A point has coordinates (−3,4,5). What is its perpendicular distance from the yz-plane
A 4
B 5
C 3
D √50
The yz-plane is x = 0. The perpendicular distance of (x,y,z) from yz-plane equals |x|. Here x = −3, so distance is |−3| = 3 units.
A point P(2,−1,3) is given. What is its perpendicular distance from the xz-plane
A 1
B 2
C 3
D √14
The xz-plane is y = 0. Distance of point (x,y,z) from xz-plane equals |y|. Here y = −1, so the distance is |−1| = 1 unit.
Point divides A(0,1,2) and B(6,4,8) internally in ratio 1:2. What are its coordinates
A (4,3,6)
B (3,2,5)
C (2,2,4)
D (2,3,4)
For internal ratio m:n (AP:PB = 1:2), coordinates are ((n x1 + m x2)/(m+n), similarly for y and z). This gives (2,2,4).
A line has direction ratios (3, −4, 12). What are its direction cosines
A (3/12,−4/12,1)
B (3/13,−4/13,12/13)
C (3/7,−4/7,12/7)
D (3/14,−4/14,12/14)
Direction cosines are obtained by dividing direction ratios by their magnitude. Magnitude is √(3²+(-4)²+12²)=√169=13. So cosines are (3/13, −4/13, 12/13).
A line has direction cosines (1/3, 2/3, 2/3). What is its angle with the z-axis
A cos⁻¹(2/3)
B cos⁻¹(1/3)
C cos⁻¹(−2/3)
D cos⁻¹(0)
Direction cosine n equals cosγ, where γ is the angle with z-axis. Here n = 2/3, so γ = cos⁻¹(2/3). Signs decide octant, not the method.
A line is x=1+2t, y=t, z=2−t. For which value of t does the point lie on the xy-plane
A 0
B 1
C −2
D 2
A point lies on the xy-plane when z = 0. Here z = 2 − t. Setting 2 − t = 0 gives t = 2. Then the corresponding point is on xy-plane.
Check whether point (5,2,−1) lies on line (x−1)/2 = (y−0)/1 = (z−2)/(−1)
A Lies on line
B Parallel to line
C Not on line
D Perpendicular line
A point lies on the line if the three ratios give the same parameter value. Here (5−1)/2 = 2, (2−0)/1 = 2, but (−1−2)/(−1) = 3, so it does not lie.
Find distance between parallel planes 3x−4y+12z+5=0 and 3x−4y+12z−7=0
A 12/13
B 13/12
C 12
D 13
Distance between parallel planes ax+by+cz+d1=0 and ax+by+cz+d2=0 is |d1−d2|/√(a²+b²+c²). Here |5−(−7)|=12 and √(9+16+144)=13.
A plane is parallel to xz-plane and passes through (2,−3,5). What is its equation
A x = 2
B y = −3
C z = 5
D y = 3
Any plane parallel to xz-plane has equation y = constant, since xz-plane itself is y = 0. Passing through y = −3 fixes the constant as −3.
A plane is perpendicular to the y-axis and passes through (0,4,0). Which equation fits
A x = 4
B z = 4
C y = 4
D x+y = 4
A plane perpendicular to y-axis has normal along y-axis, so it must be of form y = constant. Passing through (0,4,0) gives y = 4.
Two planes x+2y+2z=0 and 2x+y−2z=0 are at what angle
A 90°
B 0°
C 45°
D 60°
Angle between planes equals angle between normals. Normals are (1,2,2) and (2,1,−2). Dot product = 2+2−4=0, so normals are perpendicular, hence planes are perpendicular.
Plane has normal (2,−1,1) and passes through (1,2,3). Which equation matches
A 2x+y+z−3=0
B x−2y+z−3=0
C 2x−y−z−3=0
D 2x−y+z−3=0
Use point-normal form: 2(x−1)−1(y−2)+1(z−3)=0. Expanding gives 2x−2−y+2+z−3=0, so 2x−y+z−3=0.
