Dot product zero means cosθ=0, so θ=90°. This holds when neither vector is zero, because zero vector would make dot product zero without giving a definite angle.
If a×b ≠ 0, then angle between them is not
A 90°
B 60°
C 120°
D 0°
a×b ≠ 0 implies sinθ ≠ 0, so θ cannot be 0° or 180°. Angles like 60°, 90°, 120° are possible because they give nonzero sine.
Subtract component-wise: (2−1)i + (3−(−1))j + (−1−2)k = i+4j−3k. Subtraction is adding the negative vector.
If OA=a and OB=b, then position vector of point on AB is
A a+tb
B ta+tb
C (1−t)a+tb
D a×b
Any point P on segment AB can be written as OP = (1−t)OA + tOB with 0≤t≤1. This is a standard parametric form using endpoints.
If t=1/2 in OP=(1−t)a+tb, point is
A A itself
B midpoint
C B itself
D external point
Putting t=1/2 gives OP = (a+b)/2, which is midpoint formula. It lies halfway between A and B and divides the segment in 1:1 ratio.
If P divides AB internally in 3:2, then OP is
A (2a+3b)/5
B (3a+2b)/5
C (3b−2a)/5
D (2b−3a)/5
Internal division formula: OP = (na+mb)/(m+n) for AP:PB=m:n. Here m=3, n=2, so OP = (2a+3b)/5.
In internal division AP:PB = m:n, P is closer to
A A when m>n
B A when m=n
C outside segment
D B when m>n
If AP:PB = m:n and m>n, AP is longer than PB, so P lies nearer to B because the smaller part PB is adjacent to B.
If three points A,B,C are collinear, then vectors satisfy
A AB·AC=0
B AB+AC=0
C AB×AC=0
D AB=AC always
If A, B, C lie on same line, AB and AC are parallel. Cross product of parallel vectors is zero, so AB×AC=0 tests collinearity.
If a=(1,0,0), b=(0,1,0), c=(0,0,1), then a·(b×c) is
A 1
B 0
C −1
D 2
b×c = j×k = i. Then a·(b×c)= i·i =1. This also matches unit cube volume formed by orthogonal unit vectors.
If vectors a,b,c are coplanar, then
A a×b=0
B a·b=0
C a·(b×c)=0
D |a|=|b|
Scalar triple product gives volume. Coplanar vectors form zero volume parallelepiped, so a·(b×c)=0 is the standard coplanarity condition.
If a×b = b×a, then condition is
A a=b
B a×b=0
C a·b=0
D |a|=|b|
Since b×a = −a×b, equality implies a×b = −a×b, so 2(a×b)=0, hence a×b=0. This means vectors are parallel (or one is zero).
If a×b is a unit vector, then |a×b| is
A 0
B √2
C |a||b|
D 1
A unit vector has magnitude 1 by definition. So if a×b itself is a unit vector, its magnitude |a×b| must be 1.
The vector perpendicular to both a and b is
A a+b
B a−b
C a×b
D a·b
Cross product a×b produces a vector normal to plane containing a and b. This perpendicular direction is important for area, torque, and normal vectors of planes.
For vectors a,b, |a×b| equals area of
A parallelogram
B circle
C rectangle only
D triangle only
Magnitude of cross product equals |a||b|sinθ, which is base×height of parallelogram formed by a and b. Triangle area becomes half of this.
Area of triangle with vertices A,B,C can be computed by
A |AB·AC|/2
B |AB+AC|/2
C |AB×AC|/2
D |AB−AC|/2
If two sides from same vertex are AB and AC, then cross product gives parallelogram area. Triangle is half of it, so area = |AB×AC|/2.
If a=(1,2,0), b=(2,4,0), then a×b is
A zero vector
B unit vector
C i+j+k
D perpendicular to plane
b is 2a, so they are parallel. Cross product of parallel vectors is zero. Both lie in xy-plane, so no perpendicular area is formed.
If a=(1,0,0) and b=(0,0,1), then a×b equals
A j
B i
C −j
D k
i×k = −(k×i). Since k×i = j, we get i×k = −j. Direction follows right-hand rule and standard basis multiplication.
If b×c is parallel to a, then a·(b×c) is
A 0 always
B negative always
C undefined
D maximum value
a·(b×c)=|a||b×c|cosθ. If a is parallel to (b×c), θ=0 so cosθ=1, giving maximum possible magnitude for given lengths.
If a is perpendicular to (b×c), then a·(b×c) is
A 1
B 0
C −1
D |a||b||c|
Dot product is zero when vectors are perpendicular. So if a ⟂ (b×c), then scalar triple product becomes zero, meaning no volume in direction of a.
If direction cosines are (l,m,n), then direction ratios can be
A l²,m²,n²
B 1/l,1/m,1/n
C l,m,n
D l+m+n
Direction ratios are any proportional set to direction cosines. Since (l,m,n) already represent unit direction components, they themselves can serve as direction ratios.
If DR are proportional, lines are
A parallel
B perpendicular
C intersecting always
D skew always
Two lines in 3D are parallel when their direction vectors are scalar multiples. Proportional direction ratios indicate same direction, which is the condition for parallel lines.
Angle between lines depends on angle between
A their points
B their midpoints
C their direction vectors
D their lengths
The angle between two lines is defined as the angle between their direction vectors. This can be found using dot product formula with the two direction vectors.
