Vector Components and Products
Core Definitions
Physical quantities are distinguished by their directional requirements. Scalar quantities depend only on magnitude, while Vector quantities require both magnitude and direction.
$$\vec{V} = |V|\hat{u}$$
A vector is defined by its length multiplied by a unit vector indicating its orientation.
$\vec{V}$=Vector quantity(Variable)
$|V|$=Magnitude (Scalar length)(Variable)
$\hat{u}$=Unit vector (direction)(Dimensionless)
$|V| = 0$
→The quantity becomes a Null Vector with undefined direction.
Geometric Representation: Vectors are drawn as arrows where length represents magnitude and the tip indicates direction.
Equality: Two vectors are equal only if they share identical magnitude AND direction, regardless of starting position.
Scalar Multiplication: Multiplying a vector by a positive scalar $n$ gives a new vector $n\vec{A}$ with the same direction but $n$ times the magnitude. A negative scalar reverses the direction.
Dimensional Change: When a vector is multiplied by a scalar with its own dimensions, the result is a new physical quantity. For example, $m\vec{v}$ (mass × velocity) gives momentum — a vector with different dimensions than velocity.
A Unit Vector is a vector with magnitude exactly 1, used purely to specify direction. The unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$ point along the x, y, and z axes respectively.
$$\hat{A} = \frac{\vec{A}}{|A|}$$
Divide any vector by its own magnitude to extract its direction as a unit vector.
$\hat{A}$=Unit vector in the direction of A(Dimensionless)
$\vec{A}$=Original vector(Variable)
$|A|$=Magnitude of A(Variable)
Any Direction: Unit vectors are not restricted to axes — $\hat{r}$ denotes direction along a position vector, $\hat{n}$ denotes a surface normal.
Shortcut: To find a unit vector, divide each component by the magnitude: $\hat{A} = \frac{A_x\hat{i} + A_y\hat{j} + A_z\hat{k}}{\sqrt{A_x^2 + A_y^2 + A_z^2}}$.
Vector Addition and Subtraction
The Resultant vector represents the single effect of multiple vectors combined via the Head-to-Tail Method or the Parallelogram Law.
$$\vec{R} = \vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$$
Resultant components are the algebraic sum of the individual vector components.
$R_x, R_y$=Resultant components(Variable)
$A_x, B_x$=x-components of vectors A and B(Variable)
$\hat{i}, \hat{j}$=Unit vectors along x and y axes(Dimensionless)
$\vec{A} = -\vec{B}$
→The resultant is a Null Vector.
$\vec{A} \parallel \vec{B}$ (same direction)
→Maximum resultant: $R = A + B$.
$\vec{A}$ antiparallel to $\vec{B}$
→Minimum resultant: $R = |A - B|$.
Commutative Property: The order of addition does not affect the resultant: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$.
Subtraction: Defined as adding a negative vector: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$.
Resultant Range: For two vectors of magnitudes $A$ and $B$, the resultant magnitude is bounded: $|A - B| \leq R \leq A + B$.
Quadrant Determination: The signs of $R_x$ and $R_y$ determine the quadrant. Use: Q1 ($\theta = \phi$), Q2 ($\theta = 180° - \phi$), Q3 ($\theta = 180° + \phi$), Q4 ($\theta = 360° - \phi$), where $\phi = \tan^{-1}|R_y/R_x|$.
Resolution into Components
Resolution is the process of splitting a vector into mutually perpendicular Rectangular Components using trigonometric principles.
$$A_x = A \cos \theta, \quad A_y = A \sin \theta$$
Components are calculated using the vector magnitude and the angle relative to the positive x-axis.
$A$=Vector magnitude(Variable)
$\theta$=Angle with positive x-axis(degrees/radians)
$A_x$=Horizontal projection(Variable)
$A_y$=Vertical projection(Variable)
$\theta = 0°$
→Entire vector lies along x-axis: $A_x = A$, $A_y = 0$.
$\theta = 90°$
→Entire vector lies along y-axis: $A_x = 0$, $A_y = A$.
