This MCQ module is based on: Vector Addition, Components and Section Formula
TOPIC 15 OF 25
Vector Addition, Components and Section Formula
🎓 Class 12
Mathematics
CBSE
Theory
Ch 10 — Vector Algebra
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Vector Addition, Components and Section Formula
Targeting Class 12 level in General Mathematics, with Advanced difficulty.
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10.4 Addition of Vectors
Triangle Law
Triangle Law
If two vectors are represented by sides \(\vec{AB}\) and \(\vec{BC}\) of a triangle taken in order (head of one at the tail of the next), then the sum is represented by the closing side \(\vec{AC}\):
\[\boxed{\;\vec{AB}+\vec{BC}=\vec{AC}\;}\]
Triangle law: head-to-tail placement; the closing side is the sum.
Parallelogram Law
Parallelogram Law
If two vectors with the same initial point form adjacent sides of a parallelogram, their sum is the diagonal of the parallelogram drawn from the common point:
\[\vec a+\vec b=\vec c\quad\text{(diagonal)}\]
The two laws are equivalent — they describe the same operation.
Properties of vector addition
Algebra of vector addition
For any vectors \(\vec a,\vec b,\vec c\):
- Commutative: \(\vec a+\vec b=\vec b+\vec a\) (parallelogram law).
- Associative: \((\vec a+\vec b)+\vec c=\vec a+(\vec b+\vec c)\).
- Identity: \(\vec a+\vec 0=\vec a\).
- Inverse: \(\vec a+(-\vec a)=\vec 0\).
- Subtraction: \(\vec a-\vec b=\vec a+(-\vec b)\).
10.5 Multiplication of a Vector by a Scalar
Scalar multiplication
For a scalar \(\lambda\in\mathbb R\) and vector \(\vec a\), the scalar multiple \(\lambda\vec a\) has• magnitude \(|\lambda|\,|\vec a|\),
• direction same as \(\vec a\) if \(\lambda>0\), opposite if \(\lambda<0\), and \(\lambda\vec a=\vec 0\) if \(\lambda=0\).
Properties: \(\lambda(\vec a+\vec b)=\lambda\vec a+\lambda\vec b\); \((\lambda+\mu)\vec a=\lambda\vec a+\mu\vec a\); \(\lambda(\mu\vec a)=(\lambda\mu)\vec a\); \(1\vec a=\vec a\).
Unit vector
\(\hat a=\dfrac{\vec a}{|\vec a|}\) is the unit vector along \(\vec a\). It captures the direction of \(\vec a\) only; the magnitude is normalised to 1.
10.5.1 Components of a Vector
Components
Any vector \(\vec r\) in 3-D can be uniquely written as
\[\vec r=a\,\hat\imath+b\,\hat\jmath+c\,\hat k,\]
where \(a, b, c\) are the scalar components (real numbers) and \(a\hat\imath, b\hat\jmath, c\hat k\) are the vector components along the three coordinate axes.\(|\vec r|=\sqrt{a^2+b^2+c^2}\). Two vectors are equal iff all three components match.
10.5.2 Vector Joining Two Points
PQ formula
If \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\) have position vectors \(\vec a,\vec b\) respectively, then
\[\boxed{\;\vec{PQ}=\vec b-\vec a=(x_2-x_1)\hat\imath+(y_2-y_1)\hat\jmath+(z_2-z_1)\hat k\;}\]
"Head minus tail" — direction goes FROM \(P\) TO \(Q\). Magnitude \(|\vec{PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\) is the distance from \(P\) to \(Q\).
10.5.3 Section Formula
Section formula — internal and external division
If \(R\) divides segment \(PQ\) (with position vectors \(\vec a,\vec b\)) in the ratio \(m:n\),Internal division: \(\vec{OR}=\dfrac{m\vec b+n\vec a}{m+n}\).
External division: \(\vec{OR}=\dfrac{m\vec b-n\vec a}{m-n}\).
Midpoint (\(m=n=1\)): \(\vec{OR}=\dfrac{\vec a+\vec b}{2}\).
Worked Examples
Example 6. \(\vec a=2\hat\imath+3\hat\jmath\), \(\vec b=\hat\imath-\hat\jmath\). Find \(\vec a+\vec b,\ \vec a-\vec b,\ 2\vec a+3\vec b\).
\(\vec a+\vec b=3\hat\imath+2\hat\jmath\). \(\vec a-\vec b=\hat\imath+4\hat\jmath\). \(2\vec a+3\vec b=4\hat\imath+6\hat\jmath+3\hat\imath-3\hat\jmath=7\hat\imath+3\hat\jmath\).
Example 7. Find \(\vec{PQ}\) where \(P(2, 1, 3)\) and \(Q(5, 4, -1)\). Magnitude.
\(\vec{PQ}=(5-2)\hat\imath+(4-1)\hat\jmath+(-1-3)\hat k=3\hat\imath+3\hat\jmath-4\hat k\). \(|\vec{PQ}|=\sqrt{9+9+16}=\sqrt{34}\) units.
Example 8. Position vectors \(\vec a=2\hat\imath+\hat\jmath-3\hat k,\ \vec b=4\hat\imath+5\hat\jmath+\hat k\). Find the position vector of the point dividing AB in ratio 2:3 internally.
\(\vec{OR}=\dfrac{2\vec b+3\vec a}{2+3}=\dfrac{2(4,5,1)+3(2,1,-3)}{5}=\dfrac{(8+6,10+3,2-9)}{5}=\dfrac{14\hat\imath+13\hat\jmath-7\hat k}{5}\).
Example 9. Find the midpoint of \(P(2,3,4)\) and \(Q(4,1,-2)\).