If a=(2,−1,2), then a is perpendicular to
A (1,2,0)
B (1,2,1)
C (2,−1,2)
D (1,0,−1)
Check dot products: a·(1,0,−1)=2·1+(−1)·0+2·(−1)=2−2=0. Zero dot product confirms perpendicularity.
If a=(1,2,3), projection length on x-axis is
A 2
B 1
C 3
D √14
Projection on x-axis equals x-component when axis uses unit vector i. Dot with i gives a·i = 1, so scalar projection on x-axis is 1.
If a=(1,2,3), projection vector on x-axis is
A 2i
B 3i
C i
D (1)i
Vector projection onto x-axis is (a·i)i. Since a·i = 1 and i is unit, projection vector equals 1·i = i.
A plane through origin with normal n has equation
A r·n = 0
B r = a+tb
C r×n = 0
D r+n = 0
For a plane passing through origin, any position vector r in plane is perpendicular to normal n. Therefore r·n = 0 represents that plane in vector form.
Plane through point a with normal n is
A (r+a)·n=0
B (r×a)·n=0
C (r−a)·n=0
D r·(a×n)=0
Vector r−a lies in the plane, and it is perpendicular to the normal n. So their dot product is zero, giving the plane equation.
If n=(1,2,3), then plane through origin is
A x−2y+3z=0
B 2x+y+z=0
C x+2y−3z=0
D x+2y+3z=0
r·n=0 becomes (x,y,z)·(1,2,3)=0, so x+2y+3z=0. This is standard Cartesian form from vector normal.
If line direction is parallel to plane, then direction vector is
A parallel to normal
B equal to normal
C perpendicular to normal
D opposite to normal
A line parallel to a plane lies along directions within the plane. Any direction within plane is perpendicular to plane’s normal vector, so dot product with normal is zero.
If b is direction of line and n is plane normal, line parallel plane if
A b×n=0
B b·n=0
C |b|=|n|
D b=n
Condition b·n=0 means direction b is perpendicular to normal n, so b lies in plane direction. Thus the line is parallel to the plane.
If a=(1,1,1) and b=(1,1,1), then a×b is
A i+j+k
B unit vector
C zero vector
D perpendicular vector
Same vectors have zero angle between them, so sin0=0 and cross product magnitude is zero. Therefore a×b is the zero vector.
If a=(1,1,0) and b=(1,−1,0), then |a×b| is
A 2
B 0
C 1
D √2
Compute cross product in xy-plane: (1,1,0)×(1,−1,0) = (0,0,−2). Magnitude is 2. This equals parallelogram area.
A×(B×C) lies in plane of
A A and B
B A and C
C all three
D B and C
Using identity a×(b×c)=(a·c)b−(a·b)c, the result is a linear combination of b and c. Hence it lies in the plane spanned by b and c.
Vector joining two points in 3D uses
A dot product
B cross product
C subtraction of vectors
D triple product
Vector from A to B is AB = b−a, where a and b are position vectors of A and B. This is a core method for lines, distance, and section formula.
If A(1,0,2), B(3,2,5), then AB is
A (2,2,3)
B (4,2,7)
C (−2,−2,−3)
D (3,2,5)
AB = (3−1, 2−0, 5−2) = (2,2,3). This direction vector also gives slope information in 3D and helps compute distance.
If AB=(2,2,3), then distance AB is
A √13
B √17
C √21
D √(17)
Distance is magnitude of AB: √(2²+2²+3²)=√(4+4+9)=√17. This matches 3D distance formula from coordinate differences.
If P is midpoint of A(1,0,2) and B(3,2,5), then P is
A (1,1,3.5)
B (2,2,3)
C (2,1,3.5)
D (4,2,7)
Midpoint coordinates are averages: ((1+3)/2, (0+2)/2, (2+5)/2) = (2,1,3.5). This matches vector midpoint formula.
If triangle vertices are a,b,c, centroid vector is
A (a+b)/2
B (a−b+c)/3
C (a+b−c)/3
D (a+b+c)/3
Centroid is average of the position vectors of vertices. It lies at intersection of medians and divides each median in 2:1 ratio from the vertex.
In 2:1 median division, centroid is closer to
A vertex
B outside triangle
C midpoint side
D origin always
Centroid divides each median in ratio 2:1 from the vertex. So the centroid lies closer to the midpoint of the opposite side than to the vertex.
If a,b,c are coplanar, then volume formed is
A zero
B 1
C maximum
D negative always
Coplanar vectors produce no 3D “height,” so parallelepiped volume is zero. This is why scalar triple product becomes zero for coplanar vectors.
Lagrange identity relates
A a+b and a−b
B i,j,k only
C a·b and a×b
D section formula
Lagrange identity connects dot and cross products: |a×b|² + (a·b)² = |a|²|b|². It is useful for proving angles and area relations.
If |a|=3, |b|=4, and a ⟂ b, then |a×b| is
A 7
B 1
C 0
D 12
For perpendicular vectors, sin90=1, so |a×b|=|a||b|=3×4=12. This also equals area of parallelogram formed by them.
If |a|=5, |b|=5 and a·b=0, then angle is
A 0°
B 60°
C 90°
D 180°
Dot product zero means cosθ=0, so θ=90°, regardless of equal magnitudes. Equal lengths do not decide angle; dot product decides perpendicularity.