Q.If x+3−6b−3z+4a−1−212y−700=0−62b+46−3−213y−22c+20 Find the values of a,b,c,x,y and z.
Concept understanding — Vector Equality
Vector Equality
When are two vectors the same vector? A vector carries only two pieces of information — magnitude (length) and direction. It does not carry a fixed starting point. So two vectors are equal when they agree on both magnitude and direction, no matter where each one happens to be drawn.
Precise definition
Two vectors a and b are equal, written a=b, if and only if
- they have the same magnitude, ∣a∣=∣b∣, and
- they have the same direction.
Position is irrelevant. An arrow drawn at the top-left of the page and an identical-length, identical-direction arrow at the bottom-right are the same vector. This is why such vectors are called "free" vectors — you may slide them anywhere without changing them.
In component form
Equality becomes a simple check on coordinates. If
a=a1i^+a2j^+a3k^,b=b1i^+b2j^+b3k^,
then
a=b⟺a1=b1,a2=b2,a3=b3.
Equal vectors have equal corresponding components — a single vector equation is really three scalar equations at once.
Don't confuse it with neighbouring ideas
Equal vectors are not the same as coinitial vectors (which merely share a starting point) or collinear vectors (which are parallel but may differ in length or point oppositely). Two vectors of equal length pointing in opposite directions are not equal — they are negatives of each other, a=−b.
Why it matters
Because equal vectors can be moved freely, we are allowed to shift vectors to a common tail before adding them, to compare forces acting at different points, and to represent every point by a position vector from the origin. Vector equality is what makes the whole "slide it wherever you like" freedom of vector algebra legitimate.
Vector equality is a foundational definition in the NCERT Class 12 Vector Algebra chapter, frequently tested through short conceptual CBSE board questions that distinguish it from coinitial or collinear vectors. Students revising "vector algebra class 12 important questions" for boards or JEE Main should treat this component-matching test as a quick sanity check before any vector proof.
Idea: Two matrices are equal iff corresponding entries are equal, so read off one equation per position.
- (1,1): x+3=0⇒x=−3
- (1,2): z+4=6⇒z=2
- (1,3): 2y−7=3y−2⇒−5=y⇒y=−5
- (2,2): a−1=−3⇒a=−2
- (2,3): 0=2c+2⇒c=−1
- (3,1): b−3=2b+4⇒−7=b⇒b=−7
The remaining positions (−6=−6, −21=−21, 0=0) are automatically satisfied.
a=−2, b=−7, c=−1, x=−3, y=−5, z=2.
Equating the two matrices entry by entry gives a=−2, b=−7, c=−1, x=−3, y=−5, z=2.
Two matrices of the same order are equal exactly when every entry in the same position matches. So this single matrix equation splits into nine ordinary equations; six carry the unknowns and three are automatically true.
x+3−6b−3z+4a−1−212y−700=0−62b+46−3−213y−22c+20.
Read off each position
- (1,1): x+3=0⇒x=−3.
- (1,2): z+4=6⇒z=2.
- (1,3): 2y−7=3y−2. Bring terms together: −7+2=3y−2y, so −5=y, i.e. y=−5.
- (2,1): −6=−6 — always true.
- (2,2): a−1=−3⇒a=−2.
- (2,3): 0=2c+2⇒2c=−2⇒c=−1.
- (3,1): b−3=2b+4. Then −3−4=2b−b, so b=−7.
- (3,2): −21=−21 — always true.
- (3,3): 0=0 — always true.
All nine equations are consistent, so every unknown is determined.
Mind the signs when rearranging: 2y−7=3y−2 gives y=−5 (not +5), and b−3=2b+4 gives b=−7.
a=−2, b=−7, c=−1, x=−3, y=−5, z=2.
Method: Solving unknowns from equality of two matrices
Use this whenever two matrices are set equal and you must find the unknowns inside them.
Steps
Step 1: Use the equality condition.
Two matrices of the same order are equal iff every corresponding entry is equal. This turns one matrix equation into a set of scalar equations, one per position.
Step 2: Write down each entry equation.
Match position by position. Some positions give trivially true statements (e.g. −6=−6) and can be skipped; the rest are equations in the unknowns.
Step 3: Solve each equation, watching the signs.
Many are one-line linear equations; isolate each unknown and solve.
Common Mistakes
Mistake 1: Sign errors when rearranging.