Midpoint = \((\vec a+\vec b)/2=((2+4)/2,(3+1)/2,(4-2)/2)=(3,2,1)\), or as vector \(3\hat\imath+2\hat\jmath+\hat k\).
Example 10. Show that \(A(1,2,7),\ B(2,6,3),\ C(3,10,-1)\) are collinear.
\(\vec{AB}=\hat\imath+4\hat\jmath-4\hat k\); \(\vec{BC}=\hat\imath+4\hat\jmath-4\hat k=\vec{AB}\). So \(\vec{BC}=\vec{AB}\), meaning B and C lie on the same direction from A — collinear (with B as midpoint of AC).
Activity: Force Resultant
L4 AnalyseMaterials: Pen, paper.
Predict: Two forces \(\vec F_1=3\hat\imath+4\hat\jmath\) N and \(\vec F_2=-\hat\imath+2\hat\jmath\) N act on a body. What's the magnitude of the resultant force?
- Resultant: \(\vec F=\vec F_1+\vec F_2=2\hat\imath+6\hat\jmath\) N.
- Magnitude: \(\sqrt{4+36}=\sqrt{40}=2\sqrt{10}\approx 6.32\) N.
- Direction: angle with x-axis is \(\tan^{-1}(6/2)=\tan^{-1}3\approx 71.6°\).
- Now add a third force \(\vec F_3=2\hat\imath-3\hat\jmath\) N. Find total. (\(\vec F_{\text{tot}}=4\hat\imath+3\hat\jmath\); magnitude 5.)
- Lesson: components convert vector addition into ordinary scalar arithmetic.
Vector addition by components is ALWAYS easier than triangle/parallelogram law for arbitrary numbers of vectors. The geometric laws give intuition; the algebra delivers efficiency. This is exactly why physics simulations work in components.
Competency-Based Questions
Scenario: A delivery drone goes from warehouse W(0, 0, 0) to checkpoint A(3, 4, 0) and then to destination B(7, 8, 6). Total displacement is computed via vector addition.
Q1. The vector \(\vec{WB}\) is:
L3 ApplyAnswer: \(\vec{WB}=7\hat\imath+8\hat\jmath+6\hat k\). Magnitude \(\sqrt{49+64+36}=\sqrt{149}\) units.
Q2. Verify \(\vec{WA}+\vec{AB}=\vec{WB}\) (triangle law).
L3 ApplySolution: \(\vec{WA}=3\hat\imath+4\hat\jmath\); \(\vec{AB}=4\hat\imath+4\hat\jmath+6\hat k\). Sum \(=7\hat\imath+8\hat\jmath+6\hat k=\vec{WB}\). ✓
Q3. (T/F) "If A, B, C are collinear, then \(\vec{AB}\) and \(\vec{BC}\) are parallel." Justify.
L5 EvaluateTrue. Collinear ⇒ A, B, C lie on a single line, so \(\vec{AB}\) and \(\vec{BC}\) point along this line — parallel (same or opposite direction). Algebraic test: \(\vec{BC}=k\vec{AB}\) for some scalar k.
Q4. Apply: find the point dividing \(P(1,−2,3)\) and \(Q(3,4,−5)\) in ratio 2:3 externally.
L4 AnalyseSolution: External: \(\vec{OR}=(2\vec b-3\vec a)/(2-3)=-(2\vec b-3\vec a)=3\vec a-2\vec b=3(1,-2,3)-2(3,4,-5)=(3-6,-6-8,9+10)=(-3,-14,19)\).
Q5. Design: in 3-D space, find a unit vector that is the sum of \(\hat\imath+2\hat\jmath+3\hat k\) and \(2\hat\imath-\hat\jmath+\hat k\), divided by the magnitude.
L6 CreateSolution: Sum = \(3\hat\imath+\hat\jmath+4\hat k\). Magnitude = \(\sqrt{9+1+16}=\sqrt{26}\). Unit vector: \(\dfrac{1}{\sqrt{26}}(3\hat\imath+\hat\jmath+4\hat k)\).
Assertion–Reason Questions
Assertion (A): Vector addition is commutative.
Reason (R): The parallelogram law gives the same diagonal regardless of which side is labelled \(\vec a\) or \(\vec b\).
Reason (R): The parallelogram law gives the same diagonal regardless of which side is labelled \(\vec a\) or \(\vec b\).
Answer: (a). Geometric symmetry of the parallelogram is the visual proof.
Assertion (A): The position vector of the midpoint of segment PQ (position vectors a, b) is (a + b)/2.
Reason (R): The midpoint divides PQ internally in ratio 1:1, so by section formula it equals (1·b + 1·a)/(1+1).
Reason (R): The midpoint divides PQ internally in ratio 1:1, so by section formula it equals (1·b + 1·a)/(1+1).
Answer: (a). R is the section-formula derivation of A.
Assertion (A): If \(\vec a\) and \(\vec b\) are non-zero non-collinear vectors, then \(\alpha\vec a+\beta\vec b=\vec 0\) implies \(\alpha=\beta=0\).
Reason (R): Non-collinear vectors are linearly independent.
Reason (R): Non-collinear vectors are linearly independent.
Answer: (a). R is the technical name for the property in A.
Frequently Asked Questions — Vector Addition, Components and Section Formula
What is the triangle law of vector addition?
Two vectors a and b placed head-to-tail; the sum is the closing side.
What is the parallelogram law of vector addition?
Two vectors from the same point form sides of a parallelogram; the sum is the diagonal from the same starting point.
How do you add vectors in component form?
Add corresponding components.
What is the section formula?
Internal: (m·b + n·a)/(m+n) for ratio m:n. External: (m·b − n·a)/(m−n).
How is the vector joining two points found?
PQ = position vector of Q − position vector of P (head minus tail).
What is the midpoint formula in vectors?
(a + b)/2 — section formula with m = n = 1.
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