A line has direction (1,2,2). A plane has normal (2,−1,0). What is the relation
A Perpendicular
B Coincident
C Parallel
D No relation
A line is parallel to a plane when its direction vector is perpendicular to the plane’s normal. Here (1,2,2)·(2,−1,0)=2−2+0=0, so the line is parallel.
Two lines have direction ratios (1,1,0) and (1,−1,0). What is the angle between them
A 0°
B 90°
C 60°
D 180°
Treat direction ratios as vectors. Dot product is 1·1 + 1·(−1) + 0·0 = 0. Zero dot product means the lines’ directions are perpendicular, so the angle is 90°.
Lines L1: (1,2,t) and L2: (4,6,t) are parallel. What is their shortest distance
A 5
B √41
C √17
D 1
Both lines have same z-parameter, so separation comes from x and y only: Δx=3, Δy=4. Shortest distance is √(3²+4²)=5, measured perpendicular to both lines.
Lines r=(1,0,0)+t(1,1,1) and r=(0,1,0)+s(1,1,1) are
A Coincident
B Intersecting
C Parallel distinct
D Perpendicular
Both have the same direction (1,1,1), so they are parallel. The vector between given points (−1,1,0) is not a multiple of (1,1,1), so they are not the same line.
Find distance of origin from plane 2x+2y+z−6=0
A 3
B 6
C 1
D 2
Distance from origin to plane ax+by+cz+d=0 is |d|/√(a²+b²+c²). Here d=−6 and √(4+4+1)=3, so distance is 6/3=2.
A line is equally inclined to x and y axes and makes 60° with z-axis. Direction cosines are
A (√6/4,√6/4,1/2)
B (1/2,1/2,1/2)
C (√3/3,√3/3,√3/3)
D (√2/2,√2/2,0)
Equal inclination to x and y gives l=m. Given angle with z is 60°, so n=cos60°=1/2. Using l²+m²+n²=1 gives 2l²+1/4=1, so l=m=√6/4.
For A(1,1,1) and B(2,3,4), direction ratios of AB are
A (2,3,4)
B (3,2,1)
C (1,2,3)
D (1,1,1)
Direction ratios from A to B are (x2−x1, y2−y1, z2−z1). Here differences are (2−1, 3−1, 4−1) = (1,2,3), giving the line’s direction.
Points (1,2,3), (2,4,6), (3,6,9) are
A Not collinear
B Collinear
C Form triangle
D Form tetrahedron
Vectors between successive points are proportional: (2−1,4−2,6−3)=(1,2,3) and (3−2,6−4,9−6)=(1,2,3). Parallel direction vectors mean all three points lie on one line.
Which equation represents a sphere with center (2,−1,3) and radius 4
A x²+y²+z²=16
B (x+2)²+(y−1)²+(z+3)²=16
C (x−2)²+(y+1)²+(z−3)²=16
D (x−2)²+(y−1)²+(z−3)²=4
Sphere standard form is (x−a)²+(y−b)²+(z−c)²=r². Here a=2, b=−1 so (y+1)², c=3, and r²=16.
Which point lies on sphere x²+y²+z²=9
A (1,2,2)
B (2,2,2)
C (3,1,1)
D (0,2,2)
A point lies on the sphere if x²+y²+z²=9. For (1,2,2), sum is 1+4+4=9, so it satisfies the equation exactly.
For line (x−1)/1=(y−2)/2=(z−3)/3, the point at parameter t=2 is
A (2,4,6)
B (1,2,3)
C (3,6,9)
D (4,8,12)
Write parametric form: x=1+t, y=2+2t, z=3+3t. Substituting t=2 gives x=3, y=6, z=9. This is a specific point on the line.
A line has direction ratios (2,−1,2). What is its angle with x-axis
A cos⁻¹(1/3)
B cos⁻¹(2/3)
C cos⁻¹(−1/3)
D cos⁻¹(0)
Angle with x-axis has cosine l = a/√(a²+b²+c²). Here √(4+1+4)=3, so l=2/3. Therefore the required angle is cos⁻¹(2/3).