$\theta = 45°$
→Equal components: $A_x = A_y = \frac{A}{\sqrt{2}}$.
Unit Vectors: The symbols $\hat{i}, \hat{j}, \hat{k}$ denote unit vectors along x, y, and z axes respectively.
Magnitude Recovery: Magnitude is found via Pythagorean theorem: $A = \sqrt{A_x^2 + A_y^2}$.
Direction Recovery: $\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$ — but check the quadrant using signs of $A_x$ and $A_y$.
3D Extension: In three dimensions, $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ with $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$.
Position Vector
A Position Vector $\vec{r}$ describes the location of a point relative to the origin. Its rectangular components are the coordinates of the point itself.
$$\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}, \quad r = \sqrt{a^2 + b^2 + c^2}$$
The position vector points from the origin to the point P(a, b, c).
$a, b, c$=Coordinates of the point along x, y, z axes(Variable (m, km, etc.))
$r$=Distance from origin to the point(Variable)
Displacement Vector: The displacement from point A to point B is $\vec{AB} = \vec{r}_B - \vec{r}_A$.
Distance Between Points: $|AB| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2}$.
Scalar (Dot) Product
The Dot Product of two vectors yields a scalar quantity. It measures how much two vectors are aligned — geometrically, it is the product of one vector's magnitude with the Projection of the other onto it.
$$\vec{A} \cdot \vec{B} = AB\cos\theta$$
The scalar product equals the product of both magnitudes times the cosine of the angle between them.
$A, B$=Magnitudes of vectors A and B(Variable)
$\theta$=Angle between the two vectors (0° to 180°)(degrees/radians)
$\theta = 0°$ (parallel)
→$\vec{A} \cdot \vec{B} = AB$ — maximum positive value.
$\theta = 90°$ (perpendicular)
→$\vec{A} \cdot \vec{B} = 0$ — the key test for orthogonality.
$\theta = 180°$ (antiparallel)
→$\vec{A} \cdot \vec{B} = -AB$ — maximum negative value.
Commutative: $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$ — order doesn't matter.
Component Form: $\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$ — multiply matching components and add.
Unit Vector Products: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$.
Physical Application: Work done by a force: $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$.
Finding Angle: $\cos\theta = \frac{A_xB_x + A_yB_y + A_zB_z}{AB}$ — equate the two forms of the dot product.
The Projection of vector $\vec{A}$ onto vector $\vec{B}$ gives the component of $\vec{A}$ in the direction of $\vec{B}$. It is a scalar value that can be positive or negative.
$$\text{Proj}_{B} A = \frac{\vec{A} \cdot \vec{B}}{|B|} = A\cos\theta$$
The projection gives the scalar component of one vector along the direction of another.
$\vec{A} \cdot \vec{B}$=Dot product of the two vectors(Variable)
$|B|$=Magnitude of the direction vector (the one you project onto)(Variable)
$A\cos\theta$=Component of A along the direction of B(Variable)
$\theta = 0°$
→Projection equals the full magnitude $A$ — entire vector lies along $\vec{B}$.
$\theta = 90°$
→Projection is zero — no component along $\vec{B}$.
$\theta = 180°$
→Projection is $-A$ — vector points entirely opposite to $\vec{B}$.
Using Unit Vector: Projection of $\vec{A}$ on $\vec{B}$ equals $\vec{A} \cdot \hat{B}$, where $\hat{B} = \frac{\vec{B}}{|B|}$.
Sign Meaning: Positive projection means same general direction; negative means opposite.
Perpendicular Case: Projection is zero when vectors are perpendicular ($\cos 90° = 0$).
Vector (Cross) Product
The Cross Product of two vectors yields a new vector that is perpendicular to the plane containing both original vectors. Its direction follows the Right-Hand Rule.
$$\vec{A} \times \vec{B} = AB\sin\theta\,\hat{n}$$
The vector product has magnitude equal to the product of both magnitudes times the sine of the angle, directed perpendicular to the plane of A and B.
