Kit Library / Mathematics / Vectors
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One gives a scalar, the other a vector.
Remember: dot is for scalar results (angle, projection), cross is for vector results (area, perpendicular).
↳ Dot product answers 'how much in the same direction?' while cross product answers 'how much perpendicular?'.
The cosine formula you must memorize.
↳ Rearrange to find angle: θ = cos⁻¹( (a·b)/(|a||b|) ).
Projection is just dot product divided by magnitude.
↳ Projection is a scalar length; the projection vector is that length times the unit vector of b.
Cross product magnitude gives area directly.
↳ For triangle with vertices A, B, C, area = ½ |AB × AC|.
Cross product zero? They're parallel.
↳ Zero cross product means parallel; zero scalar triple product means coplanar.
Scalar vs vector: know the difference.
↳ Projection is a number; projection vector is a vector along b.
Order matters for cross product.
↳ Cross product is anti-commutative; reversing order reverses direction.
Use vectors to compute area easily.
↳ Area = ½ |AB × AC|. Compute cross product, then magnitude, then halve.
Cross product gives perpendicular direction.
↳ Unit vector perpendicular to both a and b is (a × b) / |a × b|.
A common exam shortcut.
If |a·b| = |a×b|, then |a||b|cosθ = |a||b|sinθ → tanθ = 1 → θ = 45° (or π/4).
↳ Equality of dot and cross magnitudes means angle is 45°.
Two products, two purposes.
Dot → Direction (angle) · Cross → Cover (area).
↳ Use dot for angle/projection, cross for area/perpendicular.
Zero triple product? Coplanar.
Three vectors a, b, c are coplanar if their scalar triple product [a b c] = a·(b×c) = 0.
↳ Coplanarity ⇔ scalar triple product = 0.
Cosines of angles with axes.
For vector a = x i + y j + z k, direction cosines are l = x/|a|, m = y/|a|, n = z/|a|, and l² + m² + n² = 1.
↳ Direction cosines are components divided by magnitude.
Unit vector needs magnitude 1.
↳ Unit vector = vector divided by its length.
Set up a 3x3 determinant.
For a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k, a × b = determinant of [[i, j, k], [a₁, a₂, a₃], [b₁, b₂, b₃]].
↳ Determinant method avoids sign errors in cross product.
Vector algebra covers operations on vectors, dot and cross products, and their applications to geometry, area, and collinearity. This set of Veda Bites distills the essential insights from the chapter, focusing on exam-r…
Full explanation, worked examples & memory tricks in the app
All 17 questions, laddered from warm-up to topper-level, each with its answer and the reasoning behind it.
Q1. Which of the following operations always returns a scalar?
BeginnerA · $\vec{a}\cdot\vec{b}$
The dot product (scalar product) returns a scalar. Addition, subtraction, and cross product return vectors.
Q2. If $\vec{a}\cdot\vec{b}=0$ for nonzero vectors $\vec{a}$ and $\vec{b}$, what is the angle between them?
BeginnerD · $90^\circ$
$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta=0$ implies $\cos\theta=0$, so $\theta=90^\circ$ (perpendicular).
Q3. What is the area of a triangle whose adjacent sides are given by vectors $\vec{a}$ and $\vec{b}$?
BeginnerB · $\tfrac12|\vec{a}\times\vec{b}|$
The area of a triangle is half the magnitude of the cross product of its adjacent sides.
Q4. Three vectors $\vec{a},\vec{b},\vec{c}$ are coplanar if and only if:
BeginnerA · $\vec{a}\cdot(\vec{b}\times\vec{c})=0$
Coplanarity is equivalent to the scalar triple product being zero: $[\vec{a}\ \vec{b}\ \vec{c}]=0$.
Q5. Which of the following is NOT a vector?
BeginnerA · $\vec{a}\cdot\vec{b}$
The dot product is a scalar, while the others are vectors.
Q6. The dot product of two vectors is always a scalar, while the cross product is always a vector.
BeginnerA · True
The dot product of two vectors is always a scalar, while the cross product is always a vector.
Q7. If $|\vec{a}\cdot\vec{b}|=|\vec{a}\times\vec{b}|$, what is the acute angle between $\vec{a}$ and $\vec{b}$?
IntermediateA · $45^\circ$
The condition gives $|\cos\theta|=|\sin\theta|$, so $\tan\theta=1$ for the acute angle, hence $\theta=45^\circ$.
Q8. What is the projection of $\vec{a}=\hat{i}-\hat{j}$ on $\vec{b}=\hat{i}+\hat{j}$?
IntermediateB · $0$
$\vec{a}\cdot\vec{b}=1-1=0$, so the projection is $0/\sqrt{2}=0$. The vectors are perpendicular.
Q9. A unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by:
IntermediateD · $\pm\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$
The cross product $\vec{a}\times\vec{b}$ is perpendicular to both; normalizing gives the unit vector, with two possible signs.
Q10. If $\vec{a}$ and $\vec{b}$ are parallel, what is $\vec{a}\times\vec{b}$?
IntermediateB · $\vec{0}$
For parallel vectors, $\sin\theta=0$, so $\vec{a}\times\vec{b}=\vec{0}$.
Q11. What is the volume of the parallelepiped with edges $\hat{i}+\hat{j}$, $\hat{j}+\hat{k}$, $\hat{k}+\hat{i}$?
IntermediateA · $2$
The scalar triple product is $2$, so the volume is $|2|=2$.
Q12. If $\vec{a}=2\hat{i}+\hat{j}-3\hat{k}$ and $\vec{b}=m\hat{i}+3\hat{j}-\hat{k}$ are perpendicular, what is $m$?
IntermediateB · $0$
$\vec{a}\cdot\vec{b}=2m+3-3=2m=0$, so $m=0$.
Q13. If $\vec{a},\vec{b},\vec{c}$ are unit vectors with $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, what is $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}$?
AdvancedA · $-3/2$
Square the sum: $0 = |\vec{a}+\vec{b}+\vec{c}|^2 = 3 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})$, so the sum is $-3/2$.
Q14. If $\vec{a},\vec{b},\vec{c}$ are coplanar, which of the following is necessarily true?
AdvancedB · $\vec{a}+\vec{b}$, $\vec{b}+\vec{c}$, $\vec{c}+\vec{a}$ are coplanar
Linear combinations of coplanar vectors remain coplanar, so the three sums are coplanar.
Q15. If $|\vec{a}|=3$, $|\vec{b}|=4$, $|\vec{c}|=4$ and each vector is perpendicular to the sum of the other two, what is $|\vec{a}+\vec{b}+\vec{c}|$?
AdvancedB · $\sqrt{41}$
The conditions give $\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}=0$, etc. Solving yields $|\vec{a}+\vec{b}+\vec{c}|^2=41$, so the magnitude is $\sqrt{41}$.
Q16. What is the value of $\hat{i}\cdot(\hat{j}\times\hat{k}) + \hat{j}\cdot(\hat{i}\times\hat{k}) + \hat{k}\cdot(\hat{i}\times\hat{j})$?
AdvancedA · $1$
$\hat{i}\cdot(\hat{j}\times\hat{k})=1$, $\hat{j}\cdot(\hat{i}\times\hat{k})=-1$, $\hat{k}\cdot(\hat{i}\times\hat{j})=1$; sum is $1$.
Q17. Given vectors \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + 4\hat{k} \), find the area of the triangle formed by these vectors.
AdvancedB · 5
The area of the triangle formed by vectors \( \vec{a} \) and \( \vec{b} \) is given by \( \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| \). First, we calculate \( \vec{a} \times \vec{b} \):…
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