A plane passes through A(1,0,2), B(2,1,0), C(0,2,1). Which equation represents that plane
A x+y+z=2
B x+y+z=3
C x−y+z=3
D 2x+y+z=3
Form vectors AB and AC and take their cross product to get a normal. Here the normal is proportional to (1,1,1). Substituting any given point gives x+y+z=3.
The image of point P(1,2,3) in plane x+y+z=3 is
A (1,0,−1)
B (0,−1,1)
C (−1,1,0)
D (−1,0,1)
For plane n·r=3 with n=(1,1,1), reflection uses P′=P−2((n·P−3)/|n|²)n. Here n·P=6, so P′=(1,2,3)−2(1,1,1)=(−1,0,1).
A line has direction ratios (1,1,1) and a plane has equation x+y+z=0. What is the angle between the line and the plane
A sin⁻¹(1)
B sin⁻¹(1/√3)
C 90°
D cos⁻¹(1/√3)
The plane’s normal is (1,1,1), which is parallel to the line’s direction. A line parallel to a plane’s normal is perpendicular to the plane, so the angle is 90°.
Find the shortest distance between skew lines r=(1,0,1)+t(1,2,3) and r=(0,1,2)+s(2,1,1)
A √35/3
B 3/√35
C 3/35
D √35
Use d = |(r2−r1)·(d1×d2)| / |d1×d2|. Here d1×d2=(−1,5,−3), |×|=√35, and (r2−r1)·(×)=3, giving 3/√35.
Line x=2+t, y=1−2t, z=3+t meets plane x−y+z=5 at which point
A (9/4,1/2,13/4)
B (2,1,3)
C (5/2,0,7/2)
D (3,−1,4)
Substitute x,y,z into x−y+z=5: (2+t)−(1−2t)+(3+t)=4+4t=5, so t=1/4. Then x=9/4, y=1/2, z=13/4.
Lines L1: r=(1,0,2)+t(1,1,0) and L2: r=(0,1,1)+s(2,−1,1) are
A Parallel lines
B Intersecting lines
C Perpendicular lines
D Skew lines
Check coplanarity using scalar triple product (r2−r1)·(d1×d2). Here it is nonzero, so they are not coplanar. Non-parallel and non-coplanar lines are skew.
Find perpendicular distance from P(1,2,0) to line through origin with direction (2,1,2)
A √29
B √29/3
C 3/√29
D √(29/9)
Distance from point to line through origin is |P×d|/|d|. Here P×d=(4,−2,−3) so |P×d|=√29 and |d|=3, giving √29/3.
Angle between planes 2x−y+2z=3 and x+2y−2z=1 equals
A sin⁻¹(4/9)
B 60°
C cos⁻¹(4/9)
D 90°
Normals are n1=(2,−1,2) and n2=(1,2,−2). Dot product is −4, magnitudes are 3 and 3. So cosθ=|−4|/(3·3)=4/9, hence θ=cos⁻¹(4/9).
Direction ratios of the line of intersection of planes x+2y+z=1 and 2x−y+3z=4 are
A (1,2,1)
B (2,−1,3)
C (−7,−1,5)
D (7,−1,−5)
Direction of intersection line is n1×n2 where n1=(1,2,1), n2=(2,−1,3). Cross product gives (7,−1,−5). Any nonzero multiple is also correct.
Foot of perpendicular from P(1,2,3) on plane x+2y+2z=9 is
A (7/9,14/9,23/9)
B (1,2,3)
C (2,2,2)
D (0,0,0)
For plane ax+by+cz+d=0, foot is P − ((a x1+b y1+c z1+d)/(a²+b²+c²))·(a,b,c). Here value=2, denom=9, so subtract (2/9)(1,2,2) from P.
Planes 2x+3y+4z=5 and 4x+6y+8z=1 are
A Coincident planes
B Perpendicular planes
C Parallel distinct
D Intersecting planes
Normals are proportional: (4,6,8)=2(2,3,4). But constants are not in the same ratio because 1 ≠ 2·5. So planes are parallel but not the same plane.
Distance between planes 2x+3y+4z=5 and 4x+6y+8z=1 equals
A 9/√29
B 9/(2√29)
C √29/9
D 2√29/9
Make coefficients same: second plane ÷2 gives 2x+3y+4z=1/2. Distance = |5−1/2|/√(2²+3²+4²) = (9/2)/√29 = 9/(2√29).
Angle between lines with direction ratios (1,1,2) and (2,−1,1) is
A 45°
B 90°
C 30°
D 60°
Use cosθ=(u·v)/(|u||v|). Dot: 1·2+1·(−1)+2·1=3. |u|=√6 and |v|=√6, so cosθ=3/6=1/2, hence θ=60°.
