Chapter 6 · Mathematics · 3 questions in the paper · 6 min read
Vectors and their Products
What the syllabus asks for 2
- Vectors in plane and space, algebra of vectors, linear combination of vectors, linearly dependent and independent sets of vectors
- Product of two vectors, scalar and vector product of two vectors, scalar triple product
Vectors carry about 3 questions (our estimate). The booklet's model question — the projection of \vec b=2\hat i+3\hat j-2\hat k on \vec a=\hat i+2\hat j+3\hat k — is typical: a dot product, a cross product, or a scalar triple product, computed from components. The whole chapter reduces to four formulas used fluently.
- Components, magnitude, unit vector, section formula, collinearity.
- Dot product: \vec a\cdot\vec b=ab\cos\theta; projection of \vec b on \vec a is \dfrac{\vec a\cdot\vec b}{|\vec a|}.
- Cross product: |\vec a\times\vec b|=ab\sin\theta, area of a triangle \frac{1}{2}|\vec a\times\vec b|.
- Scalar triple product: volume, coplanarity ([\vec a\,\vec b\,\vec c]=0).
- Linear dependence: two vectors are dependent iff parallel; three iff coplanar.
6.1 Vectors in plane and space, algebra, linear combination, dependence
A vector has magnitude and direction. For \vec a=a_1\hat i+a_2\hat j+a_3\hat k: |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2} and the unit vector is \hat a=\vec a/|\vec a|. Direction cosines \frac{a_1}{|\vec a|},\frac{a_2}{|\vec a|},\frac{a_3}{|\vec a|}. The vector \overrightarrow{AB}=\vec b-\vec a (head minus tail).
- Triangle law: \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}; the parallelogram law; |\vec a\pm\vec b|^2=a^2+b^2\pm2\vec a\cdot\vec b.
- Section formula: point dividing AB in the ratio m:n internally is \dfrac{m\vec b+n\vec a}{m+n} (externally \dfrac{m\vec b-n\vec a}{m-n}); midpoint \frac{\vec a+\vec b}{2}; centroid \frac{\vec a+\vec b+\vec c}{3}.
- Collinear points A,B,C: \overrightarrow{AC}=k\,\overrightarrow{AB}.
- Linear combination: \vec r=x\vec a+y\vec b+z\vec c. A set is linearly dependent if some non-trivial combination is zero: two vectors iff parallel; three iff coplanar ([\vec a\,\vec b\,\vec c]=0); any four in space always. Three non-coplanar vectors form a basis of space.
6.2 Product of two vectors, scalar and vector product, scalar triple product
Scalar (dot) product
\vec a\cdot\vec b=ab\cos\theta=a_1b_1+a_2b_2+a_3b_3; commutative and distributive. \hat i\cdot\hat i=1, \hat i\cdot\hat j=0. \vec a\cdot\vec a=|\vec a|^2. Perpendicular iff \vec a\cdot\vec b=0. \cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}. Projection of \vec b on \vec a: \dfrac{\vec a\cdot\vec b}{|\vec a|} (a scalar); the vector component is that times \hat a. Work =\vec F\cdot\vec d. (\vec a+\vec b)\cdot(\vec a-\vec b)=a^2-b^2.
Vector (cross) product
\vec a\times\vec b=ab\sin\theta\,\hat n (right-hand rule), anticommutative; \hat i\times\hat j=\hat k, \hat j\times\hat k=\hat i, \hat k\times\hat i=\hat j.
|\vec a\times\vec b|^2=a^2b^2-(\vec a\cdot\vec b)^2 (Lagrange). Parallel iff \vec a\times\vec b=\vec 0. Area of a parallelogram with sides \vec a,\vec b is |\vec a\times\vec b|; of the triangle, half. A unit vector perpendicular to both: \pm\dfrac{\vec a\times\vec b}{|\vec a\times\vec b|}. Moment of a force: \vec r\times\vec F.