A line lies in the xz-plane. Which condition must hold for its direction cosines (l,m,n)
A m = 0
B l = 0
C n = 0
D l = m
A line lying in xz-plane has no y-direction component. Direction cosine m represents cosine with y-axis, so the y-component of unit direction vector is zero, giving m=0.
A plane contains the z-axis and also passes through point (1,2,0). Which equation matches
A y = −2x
B x = 2y
C x+y = 0
D y = 2x
A plane containing z-axis passes through all (0,0,z), so it passes through origin and includes direction (0,0,1). With point (1,2,0), direction (1,2,0) also lies in plane. Normal from cross product gives equation y=2x.
A plane parallel to yz-plane passes through (2,−1,4). What is its equation
A y = −1
B z = 4
C x = 2
D x+y = 1
Any plane parallel to yz-plane has form x = constant because yz-plane is x=0. Passing through x=2 fixes the constant as 2, so equation is x = 2.
Find distance of point (1,2,3) from plane x+2y+2z=9
A 2/3
B 3/2
C 2
D 3
Write plane as x+2y+2z−9=0. Substitute point: 1+4+6−9=2. Denominator √(1²+2²+2²)=3. Distance is |2|/3 = 2/3 units.
Two lines have direction vectors (1,2,3) and (2,4,6). What is the acute angle between them
A 60°
B 0°
C 90°
D 45°
Vectors (2,4,6) are a multiple of (1,2,3), so directions are parallel. The acute angle between parallel directions is 0°. (They could be same or distinct lines.)
Two lines have direction vectors (1,0,1) and (1,0,−1). What is their angle
A 0°
B 60°
C 90°
D 180°
Dot product is 1·1 + 0·0 + 1·(−1) = 0. Zero dot product means perpendicular directions, so the angle between the two lines is 90°.
In the shortest distance formula between two skew lines, the numerator uses
A Dot product
B Midpoint formula
C Section formula
D Scalar triple product
The shortest distance between skew lines is found using a scalar triple product involving the vector between points and the cross product of direction vectors. This captures the perpendicular separation in 3D.
L1 is x-axis. L2 passes through (0,1,1) with direction (0,1,0). What is the shortest distance between L1 and L2
A 1
B √2
C 2
D √3
L2 is parallel to y-axis with x=0 and z=1. The closest point on x-axis to x=0 is (0,0,0). Closest point on L2 is (0,0,1). Distance between them is 1.
Planes x+2y+3z=4 and 2x+4y+6z=8 are
A Parallel distinct
B Coincident
C Perpendicular
D Intersecting
The second equation is exactly 2 times the first, including the constant term. That means both represent the same plane, so the planes are coincident.
Planes x=1 and y=2 are at what angle
A 0°
B 45°
C 90°
D 60°
Plane x=1 has normal (1,0,0) and plane y=2 has normal (0,1,0). Their dot product is 0, so normals are perpendicular and the planes meet at 90°.
Two planes are x+y+z=1 and x−y+z=3. On their intersection line, the y-value is
A −1
B 1
C 0
D 2
Subtract equations: (x+y+z)−(x−y+z)=2y=−2, so y=−1. This constant y holds for every point on the intersection line of the two planes.
A line has direction ratios (3,4,12). What is cos(angle with x-axis)
A 4/13
B 12/13
C 3/13
D 13/3
cosα = a/√(a²+b²+c²) for direction ratios (a,b,c). Here √(9+16+144)=13, so cosα = 3/13. This is the direction cosine along x-axis.
A point is equally distant from xy-plane and xz-plane. Which condition must always be true
A y = z
B y = −z
C y+z = 0
D |y| = |z|
Distance from xy-plane is |z| and from xz-plane is |y|. Equal distances require |y|=|z|. This allows both y=z and y=−z cases, so absolute equality is the must.
