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.
- CBSE 2024Set 65/1/11 markMCQQ.If [x+y52xy]=[6528], then the value of (x24+y24) is: (A) 7 (B) 6 (C) 8 (D) 18
›Reveal solutionSolution
Two matrices are equal only when their corresponding entries are equal. Equating the entries gives x+y=6 and xy=8, which are the sum and product of the roots of a quadratic. The required expression x24+y24 simplifies to 24⋅xyx+y=24⋅86=18.
When two matrices are equal, every entry in the same position must match exactly. This is the fundamental idea of vector equality extended to matrices — think of it as comparing two tables of numbers cell by cell. No shortcuts, no tricks: if the top-left corner of the first matrix is x+y, and the top-left corner of the second matrix is 6, then x+y must equal 6. The same logic applies to every other position.
Here, both matrices are 2×2, so we get four equations. But notice that two of the entries are already identical (2 and 5), so they give no new information. The real constraints come from the other two positions.
-
Equate the (1,1) entries — the top-left corner:
x+y=6
-
Equate the (2,2) entries — the bottom-right corner:
xy=8
We now have two equations in two unknowns. You could solve for x and y individually (they are the roots of t2−6t+8=0, so t=2 or 4), but the question doesn't ask for x and y separately — it asks for x24+y24.
TipWhen you see a sum of reciprocals like x24+y24, always try to rewrite it using x+y and xy:
x24+y24=24(x1+y1)=24(xyx+y)
This avoids solving for x and y individually — a huge time-saver in exams.
- Substitute the known values: x24+y24=24⋅xyx+y=24⋅86=24⋅43=18
Watch outA common mistake is to forget that xy=8 and instead try to solve x+y=6 and xy=8 by guessing. Some students pick x=2,y=4 (correct) but then compute 224+424=12+6=18 — which works here. But if the numbers were less friendly, guessing would fail. Always use the symmetric sum-and-product approach; it's foolproof.
✓Final answerThe value is 18, which corresponds to option (D).
-
- CBSE 2025Set 65/1/11 markMCQQ.If vector a=3i^+2j^−k^ and vector b=i^−j^+k^, then which of the following is correct? (A) a∥b (B) a⊥b (C) ∣b∣>∣a∣ (D) ∣a∣=∣b∣
›Reveal solutionSolution
Two vectors are perpendicular when their dot product is zero. Computing a⋅b=0 confirms orthogonality, so the answer is (B).
When you're given two vectors and asked about their relationship, you need to check three fundamental properties: parallelism (same or opposite direction), perpendicularity (right angles), and magnitude (length). Each has a clear algebraic test.
Two vectors are parallel if one is a scalar multiple of the other: a=λb for some real λ. Two vectors are perpendicular (orthogonal) if their dot product vanishes: a⋅b=0. The magnitude of a vector v=v1i^+v2j^+v3k^ is ∣v∣=v12+v22+v32.
Let's systematically check each possibility.
Checking the options
1. Are the vectors parallel?
For a=3i^+2j^−k^ and b=i^−j^+k^ to be parallel, we'd need
13=−12=1−1
The first ratio gives 3, the second gives −2, and the third gives −1. These are all different, so the vectors are not parallel. Option (A) is incorrect.
2. Are the vectors perpendicular?
Compute the dot product:
a⋅b=(3)(1)+(2)(−1)+(−1)(1)=3−2−1=0
The dot product is zero, which means the vectors are perpendicular. Option (B) looks promising.
3. Compare the magnitudes
Calculate ∣a∣:
∣a∣=32+22+(−1)2=9+4+1=14
Calculate ∣b∣:
∣b∣=12+(−1)2+12=1+1+1=3
Since 14≈3.74 and 3≈1.73, we have ∣a∣>∣b∣. So option (C) is false and option (D) is false.
TipWhen checking perpendicularity, the dot product test is immediate and definitive. If a⋅b=0, you're done — no need to compute angles or magnitudes unless the question specifically asks for them.
✓Final answerThe correct option is (B) a⊥b.
- CBSE 2026Set ANNUAL1 markMCQQ.Assertion (A): For vector \vec{a} = \hat{i} + 2\hat{j} and vector \vec{b} = 2\hat{i} + \hat{j}, |\vec{a}| = |\vec{b}|. Reason (R): Vector \vec{a} and vector \vec{b} are equal vectors.(a)(i) Both A and R are correct and R is the correct explanation of A.(b)(ii) Both A and R are correct but R is not the correct explanation of A.(c)(iii) A is correct but R is incorrect.(d)(iv) Both A and R are incorrect.
›Reveal solutionSolution
A is true (∣a∣=∣b∣=5) but R is false (the vectors are not equal) — option (iii).
Concept. The magnitude of xi^+yj^ is x2+y2. Two vectors are equal only if all their corresponding components are equal — equal magnitude alone is not enough.
Steps.
-
a=i^+2j^ ⇒ ∣a∣=12+22=5.
-
b=2i^+j^ ⇒ ∣b∣=22+12=5.
-
So ∣a∣=∣b∣: Assertion (A) is correct.
-
Components: a has (1,2) while b has (2,1); these differ, so a=b: Reason (R) is incorrect.
✓Final answerA is correct but R is incorrect — option (iii).
-
- CBSE 2023Set ANNUAL1 markMCQQ.If xi^+2j^=3i^−yj^, then find (x,y).(a) 3,−2(b) 2,3(c) −3,2(d) none of these
›Reveal solutionSolution
Two vectors are equal exactly when their corresponding components are equal.
xi^+2j^=3i^−yj^. Equate i^-components: x=3. Equate j^-components: 2=−y⇒y=−2.
So (x,y)=(3,−2).
✓Final answer(a) x=3, y=−2.
- CBSE 2020Set ANNUAL1 markQ.Define Negative of a vector.
›Reveal solutionSolution
−a has the same magnitude as a but points in the opposite direction.
Concept. A vector has both magnitude and direction. Negating a vector keeps its length unchanged but flips its direction through 180∘.
Formal statement. The negative of a given vector a is defined as the vector whose magnitude is ∣a∣ and whose direction is exactly opposite to that of a. It is written −a, and satisfies a+(−a)=0.
✓Final answerThe negative of a, written −a, is the vector with the same magnitude as a but opposite direction.
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