Triple products
[\vec a\,\vec b\,\vec c]=\vec a\cdot(\vec b\times\vec c)= the 3×3 determinant of components; cyclic invariant, changes sign when two vectors swap. Volume of the parallelepiped =|[\vec a\,\vec b\,\vec c]|; tetrahedron =\frac{1}{6}|[\vec a\,\vec b\,\vec c]|. Coplanar iff [\vec a\,\vec b\,\vec c]=0. Vector triple product: \vec a\times(\vec b\times\vec c)=\vec b(\vec a\cdot\vec c)-\vec c(\vec a\cdot\vec b) ("BAC minus CAB").
The dot product is a scalar, the cross product a vector. \vec a\times(\vec b\times\vec c)\ne(\vec a\times\vec b)\times\vec c. The projection can be negative.
Formula and fact sheet
| Topic | Result |
|---|---|
| angle | \cos\theta=\frac{\vec a\cdot\vec b}{ab} |
| projection of \vec b on \vec a | \frac{\vec a\cdot\vec b}{|\vec a|} |
| triangle area | \frac{1}{2}|\vec a\times\vec b| |
| Lagrange | |\vec a\times\vec b|^2=a^2b^2-(\vec a\cdot\vec b)^2 |
| volume | |[\vec a\,\vec b\,\vec c]|; tetrahedron \frac{1}{6} of it |
| triple product | \vec a\times(\vec b\times\vec c)=\vec b(\vec a\cdot\vec c)-\vec c(\vec a\cdot\vec b) |
Chapter summary
- Head minus tail gives \overrightarrow{AB}; the section formula weights the opposite end.
- Dot product zero means perpendicular; cross product zero means parallel; triple product zero means coplanar.
- Projection is \frac{\vec a\cdot\vec b}{|\vec a|}, a signed scalar.
- Areas come from |\vec a\times\vec b|, volumes from the scalar triple product.
Practice questions
Set A — Recall (10 questions)
- For \vec a=2\hat i+3\hat j+\hat k and \vec b=\hat i-\hat j+2\hat k, \vec a\cdot\vec b equalsa7b5c1d-1
- The magnitude of 3\hat i+4\hat j-12\hat k isa13b19c11d5
- \hat i\times\hat j equalsa0b-\hat kc1d\hat k
- The vectors 2\hat i+\lambda\hat j+\hat k and \hat i-2\hat j+3\hat k are perpendicular for \lambda=a2b5/2c3/2d1
- The area of the parallelogram with adjacent sides \hat i+\hat j and \hat j+\hat k isa2b\sqrt{2}c\sqrt{3}d3
- If [\vec a\,\vec b\,\vec c]=0, the three vectors areacoplanarbparallelcmutually perpendiculardcollinear
- The angle between \hat i+\hat j and \hat j isa30^\circb60^\circc90^\circd45^\circ
- The midpoint of A(2,4,6) and B(4,0,2) isa(6,4,8)b(3,2,4)c(1,2,2)d(3,4,2)
- The unit vector along 3\hat i-4\hat j isa\dfrac{3\hat i-4\hat j}{25}b3\hat i-4\hat jc\dfrac{3\hat i+4\hat j}{5}d\dfrac{3\hat i-4\hat j}{5}
- If |\vec a|=5 and |\vec b|=3, then (\vec a+\vec b)\cdot(\vec a-\vec b) isa16b34c4d25
Answers and explanations — Set A
1. (c) — 2-3+2=1.
2. (a) — \sqrt{9+16+144}=13.
3. (d) — right-hand rule: \hat i\times\hat j=\hat k (\hat j\times\hat i=-\hat k).
4. (b) — 2-2\lambda+3=0\Rightarrow\lambda=\frac{5}{2}.
5. (c) — \vec a\times\vec b=(1,-1,1), magnitude \sqrt{3}.
6. (a) — a zero scalar triple product means the vectors lie in one plane.
7. (d) — \cos\theta=\frac{1}{\sqrt{2}}.
8. (b) — average of the coordinates.
9. (d) — divide by the magnitude 5.