$A, B$=Magnitudes of vectors A and B(Variable)
$\theta$=Angle between A and B (0° to 180°)(degrees/radians)
$\hat{n}$=Unit vector perpendicular to the plane of A and B (via right-hand rule)(Dimensionless)
$\theta = 90°$ (perpendicular)
→$|\vec{A} \times \vec{B}| = AB$ — maximum magnitude.
$\theta = 0°$ or $180°$ (parallel/antiparallel)
→$\vec{A} \times \vec{B} = \vec{0}$ — cross product of parallel vectors is null vector.
Non-Commutative: Reversing the order flips the direction: $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$.
Right-Hand Rule: Curl fingers from $\vec{A}$ toward $\vec{B}$ through the smaller angle — thumb points in the direction of $\vec{A} \times \vec{B}$.
Unit Vector Products: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$. Self-products: $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0}$.
Component Form: $\vec{A} \times \vec{B} = (A_yB_z - A_zB_y)\hat{i} + (A_zB_x - A_xB_z)\hat{j} + (A_xB_y - A_yB_x)\hat{k}$. This can also be remembered as a 3×3 determinant with $\hat{i}, \hat{j}, \hat{k}$ in the first row.
Cyclic Order Shortcut: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$ — follow the cycle $i \to j \to k \to i$. Reversing the cycle flips the sign.
Geometric Meaning: $|\vec{A} \times \vec{B}|$ equals the area of the parallelogram formed by $\vec{A}$ and $\vec{B}$.
Physical Applications: Torque: $\vec{\tau} = \vec{r} \times \vec{F}$. Magnetic force: $\vec{F} = q(\vec{v} \times \vec{B})$.
Application Strategy
Worked Examples
Problem Statement
Find the magnitude and direction of the vector $\vec{V} = 6\hat{i} - 8\hat{j}$.
Magnitude formulaDirection (inverse tangent)
Strategic Toolkit
$|V| = \sqrt{V_x^2 + V_y^2}$
Step 1
Identify Components
Extract the horizontal and vertical components from the unit vector notation.
$V_x = 6, V_y = -8$
Step 2
Calculate Magnitude
Apply the Pythagorean theorem to find the vector length.
$|V| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = 10 \text{ units}$
Step 3
Determine Angle
Calculate the angle relative to the positive x-axis. Since $V_x > 0$ and $V_y < 0$, the vector is in the 4th quadrant, so $\theta = 360° - \phi$.
$\phi = \tan^{-1}\left(\frac{8}{6}\right) \approx 53.1°$, so $\theta = 360° - 53.1° = 306.9°$
Problem Statement
A force $\vec{F} = 2\hat{i} + 3\hat{j}$ N moves its point of application from A(1, 3) to B(5, 7). Find the work done.
Dot Product component formWork = F · dDisplacement vector
Strategic Toolkit
$W = \vec{F} \cdot \vec{d} = F_x d_x + F_y d_y$
Step 1
Find Displacement
Calculate the displacement vector from A to B by subtracting position vectors.
$\vec{d} = \vec{r}_B - \vec{r}_A = (5-1)\hat{i} + (7-3)\hat{j} = 4\hat{i} + 4\hat{j}$ m
Step 2
Apply Dot Product
Multiply corresponding components and sum them.
$W = (2)(4) + (3)(4) = 8 + 12 = 20$ J
Problem Statement
A force $\vec{F} = 3\hat{i} + 2\hat{j}$ N acts at position $\vec{r} = 4\hat{i} - \hat{j}$ m from the pivot. Find the torque.
Cross product in 2DTorque = r × F
Strategic Toolkit
$\vec{\tau} = \vec{r} \times \vec{F}$
Step 1
Set Up Cross Product
For 2D vectors in the xy-plane, the cross product has only a $\hat{k}$ component.
$\vec{\tau} = (r_x F_y - r_y F_x)\hat{k}$
Step 2
Substitute Values
Plug in the components: $r_x = 4, r_y = -1, F_x = 3, F_y = 2$.
