A vector quantity is defined by two things: how much (magnitude) and which way (direction). If only magnitude is involved, it is a scalar, not a vector.
A quantity with only magnitude is
A Vector quantity
B Unit vector
C Scalar quantity
D Position vector
Scalars have magnitude only, like mass and time. Vectors additionally require direction, like displacement and velocity, so a scalar cannot show direction.
A vector of zero magnitude is
A Unit vector
B Position vector
C Negative vector
D Zero vector
Zero vector has magnitude 0 and no defined direction. It acts as the additive identity: adding it to any vector does not change that vector.
Two vectors are equal if they have
A Same magnitude and direction
B Same start point
C Same length only
D Same direction only
Equal vectors do not need the same initial point. They are equal when their magnitudes match and their directions are exactly the same.
A unit vector has magnitude
A 0
B 1
C 2
D Depends on direction
A unit vector represents direction only. It is obtained by dividing a vector by its magnitude, so its length becomes 1 while direction remains unchanged.
The unit vector along a is
A |a|/a
B a×|a|
C a/|a|
D a·|a|
Dividing vector a by its magnitude |a| scales it to length 1. This keeps direction same and removes the size information.
Vectors in same direction are
A Like vectors
B Unlike vectors
C Coplanar only
D Negative vectors
Like vectors point in the same direction (may have different magnitudes). Unlike vectors point in opposite directions. Direction decides like/unlike, not position.
Vectors in opposite direction are
A Like vectors
B Equal vectors
C Unit vectors
D Unlike vectors
Unlike vectors are parallel but point opposite. If magnitudes are equal too, then one is the negative of the other.
Negative of vector a is
A |a|
B a/|a|
C −a
D a×a
The negative vector has the same magnitude as a but the opposite direction. Adding a and −a gives the zero vector.
A position vector is drawn from
A Origin to a point
B Any point to any point
C Point to origin only
D Along x-axis only
The position vector of point P is OP, starting at origin and ending at P. It uniquely represents the location of P in vector form.
Standard basis vectors are
A a, b, c
B i, j, k
C p, q, r
D x, y, z
In 3D, i, j, k are unit vectors along x, y, z axes. Any vector can be written as ai + bj + ck using components.
Magnitude of ai+bj+ck is
A a+b+c
B a²+b²+c²
C √(a²+b²+c²)
D √(a+b+c)
The length of a vector with components (a, b, c) follows 3D Pythagoras. Square components, add them, then take square root.
Two vectors are parallel if
A Their cross product is zero
B Their dot product is zero
C Their magnitudes are equal
D Their sum is zero
Cross product magnitude is |a||b|sinθ. It becomes zero when sinθ=0, meaning θ=0° or 180°, i.e., vectors are parallel.
Two vectors are perpendicular if
A a×b = 0
B |a| = |b|
C a = b
D a·b = 0
Dot product is |a||b|cosθ. For perpendicular vectors, θ=90° so cosθ=0, making dot product zero.
Vector addition follows
A Only circle law
B Newton law
C Triangle law
D Mirror law
Place tail of second vector at head of first; the resultant runs from tail of first to head of second. This is the triangle law of addition.
Parallelogram law gives
A Sum vector
B Difference vector
C Unit vector
D Zero vector
If two vectors form adjacent sides of a parallelogram, their sum is represented by the diagonal from the common tail point.
Vector subtraction a−b equals
A a+b
B a+(−b)
C b−a
D |a|−|b|
Subtracting b means adding the negative of b. Geometrically, reverse b to −b and then add using triangle/parallelogram law.
Scalar multiplication changes
A Only direction always
B Only magnitude always
C Magnitude and sometimes direction
D Neither magnitude nor direction
Multiplying by positive scalar changes magnitude only. Multiplying by negative scalar also reverses direction, so both magnitude and direction may change.
If b = 3a, then b is
A Perpendicular to a
B Equal to a
C Zero vector
D Parallel to a
A scalar multiple of a vector is always parallel to it. Here the magnitude becomes three times, and direction stays same because scalar is positive.
Vector from A to B is
A b−a
B a−b
C a+b
D |b|−|a|
If position vectors are a and b, then AB = OB − OA = b − a. This gives correct direction from A toward B.
Distance AB equals
A |a+b|
B |a·b|
C |b−a|
D |a×b|
The displacement from A to B is vector AB = b−a. Distance is the magnitude of this displacement, so it is |b−a|.
Midpoint position vector of A,B is
A (a+b)/2
B (a−b)/2
C 2(a+b)
D (b−a)/2
Midpoint divides the segment in ratio 1:1 internally, so its position vector is average of endpoints: (a+b)/2.
If P divides AB in m:n internally, OP is
A (ma+nb)/(m+n)
B (na+mb)/(m+n)
C (ma−nb)/(m+n)
D (na−mb)/(m+n)
Internal section formula in vector form: point closer to A gets weight of n and point closer to B gets weight of m, giving OP = (na+mb)/(m+n).
If m=n, section point becomes
A External point
B Centroid
C Origin
D Midpoint
When m=n, the ratio is 1:1. The dividing point is exactly halfway between A and B, so it is the midpoint.
External division uses denominator
A m−n
B m+n
C m×n
D m²+n²
For external division of AB in ratio m:n, the formula involves (m−n) in denominator because the point lies outside the segment.
Centroid position vector of triangle is
A (a+b)/2
B (a−b+c)/3
C (a+b+c)/3
D (a+b+c)/2
The centroid is intersection of medians. In vector terms, its position vector is the average of the three vertex position vectors: (a+b+c)/3.