10. (a) — |\vec a|^2-|\vec b|^2=25-9=16.
Set B — Understanding (7 questions)
- The projection of \vec b=2\hat i+3\hat j-2\hat k on \vec a=\hat i+2\hat j+3\hat k isa2/\sqrt{17}b2/\sqrt{14}c2/7d-2/7
- If |\vec a|=3, |\vec b|=4 and the angle between them is 30^\circ, then |\vec a\times\vec b| isa12b6\sqrt{3}c6d3
- The vectors 2\hat i-\hat j+\hat k, \hat i+2\hat j-3\hat k and 3\hat i+\lambda\hat j+5\hat k are coplanar for \lambda=a4b-1c1d-4
- The area of the triangle with vertices (1,0,0), (0,1,0), (0,0,1) isa\frac{\sqrt{3}}{2}b\sqrt{3}c\frac{1}{2}d\frac{3}{2}
- The volume of the parallelepiped with edges \hat i+\hat j, \hat j+\hat k, \hat k+\hat i isa1b4c2d3
- A unit vector perpendicular to both \hat i+\hat j and \hat j+\hat k isa\frac{\hat i+\hat j+\hat k}{\sqrt{3}}b\frac{\hat i-\hat j+\hat k}{\sqrt{3}}c\frac{\hat i-\hat j-\hat k}{3}d\frac{\hat i+\hat j-\hat k}{\sqrt{3}}
- The value of \hat i\times(\hat j\times\hat i) isa\hat jb0c-\hat jd\hat k
Answers and explanations — Set B
1. (b) — \vec a\cdot\vec b=2+6-6=2, |\vec a|=\sqrt{14}. The projection of \vec b on \vec a divides by |\vec a|.
2. (c) — 3\cdot4\sin30^\circ=6.
3. (d) — the determinant is 28+7\lambda=0.
4. (a) — \overrightarrow{AB}\times\overrightarrow{AC}=(1,1,1); half of \sqrt{3}.
5. (c) — (det | 1&1&0 ; 0&1&1 ; 1&0&1 |)=1+1=2.
6. (b) — (\hat i+\hat j)\times(\hat j+\hat k)=\hat i-\hat j+\hat k, magnitude \sqrt{3}.
7. (a) — \hat j\times\hat i=-\hat k, and \hat i\times(-\hat k)=\hat j. Check with BAC–CAB: \hat j(\hat i\cdot\hat i)-\hat i(\hat i\cdot\hat j)=\hat j.
Set C — Application (6 questions)
- The points (1,1,1), (2,3,4) and (3,5,\lambda) are collinear when \lambda=a6b7c8d5
- The work done by \vec F=2\hat i+3\hat j+\hat k in moving a particle from (1,1,1) to (3,2,4) isa10b12c8d9
- If \vec a,\vec b are unit vectors at 60^\circ to each other, |\vec a+\vec b| isa\sqrt{2}b2c1d\sqrt{3}
- If |\vec a|=2, |\vec b|=3 and \vec a\cdot\vec b=3, then |\vec a\times\vec b| isa3b6c9d3\sqrt{3}
- The vectors 2\hat i+3\hat j and \hat i+\lambda\hat j are linearly dependent for \lambda=a2/3b3/2c6d1/6
- The volume of the tetrahedron with vertices O, (1,0,0), (0,2,0), (0,0,3) isa1b6c3d2
Answers and explanations — Set C
1. (b) — \overrightarrow{AB}=(1,2,3), \overrightarrow{AC}=(2,4,\lambda-1) must be 2\overrightarrow{AB}: \lambda-1=6.
2. (a) — displacement (2,1,3): 4+3+3=10.
3. (d) — |\vec a+\vec b|^2=1+1+2\cdot\frac{1}{2}=3.
4. (d) — |\vec a\times\vec b|^2=36-9=27.
5. (b) — parallel: \frac{2}{1}=\frac{3}{\lambda}.
6. (a) — \frac{1}{6}|1\cdot2\cdot3|=1.