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.
- KCET 2019Set A-11 markMCQQ.If the angle between a and b is 32π and the projection of a in the direction of b is −2, then ∣a∣= (A) 4 (B) 3 (C) 2 (D) 1
›Reveal solutionSolution
The projection formula projba=∣a∣cosθ directly gives ∣a∣ when the projection and angle are known. Here ∣a∣=4.
The key idea is simple: the projection of one vector onto another is a measure of how much of the first vector lies along the direction of the second. It is given by the dot product divided by the magnitude of the second vector, but there is an even cleaner form when you only need the magnitude of the first vector.
The projection of a in the direction of b is defined as:
projba=∣b∣a⋅b
This is a scalar (could be positive or negative). A negative projection means the component of a along b points opposite to b's direction.
Now, recall the dot product in terms of magnitudes and the angle between them:
a⋅b=∣a∣∣b∣cosθ
Substitute this into the projection formula:
projba=∣b∣∣a∣∣b∣cosθ=∣a∣cosθ
This is the crucial simplification: the projection of a onto b depends only on ∣a∣ and the angle θ, not on ∣b∣ at all.
projba=∣a∣cosθ
Let's apply it step by step.
-
Identify the given values.
The angle θ=32π and the projection is −2.
-
Compute cosθ.
cos32π=cos(π−3π)=−cos3π=−21
- Plug into the projection formula.
−2=∣a∣⋅(−21)
- Solve for ∣a∣. Multiply both sides by −2:
∣a∣=(−2)×(−2)=4
Watch outA common mistake is to forget the sign of cos32π. Since 32π lies in the second quadrant, cosine is negative. If you mistakenly use cos32π=21, you would get ∣a∣=−4, which is impossible (magnitude is always non-negative). The negative projection already hints that cosθ must be negative.
TipNotice that the magnitude of b never entered the calculation. The projection formula simplifies to ∣a∣cosθ regardless of b's length. This is a neat shortcut for problems where only the angle and projection are given.
✓Final answerThe value is 4, which corresponds to option (A).
-
- KCET 2024Set A-11 markMCQQ.Let a and b be two unit vectors and θ is the angle between them. Then a+b is a unit vector if (A) θ=4π (B) θ=3π (C) θ=32π (D) θ=2π
›Reveal solutionSolution
Square the condition ∣a+b∣=1: it gives cosθ=−21, i.e. θ=32π.
Step 1 — Express the magnitude via the dot product.
For any vector, ∣v∣2=v⋅v. Applying that to a+b and expanding (the dot product is distributive and commutative):
∣a+b∣2=(a+b)⋅(a+b)=∣a∣2+2(a⋅b)+∣b∣2
Step 2 — Put in what is given.
a and b are unit vectors, so ∣a∣=∣b∣=1, and by the definition of the dot product
a⋅b=∣a∣∣b∣cosθ=cosθ
Hence
∣a+b∣2=1+2cosθ+1=2+2cosθ
Step 3 — Impose the condition that a+b is itself a unit vector.
∣a+b∣=1⟹∣a+b∣2=1
2+2cosθ=1⟹2cosθ=−1⟹cosθ=−21
Step 4 — Solve for θ.
The angle between two vectors lies in [0,π], and in that range
cosθ=−21⟹θ=32π (=120∘)
Step 5 — Why this is geometrically sensible.
Two unit vectors at 120∘ form, with their sum, an equilateral triangle: the resultant of two equal sides of length 1 enclosing 120∘ has length 1 as well. Checking the distractors: θ=π/2 gives ∣a+b∣=2; θ=π/3 gives 3; θ=π/4 gives 2+2≈1.85 — none is 1.
✓Final answerThe correct option is (C) — θ=32π.
ANSWER: C
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