$\vec{\tau} = [(4)(2) - (-1)(3)]\hat{k} = [8 + 3]\hat{k} = 11\hat{k}$ N·m
Step 3
Interpret
The positive $\hat{k}$ direction means the torque causes counterclockwise rotation (out of the page).
$|\tau| = 11$ N·m, counterclockwise
Problem Statement
Find the angle between $\vec{A} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{B} = \hat{i} + 2\hat{j} - 2\hat{k}$.
Dot product to find angle3D magnitude
Strategic Toolkit
$\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|A||B|}$
Step 1
Compute Dot Product
Multiply matching components and add.
$\vec{A} \cdot \vec{B} = (2)(1) + (-3)(2) + (1)(-2) = 2 - 6 - 2 = -6$
Step 2
Find Magnitudes
Calculate the magnitude of each vector.
$|A| = \sqrt{4 + 9 + 1} = \sqrt{14}$, $|B| = \sqrt{1 + 4 + 4} = 3$
Step 3
Calculate Angle
Use the dot product formula to find the angle.
$\cos\theta = \frac{-6}{3\sqrt{14}} = \frac{-2}{\sqrt{14}} \approx -0.535$, so $\theta \approx 122.3°$
Problem Statement
Two forces of magnitude 10 N and 20 N act on a body at angles 30° and 60° with the x-axis respectively. Find the magnitude and direction of the resultant.
Rectangular componentsVector addition by componentsQuadrant determination
Strategic Toolkit
$R = \sqrt{R_x^2 + R_y^2}, \quad \theta = \tan^{-1}\frac{R_y}{R_x}$
Step 1
Resolve into Components
Find x and y components of each force using trigonometry.
$F_{1x} = 10\cos 30° = 8.66$ N, $F_{1y} = 10\sin 30° = 5$ N
$F_{2x} = 20\cos 60° = 10$ N, $F_{2y} = 20\sin 60° = 17.32$ N
Step 2
Sum the Components
Add x-components together and y-components together.
$R_x = 8.66 + 10 = 18.66$ N
$R_y = 5 + 17.32 = 22.32$ N
Step 3
Find Magnitude
Use the Pythagorean theorem on the resultant components.
$R = \sqrt{18.66^2 + 22.32^2} = \sqrt{348.2 + 498.2} \approx 29$ N
Step 4
Find Direction
Both $R_x$ and $R_y$ are positive, so the resultant is in Q1.
$\theta = \tan^{-1}\frac{22.32}{18.66} = \tan^{-1}(1.196) \approx 50°$
Problem Statement
Two aeroplanes are at positions A(2, 3, 4) and B(5, 6, 7) in km. Find the distance between them.
Position vectorDisplacement vector3D distance
Strategic Toolkit
$|\vec{AB}| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2}$
Step 1
Find Displacement Vector
Subtract the position vectors to get the displacement from A to B.
$\vec{AB} = (5-2)\hat{i} + (6-3)\hat{j} + (7-4)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$
Step 2
Calculate Distance
Find the magnitude of the displacement vector.
$|AB| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3} \approx 5.2$ km
Memorize formulas
Magnitude of a 2D Vector
Use the Pythagorean theorem to find the length of a vector from its components.
Formula
$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$
Direction of a 2D Vector
Find the angle theta relative to the positive x-axis.
Formula
$\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$
Rectangular Components
Resolve a vector into x and y components using its angle with the x-axis.
Formula
$A_x = A\cos\theta, \quad A_y = A\sin\theta$
Dot Product (Angle Form)
Use to find the work done or test for perpendicularity.
Formula
$\vec{A} \cdot \vec{B} = AB\cos\theta$
Dot Product (Component Form)
Use when vector components are given directly.
Formula
$\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$
Cross Product Magnitude
Use to find torque magnitude or the area of a parallelogram.
Formula
$|\vec{A} \times \vec{B}| = AB\sin\theta$
Cross Product (Component Form)
Use when computing the full cross product vector from components.