Why it's wrong: 2y−7=3y−2 gives y=−5 (not +5), and b−3=2b+4 gives b=−7. Correct approach: move variables to one side and constants to the other, tracking each sign.
Mistake 2: Matching entries in the wrong positions.
Why it's wrong: equality is position-by-position; comparing (1,3) with (3,1) produces false equations. Correct approach: equate only entries in identical (row, column) positions.
Mistake 3: Assuming an unknown appears where it doesn't.
Why it's wrong: some positions are pure constants and give no information about the unknowns. Correct approach: extract equations only from positions that actually contain an unknown.
- AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQQ.If a,b,c are three vectors such that ∣a∣=3,∣b∣=4,∣c∣=5 and a+b+c=0, then a⋅b is equal to (A) 12 (B) 0 (C) 512 (D) 6
›Reveal solutionSolution
Three vectors summing to zero with lengths 3,4,5 is a disguised right triangle; squaring the sum condition gives the dot product directly.
Concept and Intuition
Whenever a+b+c=0, squaring (dotting with itself) converts the vector condition into a scalar relation among the magnitudes and pairwise dot products — a very common trick with "vectors summing to zero" problems.
Step-by-Step Solution
- From a+b+c=0, write c=−(a+b).
- Take magnitude squared: ∣c∣2=∣a+b∣2=∣a∣2+∣b∣2+2a⋅b.
- Substitute ∣a∣=3,∣b∣=4,∣c∣=5: 25=9+16+2a⋅b=25+2a⋅b.
- So 2a⋅b=0⇒a⋅b=0.
Common Mistakes
- Missing that 3,4,5 satisfy the Pythagorean relation, which is exactly why the dot product vanishes (the parallelogram/triangle formed is right-angled).
- Sign errors when squaring a+b=−c.
✓Final answerThe correct option is (B) — 0.
ANSWER: B
- AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQQ.aˉ,bˉ,cˉ are non-coplanar vectors. If αdˉ=aˉ+bˉ+cˉ, βaˉ=bˉ+cˉ+dˉ, then ∣aˉ+bˉ+cˉ+dˉ∣= (A) 1 (B) 2 (C) ∣aˉ−bˉ−cˉ∣ (D) 0
›Reveal solutionSolution
Substituting one relation into the other and matching coefficients (since aˉ,bˉ,cˉ are independent) forces dˉ=−(aˉ+bˉ+cˉ), so the required sum vanishes.
Concept and Intuition
When aˉ,bˉ,cˉ are non-coplanar, they form a basis for 3D space, so any vector equation expressed in this basis must match coefficient-by-coefficient — this is the key tool that turns a vector relation into ordinary scalar equations.
Step-by-Step Solution
- From αdˉ=aˉ+bˉ+cˉ: dˉ=α1(aˉ+bˉ+cˉ).
- Substitute into βaˉ=bˉ+cˉ+dˉ=bˉ+cˉ+α1(aˉ+bˉ+cˉ)=α1aˉ+(1+α1)bˉ+(1+α1)cˉ.
- Since aˉ,bˉ,cˉ are non-coplanar (linearly independent) and the left side has zero coefficient on bˉ,cˉ: 1+α1=0⇒α=−1 (and correspondingly β=α1=−1).
- Then dˉ=−11(aˉ+bˉ+cˉ)=−(aˉ+bˉ+cˉ).
- So aˉ+bˉ+cˉ+dˉ=(aˉ+bˉ+cˉ)−(aˉ+bˉ+cˉ)=0ˉ, hence ∣aˉ+bˉ+cˉ+dˉ∣=0.
Common Mistakes
- Trying to solve for dˉ numerically without using the linear-independence argument to pin down α.
- Forgetting that "non-coplanar" is exactly the condition needed to match coefficients uniquely.
✓Final answerThe correct option is (D) — 0.
ANSWER: D
- AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQQ.aˉ and bˉ are two vectors such that aˉ is not parallel to bˉ. If pˉ=(x+2y+3)aˉ+(5x−y+2)bˉ and qˉ=(2x+3y+5)aˉ+(x−5y−2)bˉ are two vectors such that pˉ=2qˉ, then x−2y= (A) 3 (B) 2 (C) −2 (D) −3
›Reveal solutionSolution
Non-parallel vectors are linearly independent, so matching coefficients of aˉ and bˉ in pˉ=2qˉ gives two linear equations in x,y; solving them yields x−2y=−3.
