Vector AB = B−A = (−1−2, 3−1, 4−0) = (−3,2,4). This gives correct displacement from A to B in 3D.
For A(2,1,0) and B(−1,3,4), distance AB is
A √29
B √21
C √25
D √33
Distance is |AB| = √((−3)²+2²+4²)=√(9+4+16)=√29. It matches standard 3D distance formula using coordinate differences.
If a=(1,2,3) and b=(3,2,1), then a·b equals
A 12
B 14
C 16
D 10
a·b = 1×3 + 2×2 + 3×1 = 3 + 4 + 3 = 10. This value helps find angle using cosθ = (a·b)/(|a||b|).
If a=(1,2,3) and b=(3,2,1), then |a| equals
A √12
B √14
C √16
D √10
Magnitude |a| = √(1²+2²+3²) = √(1+4+9)=√14. Magnitude depends only on components, not on the vector’s position.
If a=(1,2,3) and b=(3,2,1), then |b| equals
A √12
B √16
C √14
D √10
|b| = √(3²+2²+1²)=√(9+4+1)=√14. Both vectors have same length because their components are just reversed.
Angle between a=(1,2,3) and b=(3,2,1) satisfies cosθ =
A 10/14
B 10/13
C 14/10
D 7/5
cosθ = (a·b)/(|a||b|) = 10/(√14·√14)=10/14=5/7. This gives an acute angle since cosine is positive.
If a=(2,−1,0) and b=(1,2,0), then a×b is
A 0k
B −5k
C 3k
D 5k
Cross product in xy-plane gives z-component: a×b = (0,0, 2·2 − (−1)·1) = (0,0,4+1)=5k. Direction is +k by right-hand rule.
If a×b = 5k, then area of parallelogram is
A 10
B 2.5
C 5
D √5
Area of parallelogram formed by a and b equals |a×b|. Here |a×b| = |5k| = 5, so area is 5 square units.
If a×b = 5k, then area of triangle is
A 5
B 2.5
C 10
D 25
Triangle area using two sides from same vertex is half the parallelogram area. So area = |a×b|/2 = 5/2 = 2.5 square units.
For a=(1,0,1) and b=(0,1,1), scalar triple with c=(1,1,0) equals
A 2
B 0
C 1
D −1
Compute b×c = (0,1,1)×(1,1,0) = (−1,1,−1). Then a·(b×c) = (1,0,1)·(−1,1,−1)=−1+0−1=−2.
For a=(1,0,1), b=(0,1,1), c=(1,1,0), scalar triple equals
A −1
B 1
C −2
D 2
First find b×c = (0,1,1)×(1,1,0) = (−1,1,−1). Then a·(b×c) = (1,0,1)·(−1,1,−1)=−1+0−1=−2.
If a·(b×c) = 0, then vectors are
A parallel pairs
B perpendicular always
C equal magnitudes
D coplanar vectors
Scalar triple product magnitude gives volume of parallelepiped. If it is zero, volume is zero, meaning vectors lie in the same plane (coplanar).
For line through A(a) and B(b), vector form is
A r=(1−t)a+tb
B r=a+tb
C r=a×b
D r=a·b
Line through endpoints A and B can be expressed as convex combination r=(1−t)a+tb. Any real t gives full line; 0≤t≤1 gives segment AB.
Point P divides AB in ratio 4:1 internally; OP equals
A (4a+b)/5
B (4b−a)/5
C (a+4b)/5
D (b−4a)/5
Internal section: OP=(na+mb)/(m+n) with AP:PB=m:n. Here m=4, n=1, so OP=(1a+4b)/5=(a+4b)/5.
For internal division 4:1, point P is nearer to
A A
B B
C midpoint
D outside segment
AP:PB=4:1 means AP is larger and PB smaller. The point is closer to the end with the smaller segment length, so P is nearer to B.
If P divides AB externally in 2:1, OP equals
A (a+2b)/3
B (a+2b)/1
C (a+2b)/(1)
D (a+2b)/(2−1)
External section formula: OP=(na+mb)/(m−n). For m=2, n=1, OP=(1a+2b)/(2−1)=a+2b. This lies beyond B on extension.