A line has equal angles with x and y axes and makes 30° with z-axis. A correct set of direction cosines is
A (1/2,1/2,1/2)
B (1/(2√2),1/(2√2),√3/2)
C (√3/2,√3/2,1/2)
D (1/√3,1/√3,1/√3)
30° with z-axis gives n=cos30°=√3/2. Equal angles with x and y give l=m. Then l²+m²+n²=1 ⇒ 2l²+3/4=1 ⇒ l²=1/8 ⇒ l=m=1/(2√2).
A plane passes through (1,−1,2) and is perpendicular to line with direction (2,1,−2). Which is correct
A 2x+y−2z+3=0
B 2x−y+2z+3=0
C x+2y−2z+3=0
D 2x+y+2z−3=0
Plane perpendicular to the line has normal parallel to the line direction (2,1,−2). Using point-normal form: 2(x−1)+(y+1)−2(z−2)=0 gives 2x+y−2z+3=0.
Find shortest distance between lines (x−1)/1=(y−0)/2=(z+1)/(−1) and (x−0)/2=(y−1)/1=(z−2)/1
A √3/5
B 5/3
C √15
D 5/√3
Use d = |(r2−r1)·(d1×d2)| / |d1×d2|. Here d1×d2=(3,−3,−3) with magnitude 3√3, and triple product magnitude is 15, so d=15/(3√3)=5/√3.
Lines x=1+t, y=2+2t, z=3−t and x=3−2s, y=6−4s, z=1+s intersect at
A (1,2,3)
B (3,4,5)
C (3,6,1)
D (2,4,2)
Solve coordinate equations: 1+t=3−2s, 2+2t=6−4s, 3−t=1+s. The solution is t=2 and s=0. Substituting gives the common point (3,6,1).
A line has direction (1,2,2). The plane 2x−y+2z=5 makes what angle with the line
A cos⁻¹(4/9)
B sin⁻¹(4/9)
C 90°
D 60°
Angle φ between a line and a plane satisfies sinφ=|n·d|/(|n||d|). Here n=(2,−1,2), d=(1,2,2). Dot=4, |n|=3, |d|=3, so sinφ=4/9.
Lines r=(1,0,0)+t(1,2,3) and r=(2,1,2)+s(2,−1,1) are
A Skew lines
B Parallel lines
C Perpendicular lines
D Intersecting lines
First check coplanarity: (r2−r1)·(d1×d2)=0, so they are coplanar. Solving the parametric equations gives a common point, so coplanar non-parallel lines intersect.
Distance from origin to line r=(1,2,3)+t(2,1,−1) equals
A √83/6
B √(83/3)
C √(83/6)
D 83/6
Distance from origin to a line through A with direction d is |A×d|/|d|. Here A×d=(−5,7,−3) so |A×d|=√83, and |d|=√6, giving √83/√6 = √(83/6).
The point on plane x+y+z=6 closest to origin is
A (6,0,0)
B (2,2,2)
C (0,6,0)
D (0,0,6)
The closest point is the foot of the perpendicular from origin along the plane’s normal (1,1,1). Write point as k(1,1,1) and satisfy x+y+z=6 ⇒ 3k=6 ⇒ k=2.
A plane contains the z-axis and also contains point (2,−1,0). Which equation fits
A x−2y=0
B 2x+y=0
C x+y=0
D x+2y=0
A plane containing the z-axis contains direction (0,0,1) and passes through origin. With point (2,−1,0), it contains vector (2,−1,0). A normal is (0,0,1)×(2,−1,0)=(1,2,0), giving x+2y=0.
The perpendicular bisector plane of segment joining P(1,2,3) and Q(5,0,1) is
A 2x−y−z−3=0
B 2x+y+z−3=0
C x−2y+z−3=0
D 2x−y+z+3=0
The bisector plane passes through midpoint M(3,1,2) and is perpendicular to PQ=(4,−2,−2), same as normal (2,−1,−1). Using point-normal form gives 2(x−3)−(y−1)−(z−2)=0 ⇒ 2x−y−z−3=0.
Perpendicular distance from point (2,1,0) to plane 2x−y+2z=4 equals
A 1
B 2/3
C 1/3
D 3
Write plane as 2x−y+2z−4=0. Distance is |2·2−1+0−4|/√(2²+(−1)²+2²)=|−1|/3=1/3. It uses absolute value and normal magnitude.
Point (1,1,1) is tested with planes x+y+z=3 and x−y+z=1. Does it lie on their intersection line
A No, lies not
B Yes, lies
C Only on first
D Only on second
A point lies on the intersection line only if it satisfies both plane equations. For (1,1,1): 1+1+1=3 and 1−1+1=1, so it satisfies both, hence lies on the intersection line.