Concept and Intuition
If aˉ and bˉ are non-parallel (and non-zero), they are linearly independent, meaning the only way maˉ+nbˉ=0 is m=n=0. So an equation like pˉ=2qˉ (both expressed in the same aˉ,bˉ basis) forces the aˉ-coefficients to be equal on both sides, and likewise for the bˉ-coefficients — independently.
Step-by-Step Solution
- pˉ=(x+2y+3)aˉ+(5x−y+2)bˉ, qˉ=(2x+3y+5)aˉ+(x−5y−2)bˉ.
- pˉ=2qˉ gives: coefficient of aˉ: x+2y+3=2(2x+3y+5)=4x+6y+10.
- Simplify: x+2y+3−4x−6y−10=0⇒−3x−4y−7=0⇒3x+4y+7=0. (I)
- Coefficient of bˉ: 5x−y+2=2(x−5y−2)=2x−10y−4.
- Simplify: 5x−y+2−2x+10y+4=0⇒3x+9y+6=0⇒x+3y+2=0. (II)
- From (II): x=−3y−2. Substitute into (I): 3(−3y−2)+4y+7=0⇒−9y−6+4y+7=0⇒−5y+1=0⇒y=51.
- x=−3(51)−2=−53−2=−513.
- x−2y=−513−52=−515=−3.
Common Mistakes
- Trying to solve pˉ=2qˉ as a single vector equation without splitting into independent aˉ and bˉ components — this only works because aˉ,bˉ are non-parallel.
- Arithmetic slip while eliminating x or y between the two linear equations.
✓Final answerThe correct option is (D) — −3.
ANSWER: D
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.Let aˉ,bˉ be two non-collinear vectors. If lˉ=(x+2y−3)aˉ+(2x−y+1)bˉ and mˉ=(3x−y−2)aˉ+(x+3y+2)bˉ are two vectors such that 2lˉ=mˉ, then x+5y= (A) 4 (B) 6 (C) 9 (D) 8
›Reveal solutionSolution
Since aˉ,bˉ are non-collinear (linearly independent), matching their coefficients on both sides of 2lˉ=mˉ gives two linear equations in x,y; solving them gives x+5y=8.
Concept and Intuition
Two non-collinear vectors aˉ,bˉ form a basis for their plane, so any vector equation paˉ+qbˉ=p′aˉ+q′bˉ forces p=p′ and q=q′ independently — coefficients cannot mix. This converts the vector equation 2lˉ=mˉ into an ordinary pair of simultaneous linear equations.
Step-by-Step Solution
- lˉ=(x+2y−3)aˉ+(2x−y+1)bˉ, so 2lˉ=2(x+2y−3)aˉ+2(2x−y+1)bˉ.
- mˉ=(3x−y−2)aˉ+(x+3y+2)bˉ.
- Equate the aˉ-coefficients: 2(x+2y−3)=3x−y−2⇒2x+4y−6=3x−y−2⇒x−5y+4=0 … (i)
- Equate the bˉ-coefficients: 2(2x−y+1)=x+3y+2⇒4x−2y+2=x+3y+2⇒3x−5y=0 … (ii)
- From (ii): x=35y. Substitute into (i): 35y−5y+4=0⇒ multiply by 3: 5y−15y+12=0⇒−10y=−12⇒y=56.
- Then x=35⋅56=2.
- x+5y=2+5(56)=2+6=8.
Common Mistakes
- Trying to solve the vector equation without first splitting into coefficient equations.
- Sign errors while equating and simplifying the coefficient equations.
✓Final answerThe correct option is (D) 8.
ANSWER: D
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.If aˉ=xiˉ+2jˉ−kˉ (x>0) and bˉ=2iˉ−jˉ+2kˉ are two vectors such that ∣aˉ−2bˉ∣=∣2aˉ+bˉ∣, then x= (A) 4 (B) 1 (C) 3 (D) 2
›Reveal solutionSolution
Expand both magnitude-squared expressions componentwise, set them equal, and solve the resulting quadratic for x, keeping the positive root. Answer: x=2.
Concept and Intuition
When an equation involves magnitudes of vector combinations, the cleanest approach is to compute each combination's components directly (rather than expanding dot products symbolically), square each component, sum, and equate — this avoids sign errors from expanding ∣uˉ∣2=uˉ⋅uˉ term by term.
Step-by-Step Solution
- aˉ=(x,2,−1), bˉ=(2,−1,2).