If OP = a+2b, the point is
A midpoint
B internal point
C external point
D centroid point
When denominator becomes 1 in external division, the point is outside the segment, beyond one endpoint. External section always places the dividing point on extension.
If a and b are nonzero and a×b=0, then
A a ∥ b
B a ⟂ b
C |a|=|b|
D a=b always
Zero cross product implies sinθ=0, so θ=0° or 180°. With nonzero vectors, that means they are parallel or anti-parallel, i.e., collinear directions.
If a·b=|a||b|/2, then angle is
A 30°
B 60°
C 45°
D 90°
a·b=|a||b|cosθ. If it equals |a||b|/2, then cosθ=1/2, giving θ=60°. This indicates an acute angle.
If |a|=|b| and a·b=0, then resultant |a+b| is
A |a|
B 2|a|
C √2|a|
D 0
|a+b|²=|a|²+|b|²+2a·b. With |a|=|b| and a·b=0, |a+b|²=2|a|², so |a+b|=√2|a|.
If |a|=|b| and a·b=|a|², then vectors are
A same direction
B perpendicular
C opposite
D coplanar only
a·b=|a||b|cosθ. Here |a|=|b|, so a·b=|a|² implies cosθ=1, hence θ=0°. They point in same direction.
If a·b=−|a||b|, vectors are
A perpendicular
B same direction
C random angle
D parallel opposite
a·b=|a||b|cosθ. If it equals −|a||b|, then cosθ=−1 so θ=180°. They are parallel but in opposite directions.
If l,m,n are direction cosines, then m is cosine with
A x-axis
B z-axis
C y-axis
D line segment
Direction cosines are (l,m,n) corresponding to angles with x, y, z axes respectively. So m=cosβ where β is angle with y-axis.
If DR are (2,3,6), direction cosines are
A (2,3,6)
B (2/√49,3/√49,6/√49)
C (2/7,3/7,6/7)
D (2/11,3/11,6/11)
Normalize direction ratios by dividing by √(2²+3²+6²)=√(4+9+36)=√49=7. So direction cosines are (2/7,3/7,6/7).
If a=(1,1,1), then direction cosines are
A (1/√3,1/√3,1/√3)
B (1,1,1)
C (1/3,1/3,1/3)
D (√3,√3,√3)
|a|=√(1+1+1)=√3. Unit direction components are (1/√3,1/√3,1/√3). They satisfy l²+m²+n²=1.
If r=(x,y,z) lies on plane x+2y+3z=6, then vector form can be
A r×(1,2,3)=6
B r·(1,2,3)=0
C r·(1,2,3)=6
D r+(1,2,3)=6
Dot product (x,y,z)·(1,2,3)=x+2y+3z. Setting it equal to 6 gives plane equation. (1,2,3) is normal vector.
If plane passes through origin with normal (2,−1,1), equation is
A 2x+y+z=0
B x−2y+z=0
C 2x−y−z=0
D 2x−y+z=0
Plane through origin satisfies r·n=0. With n=(2,−1,1), we get 2x−y+z=0. This is standard conversion from vector to Cartesian.
Line is perpendicular to plane when line direction is
A perpendicular to normal
B parallel to normal
C parallel to plane
D zero vector
A plane’s normal is perpendicular to the plane. A line perpendicular to the plane must go along the normal direction, so its direction vector is parallel to the normal.
If b is line direction and n is plane normal, line ⟂ plane if
A b×n=0
B b·n=0
C |b|=|n|
D b+n=0
Line direction must be parallel to plane normal for perpendicularity. Parallel vectors have cross product zero, so b×n=0 indicates the line is perpendicular to the plane.
If a,b,c are vectors, cyclic permutation keeps
A a×(b×c) value
B a·b always
C a·(b×c) value
D |a| always
Scalar triple product is cyclic: a·(b×c)=b·(c×a)=c·(a×b). Cyclic change keeps value same, but swapping two vectors changes sign.
If a=(1,2,0), b=(0,1,2), c=(2,0,1), then scalar triple is
A 0
B 5
C −3
D 3
Compute b×c = (0,1,2)×(2,0,1) = (1,4,−2). Then a·(b×c) = (1,2,0)·(1,4,−2)=1+8+0=9. Not listed.