Dot product a·b equals
A |a||b|cosθ
B |a||b|sinθ
C |a×b|
D |a|+|b|
Dot product measures how much one vector points in direction of the other. It depends on cosine of angle between them.
If a·b is positive, angle θ is
A Right
B Obtuse
C 180°
D Acute
Since a·b = |a||b|cosθ, positive dot product means cosθ>0, which happens for acute angles (0° to 90°).
If a·b is negative, angle is
A Acute
B Right
C Obtuse
D Zero
Negative dot product means cosθ<0. Cosine is negative between 90° and 180°, so the angle between vectors is obtuse.
Projection of a on b (scalar) is
A (a·b)/|b|
B a·b
C (a×b)/|b|
D |a||b|
Scalar projection gives component of a along b. It equals |a|cosθ, and also equals (a·b)/|b|.
Cross product magnitude equals
A |a||b|cosθ
B |a||b|sinθ
C |a|+|b|
D |a|−|b|
Cross product measures perpendicular “spread” between vectors. Its magnitude equals area of parallelogram formed by them: |a||b|sinθ.
Direction of a×b is given by
A Right-hand rule
B Left-hand rule
C Parallel rule
D Mirror rule
Curl fingers of right hand from a toward b; thumb gives direction of a×b. It is perpendicular to the plane of a and b.
Area of parallelogram is
A |a·b|
B |a|+|b|
C |a×b|
D |a−b|
The area spanned by two vectors equals magnitude of their cross product. For triangle area, it becomes half of |a×b|.
Area of triangle with sides a,b is
A |a×b|
B |a·b|/2
C |a·b|
D |a×b|/2
A triangle is half of the parallelogram formed by two side vectors from the same vertex, so its area is half of cross product magnitude.
Scalar triple product a·(b×c) gives
A Volume of parallelepiped
B Area of triangle
C Length of vector
D Angle between axes
The absolute value of scalar triple product equals volume of parallelepiped formed by vectors a, b, c. Sign indicates orientation (right/left handed).
If a·(b×c)=0, vectors are
A Always equal
B Perpendicular
C Coplanar
D Unit vectors
Zero scalar triple product means volume is zero, so the three vectors lie in the same plane. This is a standard test for coplanarity.
Vector triple product a×(b×c) equals
A (a·b)c−(a·c)b
B (b·c)a−(a·b)c
C (a·a)(b×c)
D (a×b)×c
This identity helps simplify nested cross products. Result is a linear combination of b and c, lying in plane of b and c.
Direction ratios of vector ai+bj+ck are
A √a, √b, √c
B a, b, c
C a², b², c²
D 1/a, 1/b, 1/c
Direction ratios (DRs) are any proportional set of components of direction vector. For ai+bj+ck, the components a, b, c serve as DRs.
Direction cosines (l,m,n) satisfy
A l+m+n=1
B lm+n=0
C lmn=1
D l²+m²+n²=1
Direction cosines are cosines of angles with x, y, z axes. Squaring and adding gives 1 because they represent components of a unit direction vector.
Unit direction vector from DR (a,b,c) is
A (ai+bj+ck)/√(a²+b²+c²)
B ai+bj+ck
C √(a²+b²+c²)/(ai+bj+ck)
D (a+b+c)(i+j+k)
Normalize the direction vector by dividing by its magnitude. This converts direction ratios into direction cosines and gives a unit vector.
Angle between vectors a and b is found using
A a×b formula
B |a|−|b|
C a·b formula
D a+b magnitude
Using a·b = |a||b|cosθ, we compute cosθ = (a·b)/(|a||b|). This directly gives angle between vectors.
Vectors are collinear if
A a×b=0
B a·b=0
C |a|=|b|
D a+b=0 only
Collinear means lying on same line direction. That happens when vectors are parallel or anti-parallel, which gives cross product zero.
Coplanar points A,B,C,D satisfy
A AB·AC=0
B AB×AC=0
C |AB|=|CD|
D AB·(AC×AD)=0
Using vectors from A, if scalar triple product AB·(AC×AD)=0, then volume is zero and the four points lie in same plane.
Vector equation of line through point a with direction b is
A r = tb
B r = a + tb
C r = a×b
D r = a·b
Any point on line has position vector r = a + t b, where a is position vector of a fixed point and b gives line direction.
Distance between points with vectors a and b is
A |b−a|
B |a+b|
C |a×b|
D a·b
Difference b−a gives displacement from point A to point B. Its magnitude gives straight-line distance between the points.
Work done is computed using
A Cross product
B Triple product
C Dot product
D Vector sum
Work = F·s. Only component of force along displacement contributes, which is captured by dot product. Perpendicular component does no work.
Torque direction is along
A r×F
B r·F
C r+F
D r−F
Torque is vector product of position vector r and force F. Its direction is perpendicular to plane of r and F by right-hand rule.
Magnitude of r×F equals
A rFcosθ
B r+F
C r−F
D rFsinθ
Since |r×F| = |r||F|sinθ, torque magnitude depends on perpendicular component of force to r. Maximum torque occurs at 90°.
Identity a·(b×c) equals
A b·(a×c)
B b·(c×a)
C All are same
D c·(a×b)
Scalar triple product is cyclic: a·(b×c)=b·(c×a)=c·(a×b). This cyclic change keeps value same, but swapping two vectors changes sign.
If a×b = b×a, then
A a and b equal
B a and b parallel
C a·b=0 always
D |a|=|b|
Cross product is anti-commutative: a×b = −(b×a). For them to be equal, both must be zero, meaning vectors are parallel/collinear (or one is zero).