The reflection of point (2,−3,6) in the xy-plane is
A (2,−3,−6)
B (−2,3,6)
C (2,3,6)
D (−2,−3,−6)
Reflection in the xy-plane changes only the sign of z-coordinate because xy-plane is z=0. x and y remain the same, while z becomes −z.
Direction ratios can be obtained from direction cosines by multiplying by
A Only integers
B Only positive numbers
C Any nonzero constant
D Only prime numbers
Direction cosines define a unit direction. Direction ratios represent the same direction without normalization. Multiplying (l,m,n) by any nonzero constant gives valid direction ratios.
Direction ratios are (1,2,3). One correct set of direction cosines is
A (1/14,2/14,3/14)
B (1/√14,2/√14,3/√14)
C (1/√6,2/√6,3/√6)
D (1/3,2/3,3/3)
Direction cosines are direction ratios divided by their magnitude. Magnitude of (1,2,3) is √(1+4+9)=√14. Dividing each component by √14 gives a unit direction vector.
For plane x/2 − y/3 + z/6 = 1, the x-intercept is
A 3
B 6
C 2
D −3
In intercept form x/a + y/b + z/c = 1, the plane meets x-axis at (a,0,0). Here a=2, so x-intercept is 2. Negative sign on y term shows y-intercept is −3.
New origin is shifted to (1,2,3). What are the new coordinates of point P(4,6,5)
A (3,4,2)
B (5,8,8)
C (4,6,5)
D (−3,−4,−2)
Under translation, new coordinates are obtained by subtracting new origin coordinates from the point: (4−1, 6−2, 5−3) = (3,4,2). Translation changes coordinates but not distances.
A line’s direction vector v satisfies v·n=0 with plane normal n. What must be true
A Line perpendicular plane
B Line normal plane
C Line in plane
D Line parallel plane
If v·n=0, the line’s direction is perpendicular to the plane’s normal. That means the line has no component along the normal direction, so it is parallel to the plane (may or may not lie in it).
Planes 2x+3y−z=5 and 4x+6y−2z=10 are
A Parallel distinct
B Perpendicular
C Coincident
D Intersecting
The second equation is exactly 2 times the first, including the constant term. Therefore, both represent the same plane and are coincident.
In line (x−2)/1 = (y−1)/(−1) = (z+1)/2, direction ratios are
A (2,−1,1)
B (1,−1,2)
C (−1,1,2)
D (1,1,2)
In symmetric form (x−x1)/a = (y−y1)/b = (z−z1)/c, the denominators (a,b,c) are direction ratios. Here they are (1, −1, 2).
A plane passes through origin and has normal vector (1,2,3). Which equation fits
A x+2y+3z=0
B x+2y+3z=1
C x−2y+3z=0
D 2x+y+z=0
Plane through origin with normal (a,b,c) has equation ax+by+cz=0. Using (1,2,3) gives x+2y+3z=0. Any nonzero multiple also represents the same plane.
If |u×v| equals |u||v| for two vectors u and v, then the angle between them is
A 60°
B 45°
C 90°
D 0°
|u×v| = |u||v|sinθ. If it equals |u||v|, then sinθ=1, which happens at θ=90°. So the vectors are perpendicular.
The orthogonal projection of point (a,b,c) on the xy-plane is
A (a,0,c)
B (0,b,c)
C (0,0,c)
D (a,b,0)
Projection on xy-plane keeps x and y coordinates same but makes z=0 because xy-plane is z=0. So (a,b,c) projects to (a,b,0).
A line equally inclined to all three axes can have direction ratios
A (1,0,1)
B (1,1,1)
C (1,2,3)
D (0,1,1)
Equal inclination to x, y, z means direction cosines are equal in magnitude, so direction ratios can be proportional to (1,1,1). This represents a space diagonal direction.
L1 is x-axis. L2 passes through (3,0,2) with direction (0,1,0). What is the shortest distance between L1 and L2
A 2
B √13
C 3
D √5
L2 is parallel to y-axis with fixed x=3 and z=2. Closest point on x-axis to x=3 is (3,0,0). Closest point on L2 is (3,0,2). Distance is 2