If a=(1,2,0), b=(0,1,2), c=(2,0,1), scalar triple is
A 9
B 3
C 0
D −9
First b×c = (0,1,2)×(2,0,1) = (1,4,−2). Then a·(b×c) = (1,2,0)·(1,4,−2)=1+8+0=9.
If |a×b| gives area, then |(a×b)·c| gives
A line length
B triangle area
C parallelepiped volume
D plane equation
|(a×b)·c| equals |a·(b×c)|, the scalar triple product magnitude. It represents volume of parallelepiped formed by a, b, and c.
If vector a is resolved into parallel and perpendicular to b, tool used is
A cross only
B dot and projection
C section formula
D centroid formula
Component of a along b is found using projection based on dot product. The remaining part is perpendicular component, found by subtracting the parallel component from a.
Vector projection of a on b is zero when
A a ∥ b
B a=b
C |a|=|b|
D a ⟂ b
Projection depends on a·b. If a is perpendicular to b, then a·b=0, so scalar and vector projection of a on b becomes zero.
If a=(3,0,4), then unit vector along a is
A a/7
B a/√7
C a/5
D 5a
|a|=√(3²+0²+4²)=√25=5. Unit vector is a/|a|, so it becomes a/5. This keeps direction but makes length 1.
If a=(3,0,4), direction cosines are
A (3/5,0,4/5)
B (3,0,4)
C (3/7,0,4/7)
D (0,3/5,4/5)
Direction cosines are components of unit direction vector. Divide each component by magnitude 5, giving (3/5,0,4/5). They satisfy sum of squares 1.
Nonzero cross product means vectors are not parallel. Parallel vectors have cross product zero, so (2,3,−4) confirms they form a nonzero area.
A point P on line through A with direction d has r =
A a×td
B a·td
C a+td
D (1−t)a+td
A line through point with position vector a and direction d is r=a+td. Here t is any real number, giving all points on that infinite line.
If t is restricted 0 to 1 in r=(1−t)a+tb, it represents
A segment AB
B ray AB
C line AB
D plane ABC
When 0≤t≤1, r moves from a at t=0 to b at t=1, staying between them. This gives exactly the line segment joining A and B.
If a·(b×c) = 5, then volume is
A 0
B 5
C −5
D 25
Volume of parallelepiped is absolute value of scalar triple product. So |a·(b×c)| = |5| = 5. Negative sign would only indicate orientation.
If a·(b×c) = −5, then volume is
A 0
B −5
C 5
D 25
Volume is magnitude, so sign is ignored. |−5|=5. Negative value only means the set (a,b,c) has opposite orientation.
If b×c = 0, then b and c are
A parallel
B perpendicular
C equal always
D unit only
Cross product zero implies sinθ=0, so vectors are parallel or anti-parallel. One could also be zero vector, but in general it indicates parallel direction.
If a is perpendicular to b and c, then a is parallel to
A b+c
B b−c
C b·c
D b×c
b×c is perpendicular to both b and c. If a is also perpendicular to both b and c, then a is along the same perpendicular direction, hence parallel to b×c.
If a,b are in xy-plane, then a×b is along
A i direction
B k direction
C j direction
D xy-plane
Vectors in xy-plane have zero z-components. Their cross product is perpendicular to xy-plane, so it points along ±k direction depending on rotation.
If a=(0,2,0) and b=(3,0,0), then a×b equals
A 6k
B 6i
C −6k
D −6j
(0,2,0)×(3,0,0) = (0,0, 0·0 − 2·3) = (0,0,−6). So cross product is −6k.
If a=(0,2,0) and b=(3,0,0), then a·b equals
A 6
B −6
C 3
D 0
Dot product is 0·3 + 2·0 + 0·0 = 0. The vectors are perpendicular, one along y-axis and the other along x-axis.
If two vectors have equal magnitudes and opposite directions, their sum is
A zero vector
B unit vector
C doubled vector
D perpendicular vector
Opposite vectors satisfy b = −a with same magnitude. So a+b = a−a = 0 vector. This is used in balancing forces and symmetric geometry.
If a=(1,1,1) and b=(1,−1,0), then a·b equals
A 0
B 1
C −1
D 2
Dot product = 1·1 + 1·(−1) + 1·0 = 1−1+0 = 0. Not listed. Replace with corrected fresh MCQ: