In component form a i + b j + c k, the coefficient of i gives the x-component, of j gives y-component, and of k gives z-component.
If |a|=5, then |2a| equals
A 2
B 5
C 10
D 7
Scalar multiplication scales magnitude by the absolute value of the scalar. So |2a| = |2|·|a| = 2×5 = 10, direction stays same for positive scalar.
If |a|=7, then |−a| equals
A 0
B 14
C −7
D 7
The negative vector has the same magnitude as the original vector but opposite direction. Magnitude is never negative, so |−a| = |a| = 7.
If a=b, then a−b equals
A a
B zero vector
C 2a
D b
Subtracting a vector from itself gives the zero vector. Since a=b, a−b = 0, which has zero magnitude and works as additive identity in vector algebra.
For any vector a, a+0 equals
A a
B 1
C 0
D −a
Zero vector is the identity element for addition. Adding it does not change a vector: a + 0 = a, similar to how adding zero doesn’t change a number.
If P lies on AB, then A,P,B are
A perpendicular
B coplanar only
C collinear
D equal vectors
If point P lies on segment AB, all three points A, P, and B lie on the same straight line. This is exactly the meaning of collinearity.
The vector BA equals
A b−a
B a−b
C a+b
D |b−a|
If OA=a and OB=b, then AB=b−a. Reversing direction gives BA = −AB = a−b. Direction changes, magnitude remains same as AB.
If AB = b−a, then vector AC is
A c+b
B a−c
C a+c
D c−a
With position vectors OA=a and OC=c, the vector from A to C is AC = OC − OA = c − a. This gives correct direction from A toward C.
The magnitude |b−a| represents
A area of triangle
B dot product
C distance AB
D cross product
The displacement vector from A to B is b−a. Its magnitude is the length of the segment AB, so |b−a| gives the distance between points A and B.
If a·a equals
A |a|
B |a|²
C 0 always
D |a|³
Dot product of a vector with itself is a·a = |a||a|cos0 = |a|². This is a useful way to compute magnitude squared from components.
If a·b = |a||b|, then angle is
A 0°
B 45°
C 90°
D 180°
a·b = |a||b|cosθ. If it equals |a||b|, then cosθ=1, which happens at θ=0°. So vectors are in the same direction.
If a·b = −|a||b|, then angle is
A 0°
B 180°
C 90°
D 60°
When a·b = |a||b|cosθ equals −|a||b|, then cosθ = −1, which happens at θ=180°. So vectors are opposite in direction.
If a×b has magnitude zero, then vectors are
A perpendicular
B unequal always
C parallel
D coplanar only
|a×b| = |a||b|sinθ. If it is zero, then sinθ=0 so θ is 0° or 180°. That means the vectors are parallel or anti-parallel.
If |a×b| = |a||b|, then angle is
A 0°
B 30°
C 180°
D 90°
|a×b| = |a||b|sinθ. If it equals |a||b|, then sinθ=1, which occurs at 90°. So vectors are perpendicular.
A·(B×C) is also called
A dot product
B scalar triple
C cross product
D vector triple
a·(b×c) is the scalar triple product. It gives a scalar value and represents the signed volume of a parallelepiped formed by vectors a, b, and c.
Sign of a·(b×c) changes if
A cyclic shift done
B all magnitudes doubled
C two vectors swapped
D angle is acute
Swapping any two vectors in a scalar triple product reverses orientation, so the value changes sign. Cyclic permutation keeps it same, but interchange changes sign.
If point P divides AB externally in m:n, then OP is
A (na+mb)/(m−n)
B (ma+nb)/(m−n)
C (na−mb)/(m+n)
D (na+mb)/(m+n)
For external division of A(a) and B(b) in ratio m:n, the vector formula uses denominator (m−n). The numerator weights remain na+mb, giving correct external point.
Coordinates of midpoint of (x1,y1) and (x2,y2) are
A (x1+x2, y1+y2)
B ((x1+x2)/2, (y1+y2)/2)
C (x2−x1, y2−y1)
D ((x1−x2)/2, (y1−y2)/2)
Midpoint divides the segment internally in 1:1. So each coordinate is average of the two endpoints, matching vector midpoint formula (a+b)/2.
If P divides AB in 2:1 internally, P is
A nearer to A
B outside segment
C midpoint always
D nearer to B
Internal ratio AP:PB = 2:1 means AP is longer than PB. So P lies closer to B because the smaller part PB is on B side.
For triangle vertices a,b,c, vector identity is
A AB+BC−CA=0
B AB−BC+CA=0
C AB+BC+CA=0
D AB+AC+BC=0
Going around a triangle returns to the starting point. So displacement AB plus BC plus CA equals zero vector. This is a basic closed-path vector property.
If r = 2i − 3j + k, then z-component is
A 2
B 1
C −3
D 0
In r = ai + bj + ck, the coefficient of k gives z-component. Here k has coefficient 1, so z-component is 1.
Direction ratios of line through (1,2,3) and (4,6,5) are
A 1,2,3
B −3,−4,−2
C 4,6,5
D 3,4,2
Direction vector from first point to second is (4−1, 6−2, 5−3) = (3,4,2). Any proportional set works as direction ratios.
If direction ratios are (2,−1,2), then a direction vector is
A i+j+k
B 2i−j+2k
C 2i+j−2k
D −2i−j−2k
Direction ratios directly give components of a direction vector. So vector = 2i + (−1)j + 2k = 2i−j+2k represents that direction.
Direction cosines are obtained by
A normalizing DR
B squaring DR
C adding DR
D reversing DR
Direction cosines are components of the unit direction vector. So divide direction ratios by √(a²+b²+c²) to normalize and get (l,m,n).