Formula
$\vec{A} \times \vec{B} = (A_yB_z - A_zB_y)\hat{i} + (A_zB_x - A_xB_z)\hat{j} + (A_xB_y - A_yB_x)\hat{k}$
Unit Vector
A vector with magnitude 1 in the direction of vector A.
Formula
$\hat{A} = \frac{\vec{A}}{|A|}$
Angle Between Vectors
Find the angle between two vectors using the dot product.
Formula
$\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|A||B|}$
Resultant Magnitude
Find the resultant of two or more vectors from their summed components.
Formula
$R = \sqrt{R_x^2 + R_y^2}$
Projection of A onto B
Find the scalar component of one vector along another's direction.
Formula
$\text{Proj}_B A = \frac{\vec{A} \cdot \vec{B}}{|B|}$
Review
1
Two vectors have magnitudes 3 and 5. Which of the following CANNOT be the magnitude of their resultant?
Ans
1
Review
Tap to reveal
Apply the resultant-magnitude bounds
The magnitude of the sum of two vectors must lie between the difference and the sum of their magnitudes.
$|5-3| \le R \le 5+3 \Rightarrow 2 \le R \le 8$
Eliminate the value outside the interval
Any value below 2 is impossible, so 1 cannot be the resultant magnitude.
2
If the dot product of two vectors is zero, and the cross product of the same two vectors has a magnitude equal to the product of their individual magnitudes, what is the angle between them?
Ans
90°
Review
Tap to reveal
Use the dot-product condition
If $\vec A \cdot \vec B = 0$, then the vectors are perpendicular because $\cos\theta = 0$.
$\theta = 90^\circ$
Check the cross-product magnitude
A cross-product magnitude equal to $AB$ means $AB\sin\theta = AB$, so $\sin\theta = 1$.
$\theta = 90^\circ$
3
A force $\vec{F} = 5\hat{j}$ N is applied at position $\vec{r} = 3\hat{i}$ m from the pivot. What is the torque $\vec{\tau} = \vec{r} \times \vec{F}$?
Ans
$15\hat{k}$ N·m
Review
Tap to reveal
Multiply the magnitudes and unit vectors
The cross product keeps the scalar factor and uses the unit-vector rule $\hat i \times \hat j = \hat k$.
$\vec\tau = (3\hat i) \times (5\hat j) = 15\hat k\ \text{N·m}$
4
If $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = -\hat{i} + 2\hat{j} + 4\hat{k}$, what is $\vec{A} \cdot \vec{B}$?
Ans
0
Review
Tap to reveal
Use the component-wise dot product
Multiply matching components and add them to find the scalar product.
$\vec A \cdot \vec B = (2)(-1) + (3)(2) + (-1)(4) = 0$
5
The resultant of two forces $\vec{F}_1$ and $\vec{F}_2$ has components $R_x = -5$ N and $R_y = -12$ N. In which quadrant does the resultant lie, and what is its magnitude?
Ans
Third quadrant, 13 N
Review
Tap to reveal
Identify the quadrant from the signs
Negative x and negative y components place the vector in the third quadrant.
Compute the resultant magnitude
Use Pythagoras on the components to get the length of the vector.
$R = \sqrt{(-5)^2 + (-12)^2} = 13\ \text{N}$
6
What is the $\hat{k}$ component of $\vec{A} \times \vec{B}$ if $\vec{A} = 3\hat{i} + \hat{j}$ and $\vec{B} = -\hat{i} + 2\hat{j}$?
Ans
7
Review
Tap to reveal
Use the 2D cross-product determinant
For vectors in the $xy$-plane, the cross product reduces to the scalar $A_xB_y - A_yB_x$ in the $\hat k$ direction.
$(3)(2) - (1)(-1) = 7$
7
If $\vec{A} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{B} = 2\hat{i} + \hat{j} - \hat{k}$, what is $\vec{A} \times \vec{B}$?
Ans
$-\hat{i} + 7\hat{j} + 5\hat{k}$
Review
Tap to reveal
Expand the cross product by components
Compute each component from the determinant form, keeping the minus sign on the middle term.