- 2bˉ=(4,−2,4), so aˉ−2bˉ=(x−4,2−(−2),−1−4)=(x−4,4,−5).
- ∣aˉ−2bˉ∣2=(x−4)2+42+(−5)2=(x−4)2+16+25=(x−4)2+41.
- 2aˉ=(2x,4,−2), so 2aˉ+bˉ=(2x+2,4−1,−2+2)=(2x+2,3,0).
- ∣2aˉ+bˉ∣2=(2x+2)2+9+0=(2x+2)2+9.
- Equate: (x−4)2+41=(2x+2)2+9. Expand: x2−8x+16+41=4x2+8x+4+9, i.e. x2−8x+57=4x2+8x+13.
- Rearranged: 3x2+16x−44=0. Using the quadratic formula: x=6−16±256+528=6−16±28, giving x=2 or x=−322.
- Since x>0 is given, x=2.
Common Mistakes
- Forgetting the constraint x>0 and reporting the negative root, or reporting both roots.
- Sign slip when computing −1−4=−5 in the z-component of aˉ−2bˉ.
✓Final answerThe correct option is (D) — 2.
ANSWER: D
- AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQQ.If uˉ,vˉ,wˉ are non coplanar vectors and p,q are real numbers, then the equality [3uˉ pvˉ pwˉ]−[pvˉ wˉ quˉ]−[2wˉ qvˉ quˉ]=0 holds for (A) exactly one ordered pair of (p,q) (B) exactly two ordered pairs of (p,q) (C) all ordered pairs of (p,q) (D) no ordered pair of (p,q)
›Reveal solutionSolution
This tests manipulating scalar triple products via their multilinearity and permutation sign rules, reducing a vector identity to a scalar quadratic equation. Only the pair (p,q)=(0,0) satisfies it.
Concept and Intuition
A scalar triple product [xˉyˉzˉ] is linear in each argument, and swapping any two vectors changes its sign, while a cyclic rotation of the three vectors leaves it unchanged. This lets any triple product built from uˉ,vˉ,wˉ (each possibly scaled) be rewritten as a numeric multiple of the base value V=[uˉvˉwˉ]=0 (non-coplanar means V=0).
Step-by-Step Solution
- [3uˉ,pvˉ,pwˉ]=3⋅p⋅p[uˉvˉwˉ]=3p2V (order unchanged, so no sign flip).
- [pvˉ,wˉ,quˉ]=pq[vˉwˉuˉ]. Since (vˉ,wˉ,uˉ) is a cyclic rotation of (uˉ,vˉ,wˉ), [vˉwˉuˉ]=V. So this term is pqV.
- [2wˉ,qvˉ,quˉ]=2q2[wˉvˉuˉ]. Since (wˉ,vˉ,uˉ) is obtained from (uˉ,vˉ,wˉ) by swapping the first and third vectors (one transposition), [wˉvˉuˉ]=−V. So this term is −2q2V.
- The equation becomes 3p2V−pqV−(−2q2V)=0⇒V(3p2−pq+2q2)=0.
- Since V=0 (non-coplanar vectors), we need 3p2−pq+2q2=0.
- Viewing this as a quadratic in p: discriminant =q2−4(3)(2q2)=q2−24q2=−23q2.
- For real p, need discriminant ≥0⇒−23q2≥0⇒q2≤0⇒q=0, and then 3p2=0⇒p=0.
- So the only real ordered pair satisfying the equation is (0,0) — exactly one.
Common Mistakes
- Mis-tracking the sign when reordering the triple product's three vectors (forgetting that a single swap flips sign but a cyclic shift doesn't).
- Assuming the quadratic form 3p2−pq+2q2=0 has infinitely many real solutions (it looks like it could describe a line/conic), without checking that it's actually positive-definite (always ≥0, zero only at the origin) since its discriminant in p is negative for all q=0.
✓Final answerThe correct option is (A) — exactly one ordered pair of (p,q).
ANSWER: A
- AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQQ.If the points P=iˉ+2jˉ,Q=4iˉ+6jˉ,R=5iˉ+7jˉ,S=aiˉ+bjˉ are the consecutive vertices of a parallelogram PQRS, then (A) a = 2, b = 4 (B) a = 3, b = 4 (C) a = 2, b = 3 (D) a = 3, b = 5
›Reveal solutionSolution
Use the parallelogram property that opposite sides are equal vectors: PQ=SR.