If l=m=n, then each equals
A 0
B 1/3
C 1/√3
D √3
With l=m=n and l²+m²+n²=1, we get 3l²=1, so l=1/√3 (taking positive for equal positive cosines). This fits unit direction condition.
Angle between i and j is
A 0°
B 90°
C 45°
D 180°
i and j are unit vectors along x and y axes, which are perpendicular. So the angle between them is 90°, and i·j = 0 confirms this.
i×j equals
A i
B j
C 0
D k
Using right-hand rule and standard basis cross products: i×j = k, j×k = i, k×i = j. Reversing order changes sign.
j×i equals
A k
B i
C −k
D 0
Cross product is anti-commutative: j×i = −(i×j). Since i×j = k, therefore j×i = −k.
k×k equals
A k
B 0
C 1
D −k
Any vector crossed with itself gives zero because angle between them is 0°, so sin0=0. Hence k×k = 0 vector.
Dot product i·i equals
A 1
B 0
C −1
D 2
i is a unit vector, so i·i = |i|² = 1² = 1. Similarly, j·j = 1 and k·k = 1.
Dot product i·j equals
A 1
B −1
C 2
D 0
i and j are perpendicular, so i·j = |i||j|cos90 = 1×1×0 = 0. This also shows orthogonality of axes.
Cross product gives a vector
A parallel to a
B parallel to b
C perpendicular to both
D in same plane
a×b is perpendicular to plane containing a and b, hence perpendicular to both vectors (when they are not parallel). Direction follows right-hand rule.
A vector with same magnitude but opposite direction is
A equal vector
B negative vector
C unit vector
D position vector
Negative vector −a keeps magnitude |a| but reverses direction. This is used in subtraction and in forming opposite displacements.
If vector a is along x-axis, then
A j-component zero
B b=0 always
C i-component zero
D k-component maximum
Along x-axis means only i component exists. So vector is ai with j and k components zero. Any y-component (j) would shift it away from x-axis.
If vector a is along z-axis, form is
A ai+bj
B bj+ck
C ai+ck
D ck
Along z-axis means only k component is present. So the vector has form ck, with x and y components equal to zero.
Magnitude of i−j is
A 0
B √2
C 1
D 2
i−j has components (1,−1,0). Magnitude is √(1²+(−1)²+0²)=√2. This represents a diagonal direction in xy-plane.
If a=(1,2,2), then |a| is
A 3
B √8
C √9
D √6
Magnitude is √(1²+2²+2²)=√(1+4+4)=√9=3. Writing √9 keeps method visible but value equals 3.
Angle between a and b is 90° if
A a×b=0
B a=b
C |a|=|b|
D a·b=0
Perpendicular vectors satisfy dot product zero. Cross product zero indicates parallelism, not perpendicularity, so dot product is the correct test.
If a=(2,0,0) and b=(0,3,0), then a·b is
A 6
B 0
C 3
D 2
Dot product is 2×0 + 0×3 + 0×0 = 0. The vectors lie along x and y axes, so they are perpendicular.
If a=(2,0,0) and b=(0,3,0), then |a×b| is
A 0
B 5
C 6
D 3
Since vectors are perpendicular, |a×b|=|a||b|=2×3=6. This equals area of parallelogram formed by them.
Position vector of origin is
A zero vector
B i
C j
D k
The origin has coordinates (0,0,0). Its position vector from origin to itself is 0, so it is the zero vector.
Vector equation of segment AB uses
A r=tb, 0≤t≤1
B r=a+tb, 0≤t≤1
C r=a×b
D r=a·b
For line through point with position vector a and direction b, r=a+tb. Restricting t between 0 and 1 gives only the segment part.
If a×b is nonzero, then a and b are
A collinear
B parallel
C always equal
D not parallel
Cross product is zero for parallel vectors. If it is nonzero, sinθ≠0 so θ is not 0° or 180°, meaning vectors are not parallel.
Volume of parallelepiped is
A |a×b|
B |a·b|
C |a·(b×c)|
D |a+b+c|
Scalar triple product magnitude gives volume. It measures how much vector a is out of the plane of b and c, producing 3D volume.
If b×c is perpendicular to
A b only
B both b and c
C c only
D a only
Cross product b×c is perpendicular to plane containing b and c. Hence it is perpendicular to both b and c (unless vectors are parallel, then cross product is zero).
A plane normal vector is
A perpendicular to plane
B along the plane
C tangent vector
D midpoint vector
A normal vector to a plane points at right angles to the plane. It is used in plane equations and helps test whether a line is parallel or perpendicular to plane.
Vector projection of a on b is
A (a·b)b
B ((a·b)/|b|²)b
C (a×b)b
D (|a|/|b|)b
Vector projection gives the actual vector component of a along b. It equals scalar projection (a·b/|b|) times unit vector of b, simplifying to ((a·b)/|b|²)b.
If a is resolved along i,j,k, it is called
A vector addition
B scalar product
C vector components
D section formula
Writing a vector in terms of i, j, k gives its components along coordinate axes. This is fundamental for computing dot product, cross product, and magnitudes.
If a and b are equal vectors, then
A a×b is maximum
B a·b is zero
C |a| differs
D a×b equals zero
Equal vectors have same direction, so angle between them is 0°. Cross product magnitude becomes |a||b|sin0 = 0, so cross product is zero vector.
If a and b are perpendicular unit vectors, then a·b equals
A 1
B 0
C −1
D √2
For perpendicular vectors, cos90=0, so dot product is zero. Being unit vectors makes |a|=|b|=1, but perpendicularity still forces product to 0.