$\vec A \times \vec B = (-1)\hat i + 7\hat j + 5\hat k$
8
The position vectors of points A and B are $\vec{r}_A = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{r}_B = 4\hat{i} - 2\hat{j} + \hat{k}$ (in metres). The distance AB is:
Ans
$\sqrt{29}$ m
Review
Tap to reveal
Find the displacement vector
Subtract the coordinates of A from B to get the vector from A to B.
$\vec{AB} = (4-1)\hat i + (-2-2)\hat j + (1-3)\hat k = 3\hat i - 4\hat j - 2\hat k$
Take the magnitude
The distance is the length of that displacement vector.
$AB = \sqrt{3^2 + (-4)^2 + (-2)^2} = \sqrt{29}\ \text{m}$
9
If $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$, then the angle between $\vec{A}$ and $\vec{B}$ is:
Ans
90°
Review
Tap to reveal
Square both magnitudes
Expanding both sides makes the dot-product term visible and lets the equal terms cancel.
$A^2 + 2\vec A\cdot\vec B + B^2 = A^2 - 2\vec A\cdot\vec B + B^2$
Set the dot product to zero
The remaining equation is $4\vec A\cdot\vec B=0$, so the vectors are perpendicular.
$\vec A\cdot\vec B=0 \Rightarrow \theta = 90^\circ$
10
Which of the following is NOT a property of the cross product?
Ans
$\vec{A} \times \vec{B}$ lies in the plane of $\vec{A}$ and $\vec{B}$
Review
Tap to reveal
Use the geometric direction of the cross product
A cross product points perpendicular to the plane of the two vectors, not inside that plane.
$\vec A \times \vec B \perp$ plane of $\vec A$ and $\vec B$
11
A vector $\vec{A}$ has magnitude 13. If $\vec{A} = 5\hat{i} + n\hat{j}$ where $n > 0$, the value of $n$ is:
Ans
12
Review
Tap to reveal
Apply the magnitude formula
The length of a 2D vector is found from the square root of the sum of the squared components.
$\sqrt{5^2 + n^2} = 13$
Solve for the positive component
Square both sides and use the fact that $n>0$ to choose the positive root.
$25 + n^2 = 169 \Rightarrow n = 12$
12
If $\vec{A} \times \vec{B} = \vec{0}$ and $\vec{A} \cdot \vec{B} = -AB$, then $\vec{A}$ and $\vec{B}$ are:
Ans
Antiparallel (opposite direction)
Review
Tap to reveal
Use the zero cross product
If the cross product is zero, the vectors are parallel or antiparallel.
Use the negative dot product
The relation $\vec A\cdot\vec B=-AB$ means $\cos\theta=-1$, which selects the opposite direction.
$\theta = 180^\circ$
13
A vector has components $R_x = -3$ and $R_y = -3\sqrt{3}$. A calculator gives $\tan^{-1}\left(\frac{R_y}{R_x}\right) = 60°$. The correct direction of the vector measured from the positive x-axis is:
Ans
240°
Review
Tap to reveal
Identify the correct quadrant
Negative x and negative y components put the vector in the third quadrant, so the principal arctangent angle must be adjusted.
Add 180 degrees to the reference angle
The calculator gives the reference angle of 60°, but the true direction is measured from the positive x-axis in quadrant III.
$\theta = 180^\circ + 60^\circ = 240^\circ$
14
The area of the parallelogram formed by vectors $\vec{P} = 2\hat{i} + 3\hat{j}$ and $\vec{Q} = \hat{i} - 4\hat{j}$ is:
Ans
11 sq. units
Review
Tap to reveal
Use the 2D cross-product magnitude
The area of a parallelogram formed by two vectors is the absolute value of their 2D cross product.
$|P_xQ_y - P_yQ_x| = |(2)(-4) - (3)(1)| = 11$
Question 1 of 14
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1
Two vectors have magnitudes 3 and 5. Which of the following CANNOT be the magnitude of their resultant?