Angle between planes x+y=2 and y+z=3 equals
A 45°
B 90°
C 30°
D 60°
Normals are n1=(1,1,0) and n2=(0,1,1). Dot=1. Magnitudes are √2 and √2. So cosθ=1/2, giving θ=60°. This is the acute angle between planes.
Line (x−1)/1=(y−2)/(−1)=(z−3)/0 is related to plane x+y+z=6 as
A Parallel distinct
B Perpendicular plane
C Lies in plane
D Intersects once
Direction vector is (1,−1,0). Plane normal is (1,1,1). Dot is 0, so the line is parallel to the plane. Also point (1,2,3) satisfies 1+2+3=6, so the whole line lies in the plane.
Find distance between line through (1,2,0) with direction (1,−1,0) and plane x+y+z=6
A 3
B √3
C 3/√3
D √6
Since (1,−1,0)·(1,1,1)=0, the line is parallel to the plane, so distance is constant. Use distance from any point on line: |1+2+0−6|/√3=3/√3=√3.
A line passes through (1,0,1) and is perpendicular to plane 2x−y+2z=4. Which parametric form is correct
A x=1+2t, y=−t, z=1+2t
B x=1+t, y=−2t, z=1+t
C x=1+2t, y=t, z=1−2t
D x=1−2t, y=−t, z=1+2t
A line perpendicular to a plane has direction parallel to the plane’s normal. Here normal is (2,−1,2). Using point (1,0,1) gives x=1+2t, y=0−t, z=1+2t.
Angle between directions (1,2,0) and (2,−1,2) is
A 60°
B 30°
C 45°
D 90°
Compute dot product: (1,2,0)·(2,−1,2)=2−2+0=0. Zero dot product means the direction vectors are perpendicular, so the angle between the lines is 90°.
Find shortest distance between x-axis and line x=1, y=t, z=2+t
A 1
B 2
C √2
D √5
Treat as skew lines. x-axis has direction (1,0,0) and passes through origin. Other line passes through (1,0,2) with direction (0,1,1). Using the skew distance formula gives distance = 2/√2 = √2.
A plane contains the line x=2+t, y=1−t, z=2t and also passes through (1,1,1). Which plane is correct
A x+y+z=5
B x+3y+z=5
C 2x+3y+z=5
D x+3y−z=5
A plane containing a line and an extra point uses two independent direction vectors in the plane: the line direction (1,−1,2) and vector from a line point (2,1,0) to (1,1,1), i.e., (−1,0,1). Their cross gives normal (1,3,1), so plane is x+3y+z=5.
The line of intersection of planes x+y+z=6 and x+2y+3z=10 can be written as
A x=6+t, y=0−2t, z=t
B x=2−t, y=4+2t, z=t
C x=2+t, y=2−t, z=4
D x=2+t, y=4−2t, z=t
Solve the two plane equations together. Subtracting gives y+2z=4, so take z=t and y=4−2t. Then x from x+y+z=6 becomes x=6−(4−2t)−t=2+t.
Acute angle between planes x+y+z=6 and x+2y+3z=10 equals
A cos⁻¹(√42/7)
B cos⁻¹(6/7)
C sin⁻¹(√42/7)
D 60°
Normals are (1,1,1) and (1,2,3). Dot=6, magnitudes √3 and √14. So cosθ=6/√42 = √42/7 after rationalizing. Hence θ=cos⁻¹(√42/7).
Distance between parallel lines (1,2,3)+t(0,1,2) and (4,1,0)+t(0,1,2) equals
A √230
B 5/√230
C √230/5
D 230/5
For parallel lines with same direction d, distance is |(r2−r1)×d|/|d|. Here r2−r1=(3,−1,−3), d=(0,1,2). Cross magnitude is √230 and |d|=√5, so distance is √230/√5=√230/5·? which simplifies to √230/5.
Point divides P(2,−1,3) and Q(−4,5,1) externally in ratio 1:2. The point is
A (−2,3,2)
B (8,−7,5)
C (0,2,2)
D (2,−1,3)
External section formula for ratio m:n is ((m x2−n x1)/(m−n), similarly for y,z). With m=1, n=2: x=(−4−4)/(−1)=8, y=(5+2)/(−1)=−7, z=(1−6)/(−1)=5.
Sphere with diameter endpoints A(1,2,3) and B(3,4,5) has equation
A (x−2)²+(y−3)²+(z−4)²=12
B (x−1)²+(y−2)²+(z−3)²=3
C x²+y²+z²=3
D (x−2)²+(y−3)²+(z−4)²=3
Center is midpoint M(2,3,4). Diameter AB has length 2√3, so radius is √3 and r²=3. Using (x−a)²+(y−b)²+(z−c)²=r² gives the equation.