Concept and Intuition
In parallelogram PQRS (vertices in order), the side from P to Q is parallel and equal to the side from S to R. Writing this as a vector equation directly solves for the unknown vertex.
Step-by-Step Solution
- PQ=Q−P=(4−1,6−2)=(3,4).
- SR=R−S=(5−a,7−b).
- Since PQ=SR: 5−a=3⇒a=2, and 7−b=4⇒b=3.
- Check via diagonals bisecting each other: midpoint of PR = (3,4.5); midpoint of QS with a=2,b=3 = ((4+2)/2,(6+3)/2)=(3,4.5). ✓ Matches.
Common Mistakes
- Using PQ=RS (wrong order) instead of SR — the vertex order P→Q→R→S pairs PQ with SR, not RS.
✓Final answerThe correct option is (C) — a=2,b=3.
ANSWER: C
- AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQQ.Let ABCDEF be a regular hexagon with the vertices A,B,C,D,E,F counter clockwise. If 'O' is the center of ABCDEF, then the vector AO is to parallel/equal (A) FE (B) CD (C) CB (D) DE
›Reveal solutionSolution
Coordinatizing the hexagon on a circle shows AO and FE are literally the same vector.
Concept and Intuition
In a regular hexagon, opposite "radial" and "edge" vectors line up because the hexagon's vertices sit at 60∘ intervals on a circle whose radius equals the side length — direct coordinates make the parallelism obvious.
Step-by-Step Solution
- Place vertices counter-clockwise at angles A=0∘,B=60∘,C=120∘,D=180∘,E=240∘,F=300∘, each at distance R from center O.
- OA=R(cos0∘,sin0∘)=R(1,0), so AO=−OA=R(−1,0).
- F=R(cos300∘,sin300∘)=R(0.5,−23), E=R(cos240∘,sin240∘)=R(−0.5,−23).
- FE=E−F=R(−0.5−0.5, −23+23)=R(−1,0).
- This exactly equals AO=R(−1,0).
Common Mistakes
- Mixing up the direction of the edge vector (e.g. computing EF instead of FE), which would give the opposite sign.
- Placing vertices clockwise instead of the specified counter-clockwise order, flipping all the angles.
✓Final answerThe correct option is (A) — FE.
ANSWER: A
- AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQQ.If aˉ=tbˉ where t<0 is a scalar, then (A) aˉ,bˉ are like vectors and ∣aˉ∣>∣bˉ∣ (B) aˉ,bˉ are unlike vectors and ∣aˉ∣>∣bˉ∣ (C) aˉ,bˉ are like vectors and ∣aˉ∣<∣bˉ∣ (D) aˉ,bˉ are unlike vectors and either ∣aˉ∣≥∣bˉ∣ or ∣aˉ∣<∣bˉ∣
›Reveal solutionSolution
A negative scalar multiple always gives unlike (opposite-direction) vectors, but with no bound on ∣t∣ the relative magnitude of ∣aˉ∣ and ∣bˉ∣ cannot be pinned down either way.
Concept and Intuition
For aˉ=tbˉ: if t>0, aˉ,bˉ point the same way (like vectors); if t<0, they point opposite ways (unlike vectors) — this is independent of ∣t∣. The magnitude relationship ∣aˉ∣=∣t∣∣bˉ∣ depends entirely on whether ∣t∣ exceeds 1, which the problem never specifies (only t<0 is given, no bound on its size).
Step-by-Step Solution
- Since t<0, aˉ and bˉ point in opposite directions ⇒ they are unlike vectors. This immediately rules out any option that says "like vectors."
- ∣aˉ∣=∣t∣∣bˉ∣. Since t is just "some negative real," ∣t∣ could be >1 (giving ∣aˉ∣>∣bˉ∣) or <1 (giving ∣aˉ∣<∣bˉ∣) — both are genuinely possible.
- Only the option that states "unlike vectors" while leaving both magnitude cases open correctly describes all valid situations.
Common Mistakes
- Assuming a negative scalar automatically makes the resulting vector "smaller," which is false — magnitude depends on ∣t∣, not its sign.
- Picking an option that fixes a magnitude inequality (∣aˉ∣>∣bˉ∣) that isn't guaranteed by the given information.
✓Final answerThe correct option is (D) — unlike vectors and either ∣aˉ∣≥∣bˉ∣ or ∣aˉ∣<∣bˉ∣.
ANSWER: D
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