D |a+b+c|
Parallel planes have same normal (a,b,c). The perpendicular distance depends on difference in constants, scaled by magnitude of normal vector √(a²+b²+c²).
Projection of vector a on vector b uses
A Cross product
B Determinant only
C Section formula
D Dot product
Projection depends on component of a along b. Using dot product, scalar projection is (a·b)/|b|, and vector projection is [(a·b)/|b|²] b.
If two lines intersect, then they must be
A Coplanar lines
B Always parallel
C Always skew
D Perpendicular always
Intersecting lines always lie in a single plane because two intersecting lines determine a unique plane. Skew lines cannot intersect since they are non-coplanar.
The point-to-plane distance is always
A Negative value
B Non-negative value
C Complex number
D Zero always
Distance is a length, so it cannot be negative. The formula uses an absolute value to ensure non-negative result. It becomes zero only when the point lies on the plane.
A translation of axes in 3D changes
A Distances between points
B Angles between lines
C Coordinates of points
D Shape of figure
Translating axes shifts the origin, so coordinates of points change relative to the new origin. However, distances, angles, and shapes remain unchanged because the geometry is preserved.