For sphere x²+y²+z²=25, point (2,3,4) lies
A On sphere
B Inside sphere
C Outside sphere
D At center
Compute x²+y²+z² = 4+9+16 = 29. Since 29 is greater than 25, the point lies outside the sphere. Equality would mean on the sphere, smaller would mean inside.
If l=1/2 and m=1/3 are direction cosines (positive), then n equals
A √23/3
B √23/6
C √13/6
D √5/6
Direction cosines satisfy l²+m²+n²=1. So n²=1−1/4−1/9=1−(9/36+4/36)=23/36. Taking positive n gives n=√23/6.
A plane passes through (1,1,0) and is parallel to vectors (1,0,1) and (0,2,1). Which equation fits
A 2x−y+2z=3
B x+2y−z=3
C 2x+y+2z=3
D 2x+y−2z=3
A plane parallel to two non-parallel vectors has a normal equal to their cross product. (1,0,1)×(0,2,1)=(−2,−1,2). Using point (1,1,0) gives −2(x−1)−(y−1)+2z=0, which simplifies to 2x+y−2z=3.
Angle between lines (x−1)/2=(y+1)/(−2)=(z−0)/1 and (x−0)/1=(y−2)/2=(z−1)/2 equals
A 90°
B 60°
C 45°
D 30°
Direction vectors are (2,−2,1) and (1,2,2). Dot product is 2·1 + (−2)·2 + 1·2 = 2−4+2=0. Zero dot product means the lines are perpendicular.
Distance between parallel planes x−2y+2z=5 and x−2y+2z=−1 equals
A 1
B 2
C 3
D √3
Write as x−2y+2z−5=0 and x−2y+2z+1=0. Distance is |−5−(+1)|/√(1²+(−2)²+2²)=6/√9=2. Coefficients match, so parallel.
Value of k so that planes 2x+ky+2z=0 and x+2y+2z=0 are perpendicular is
A 3
B −2
C 2
D −3
Perpendicular planes have perpendicular normals. Normals are (2,k,2) and (1,2,2). Dot product must be zero: 2·1 + k·2 + 2·2 = 2 + 2k + 4 = 0 ⇒ 2k = −6 ⇒ k = −3.
If a line is equally inclined to all three coordinate axes in the first octant, its direction cosine l equals
A 1/3
B √3
C 1/√3
D √(2/3)
Equal inclination means l=m=n (all positive in first octant). Using l²+m²+n²=1 gives 3l²=1, so l=1/√3. The same value holds for m and n.
The point on the z-axis nearest to P(2,−1,3) is
A (0,0,0)
B (0,0,3)
C (2,0,3)
D (0,−1,3)
Any point on z-axis is (0,0,z). Distance squared from (2,−1,3) is 2²+(−1)²+(3−z)². This is minimized when z=3, so the nearest point is (0,0,3).
A line has direction ratios (2,1,2). What is the angle between the line and the xy-plane
A sin⁻¹(2/3)
B cos⁻¹(2/3)
C 60°
D 90°
Angle φ between a line and xy-plane satisfies sinφ = |z-component|/|direction| because xy-plane normal is k. Here |d|=√(4+1+4)=3 and z-component is 2, so sinφ=2/3.
A plane cuts intercepts 2 on x-axis, −3 on y-axis, and 6 on z-axis. Which equation represents it
A 3x+2y+z=6
B x−y+z=6
C 6x−3y+2z=6
D 3x−2y+z=6
Use intercept form x/a + y/b + z/c = 1 with a=2, b=−3, c=6: x/2 + y/(−3) + z/6 = 1. Multiply by 6 to get 3x−2y+z=6.
Direction ratios of the line of intersection of planes x+y+z=1 and x+y−z=0 are
A (1,1,0)
B (0,1,1)
C (1,−1,0)
D (1,0,1)
The direction of the intersection line is n1×n2. Normals are (1,1,1) and (1,1,−1). Cross product is (−2,2,0), which is proportional to (1,−1,0).
Find shortest distance between z-axis and line r=(1,2,0)+t(1,0,1)
A 1
B 2
C √2
D √5
z-axis has direction (0,0,1). Given line has direction (1,0,1). Cross of directions is (0,1,0) with magnitude 1. Vector between points (0,0,0) and (1,2,0) is (1,2,0). Dot with (0,1,0) gives 2, so distance is 2.
A line has direction ratios (2,k,−1) and is perpendicular to direction ratios (1,2,3). Value of k is
A 2
B −1/2
C −2
D 1/2
Perpendicular directions have zero dot product: (2,k,−1)·(1,2,3)=2 + 2k − 3 = 0. So 2k − 1 = 0, giving k = 1/2.